MATH 350 - DISCRETE
MATHEMATICS - Permutations and
combinations
Question Bank - Set 5
Liberty University
Question 1
Question
A committee of 5 people is to be selected from a group of 10 students and
5 professors. If the committee must contain at least 2 professors, how many
different committees can be formed?
Solution
Step 1: Find the total number of ways to select a committee of 5 people. This
can be calculated using combinations: 15
5.
Step 2: Find the number of committees with no professors. This can be
calculated by choosing all 5 members from the students, which is 10
5.
Step 3: Find the number of committees with exactly 1 professor. This can
be calculated by choosing 1 professor and 4 students, which is 5
1·10
4.
Step 4: Subtract the total number of committees with no professors and
exactly 1 professor from the total number of committees to find the number of
committees with at least 2 professors.
Therefore, the number of different committees that can be formed is:
15
5−10
5−(5
1·10
4) = 3003 −252 −1250 = 1501
Question 2
Question
A committee of 5 people is to be formed from a group of 8 women and 7 men.
How many ways can the committee be formed if it must consist of at least 3
women?
Solution
To solve this problem, we will consider the different cases based on the number
of women in the committee.
Case 1: 3 women and 2 men Step 1: Choose 3 women from 8 women:
8
3ways.
Step 2: Choose 2 men from 7 men: 7
2ways.
Step 3: Total number of ways in this case: 8
3×7
2.
Case 2: 4 women and 1 man Step 1: Choose 4 women from 8 women:
8
4ways.
Step 2: Choose 1 man from 7 men: 7
1ways.
Step 3: Total number of ways in this case: 8
4×7
1.
Case 3: 5 women and 0 men Step 1: Choose 5 women from 8 women:
8
5ways.
Step 2: Total number of ways in this case: 8
5.
Step 3: Add up the total number of ways from each case to get the final
answer:
Total ways = 8
3×7
2+8
4×7
1+8
5.
Calculate the final answer.
Question 3
Question
A committee of 5 people is to be formed from a group of 7 men and 5 women. If
the committee must consist of at least 2 men and at least 2 women, how many
different committees can be formed?
Solution
Let’s break this problem down into cases based on the number of men and
women on the committee.
Case 1: 2 men and 3 women Step 1: Choose 2 men from 7 men.
7
2= 21
Step 2: Choose 3 women from 5 women.
5
3= 10
Step 3: Multiply the number of ways for each gender group.
21 ×10 = 210
2
Case 2: 3 men and 2 women Step 1: Choose 3 men from 7 men.
7
3= 35
Step 2: Choose 2 women from 5 women.
5
2= 10
Step 3: Multiply the number of ways for each gender group.
35 ×10 = 350
Case 3: 4 men and 1 woman Step 1: Choose 4 men from 7 men.
7
4= 35
Step 2: Choose 1 woman from 5 women.
5
1= 5
Step 3: Multiply the number of ways for each gender group.
35 ×5 = 175
Case 4: 5 men and 0 women Step 1: Choose 5 men from 7 men.
7
5= 21
Step 2: Choose 0 women from 5 women.
5
0= 1
Step 3: Multiply the number of ways for each gender group.
21 ×1 = 21
Add up the possibilities from each case to find the total number of different
committees that can be formed.
210 + 350 + 175 + 21 = 756
Therefore, there are a total of 756 different committees that can be formed
from the group.
3
Question 4
Question
A committee of 5 people is to be formed from a group of 8 men and 5 women.
If the committee must have at least 3 men, how many different committees can
be formed?
Solution
Step 1: Find the number of ways to choose a committee with exactly 3 men.
Step 2: Find the number of ways to choose a committee with exactly 4 men.
Step 3: Find the number of ways to choose a committee with all 5 men. Step
4: Add the results from Step 1, Step 2, and Step 3 to get the total number of
different committees.
Step 1: Choose 3 men from 8: 8
3=8!
3!(8−3)! =8×7×6
3×2×1= 56 ways.
Step 2: Choose 4 men from 8: 8
4=8!
4!(8−4)! =8×7×6×5
4×3×2×1= 70 ways.
Step 3: Choose all 5 men from 8: 8
5=8!
5!(8−5)! =8×7×6×5×4
5×4×3×2×1= 56 ways.
Step 4: Total number of different committees = 56 + 70 + 56 = 182 commit-
tees.
Question 5
Question
A committee of 5 members is to be formed from a group of 10 students. If 3 stu-
dents refuse to serve on the committee together, how many different committees
can be formed?
Solution
Step 1: Find the total number of committees without any restrictions. There
are 10 students to choose from to form a committee of 5 members. This can be
represented by the combination formula, n
k=n!
k!(n−k)! . So, the total number
of committees without restrictions is:
10
5=10!
5!(10 −5)! =10 ×9×8×7×6
5×4×3×2×1= 252.
Step 2: Find the number of committees where the 3 students refuse to serve
together. Let’s consider those 3 students as one group. Then we have 8 students
(10 students - 2 groups of 1 and 1 group of 3) to choose the remaining 5 members
of the committee from. This can be represented by the combination formula,
n
k=n!
k!(n−k)! . So, the number of committees where the 3 students refuse to
serve together is:
8
5=8!
5!(8 −5)! =8×7×6×5×4
5×4×3×2×1= 56.
4
Step 3: Calculate the number of committees that can be formed. We subtract
the number of committees where the 3 students refuse to serve together from
the total number of committees without restrictions:
252 −56 = 196.
Therefore, there are 196 different committees that can be formed when 3
students refuse to serve on the committee together.
Question 6
Question
A group of 10 friends go out to eat at a restaurant that has 5 different appetizers,
7 different main courses, and 4 different desserts on the menu. Each person
orders one appetizer, one main course, and one dessert. In how many ways
can the group’s orders be arranged if no two people can order the exact same
combination of dishes?
Solution
Step 1: Calculate the number of ways to choose one appetizer for each person.
Since there are 5 different appetizers on the menu, the first person has 5 choices,
the second person has 4 choices (as they can’t choose the same appetizer as the
first person), the third person has 3 choices, and so on. Therefore, the total
number of ways to choose appetizers for all 10 people is 5 ×4×3×2×1.
Step 2: Calculate the number of ways to choose one main course for each
person. Following a similar logic to Step 1, there are 7 choices for the first
person, 6 choices for the second person, 5 choices for the third person, and so
on. Therefore, the total number of ways to choose main courses for all 10 people
is 7 ×6×5×4×3×2×1.
Step 3: Calculate the number of ways to choose one dessert for each person.
Similarly, there are 4 choices for the first person, 3 choices for the second person,
2 choices for the third person, and so on. Therefore, the total number of ways
to choose desserts for all 10 people is 4 ×3×2×1.
Step 4: Multiply the results from Steps 1, 2, and 3 to find the total number
of ways the group’s orders can be arranged without any two people ordering the
same combination of dishes. Therefore, the total number of ways is 5 ×4×3×
2×1×7×6×5×4×3×2×1×4×3×2×1 = 604,800. Hence, there are
604,800 ways the group’s orders can be arranged.
5
Question 7
Question
In a group of 10 people, how many ways can we choose a committee of 3 people
such that at least one person must be included?
Solution
Step 1: Calculate the total number of ways to choose a committee of 3 people
from a group of 10 people. We can do this using the combination formula
n
r=n!
r!(n−r)! .
Total ways to choose a committee of 3 from 10 = 10
3=10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
Step 2: Calculate the number of ways to choose a committee of 3 people
such that no one is included. In this case, we need to choose 3 people from the
remaining 9 people (excluding the 1 person who must be included).
Ways to choose 3 people from 9 (none from the mandatory person) = 9
3=9!
3!(9 −3)! =9×8×7
3×2×1= 84
Step 3: Subtract the number of ways we can choose a committee of 3 people
with none from the mandatory person from the total number of ways to choose
a committee of 3 people.
Number of ways to choose a committee of 3 with at least 1 person included = 120−84 = 36
Therefore, there are 36 ways to choose a committee of 3 people from a group of
10 people such that at least 1 person must be included.
Question 8
Question
How many 5-digit numbers can be formed using the digits 1, 2, 3, 4, 5, 6, 7, 8,
9, and 0 if repetition of digits is not allowed and the number cannot start with
0?
Solution
Step 1: Determine the number of ways to choose the first digit (not a zero).
There are 9 possible choices (1-9) for the first digit.
Step 2: Determine the number of ways to choose the remaining 4 digits.
Since repetition is not allowed, there are 9 choices for the second digit, 8 choices
for the third digit, 7 choices for the fourth digit, and 6 choices for the fifth digit.
6
Step 3: Calculate the total number of 5-digit numbers. Multiply the number
of choices at each step:
9×9×8×7×6 = 27216
So, there are 27,216 different 5-digit numbers that can be formed using the
digits 1-9 without repetition and not starting with 0.
Question 9
Question
A committee of 5 people is to be formed from a group of 10 mathematicians
and 6 physicists. If the committee must consist of 3 mathematicians and 2
physicists, how many different committees can be formed?
Solution
Step 1: Calculate the number of ways to choose 3 mathematicians and 2 physi-
cists from their respective groups. Step 2: Multiply the results from Step 1 to
find the total number of different committees that can be formed.
Step 1: The number of ways to choose 3 mathematicians from 10 is given
by the combination formula 10C3:
10C3=10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
The number of ways to choose 2 physicists from 6 is given by the combination
formula 6C2:
6C2=6!
2!(6 −2)! =6×5
2×1= 15
Step 2: Multiply the results from Step 1 to find the total number of different
committees that can be formed:
120 ×15 = 1800
Therefore, there are 1800 different committees that can be formed with 3
mathematicians and 2 physicists.
Question 10
Question
In how many ways can we form a committee of 5 people from a group of 10 men
and 8 women if the committee must have at least 2 men and at least 2 women?
7
Solution
Step 1: Calculate the number of ways to choose 2 men and 3 women. Since
there are 10 men and 8 women, the number of ways to choose 2 men from 10 is
10
2and the number of ways to choose 3 women from 8 is 8
3. Therefore, the
total number of ways to choose 2 men and 3 women is 10
2×8
3.
Step 2: Calculate the number of ways to choose 3 men and 2 women. Sim-
ilarly, the number of ways to choose 3 men from 10 is 10
3and the number of
ways to choose 2 women from 8 is 8
2. So, the total number of ways to choose
3 men and 2 women is 10
3×8
2.
Step 3: Add the results from Step 1 and Step 2 to find the total number of
valid committee formations. The total number of ways to form the committee
is the sum of the results from Step 1 and Step 2: Total = 10
2×8
3+10
3×8
2.
Question 11
Question
A committee of 5 students is to be formed from a group of 10 students. If 4
of the students are math majors and 6 are computer science majors, how many
different committees can be formed if each committee must have at least 2 math
majors and at least 1 computer science major?
Solution
Step 1: Calculate the number of committees with 2 math majors and 3 computer
science majors. Step 2: Calculate the number of committees with 3 math majors
and 2 computer science majors. Step 3: Add the results from Step 1 and Step
2 to find the total number of valid committees.
Step 1: To select 2 math majors from 4, we use combinations: 4
2= 6 ways.
To select 3 computer science majors from 6, we use combinations: 6
3= 20 ways.
Multiplying these choices together, we get 6 ×20 = 120 committees.
Step 2: To select 3 math majors from 4, we use combinations: 4
3= 4 ways.
To select 2 computer science majors from 6, we use combinations: 6
2= 15 ways.
Multiplying these choices together, we get 4 ×15 = 60 committees.
Step 3: Adding the results from Step 1 and Step 2, we get a total of
120 + 60 = 180 valid committees that can be formed.
Question 12
Question
A committee of 5 people is to be formed from a group of 8 women and 6 men.
In how many ways can the committee be formed if it must contain at least 2
women and 2 men?
8
Solution
Step 1: Calculate the total number of ways to form a committee with 5 people.
This can be calculated using combinations. The total number of ways to choose
5 people from 14 is 14
5.
Step 2: Calculate the number of ways to form a committee with only 2
women and 3 men. This can be calculated by choosing 2 women out of 8 and
3 men out of 6. The number of ways to choose 2 women from 8 is 8
2, and the
number of ways to choose 3 men from 6 is 6
3. Therefore, the total number of
ways to form a committee with 2 women and 3 men is 8
2×6
3.
Step 3: Calculate the number of ways to form a committee with more than
2 women and 2 men. Since the committee must have at least 2 women and
2 men, we can calculate the number of ways by subtracting the cases with
exactly 2 women and 3 men (calculated in Step 2) from the total number of
ways (calculated in Step 1). Hence, the number of ways to form a committee
with more than 2 women and 2 men is 14
5−(8
2×6
3).
Question 13
Question
In a group of 12 people, how many ways can we choose a committee of 5 people
if 2 specific people refuse to be on the committee together?
Solution
Step 1: Consider the total number of ways to choose a committee of 5 people
from 12. There are 12
5ways to choose a committee of 5 people from a group
of 12 without any restrictions.
Step 2: Consider the number of ways to choose a committee when the 2
specific people refuse to be on the committee together. Let’s first find the
number of ways to choose a committee when both of the specific people are on
the committee. This can be done in 10
3ways, choosing the remaining 3 people
from the remaining 10 people.
Step 3: Next, consider the number of ways to choose a committee when both
of the specific people are not on the committee. This can be done in 10
5ways,
since we are choosing all 5 people from the remaining 10 people. However, we
need to subtract the cases where both specific people are together. This occurs
in 10
3ways. Thus, the number of ways to choose a committee of 5 people
without the 2 specific people being together is 10
5−10
3.
Step 4: Compute the final answer. The total number of ways to choose a
committee when the 2 specific people refuse to be on the committee together is:
12
5−10
5−10
3 = 792 −(252 −120) = 660. Therefore, there are 660 ways
to choose a committee of 5 people from a group of 12 when 2 specific people
refuse to be on the committee together.
9
Question 14
Question
In how many ways can a committee of 4 people be formed from a group of 7
men and 5 women if the committee must consist of at least 2 women?
Solution
Step 1: Calculate the number of ways to choose a committee with exactly 2
women: There are 5
2ways to choose 2 women from the 5 available women,
and 7
2ways to choose the remaining 2 people from the 7 men. So, the number
of ways to choose a committee with exactly 2 women is 5
2×7
2.
Step 2: Calculate the number of ways to choose a committee with exactly
3 women: There are 5
3ways to choose 3 women from the 5 available women,
and 7
1ways to choose the remaining person from the 7 men. So, the number
of ways to choose a committee with exactly 3 women is 5
3×7
1.
Step 3: Calculate the number of ways to choose a committee with all 4
women: There is 5
4ways to choose 4 women from the 5 available women. So,
the number of ways to choose a committee with all 4 women is 5
4.
Step 4: Add up the number of ways from each case to get the total number of
ways to form the committee: Total number of ways = 5
2×7
2+5
3×7
1+5
4.
Calculating the values, we get: 5
2= 10, 7
2= 21, 5
3= 10, 7
1= 7,
5
4= 5.
Therefore, the total number of ways to form the committee is 10 ×21 + 10 ×
7 + 5 = 210 + 70 + 5 = 285.
Question 15
Question
A committee of 5 people is to be selected from a group of 10 students and
8 professors. If the committee must contain at least 3 students, how many
different committees can be formed?
Solution
Step 1: Calculate the number of committees with exactly 3, 4, or 5 students.
Choose 3 students from 10: 10
3
Choose 2 other members from the remaining 15 (5 students and 8 profes-
sors): 15
2
Multiply these to get the number of committees with exactly 3 students:
10
3·15
2
10
Choose 4 students from 10: 10
4
Choose 1 other member from the remaining 14: 14
1
Multiply these to get the number of committees with exactly 4 students:
10
4·14
1
Choose 5 students from 10: 10
5
Since we can’t have all professors in the committee, only 3 professors are
left to choose from. So, choose 3 professors from 8: 8
3
Multiply these to get the number of committees with all 5 students: 10
5·
8
3
Step 2: Add up the number of committees with exactly 3, 4, or 5 students
to find the total number of committees.
10
3·15
2+10
4·14
1+10
5·8
3
= 120 ·105 + 210 ·14 + 252 ·56
= 12600 + 2940 + 14112
= 29652
Answer: There are 29,652 different committees that can be formed.
Question 16
Question
A committee of 5 people is to be formed from a group of 8 students and 4
professors. If the committee must consist of 2 students and 3 professors, how
many different committees can be formed?
Solution
Step 1: Calculate the number of ways to choose 2 students out of 8. There are
8
2ways to choose 2 students from a group of 8.
8
2=8!
2!(8 −2)! =8×7
2×1= 28.
Step 2: Calculate the number of ways to choose 3 professors out of 4. There
are 4
3ways to choose 3 professors from a group of 4.
4
3=4!
3!(4 −3)! =4
1= 4.
11
Step 3: Find the total number of different committees that can be formed
by multiplying the number of ways to choose students and professors. Total
number of committees = 8
2×4
3= 28 ×4 = 112.
Therefore, there are 112 different committees that can be formed with 2
students and 3 professors.
Question 17
Question
A committee of 5 people is to be formed from a group of 10 men and 8 women. If
the committee must consist of at least 2 men and 2 women, how many different
committees can be formed?
Solution
Let’s break this problem down into cases: Case 1: 2 men and 3 women Case 2:
3 men and 2 women Case 3: 4 men and 1 woman Case 4: 5 men and 0 women
Case 1: 2 men and 3 women Step 1: Choose 2 men from 10 men:
10
2=10!
2!(10 −2)! = 45
Step 2: Choose 3 women from 8 women:
8
3=8!
3!(8 −3)! = 56
Step 3: Multiply the number of ways to choose men and women together:
45 ×56 = 2520
Case 2: 3 men and 2 women Step 1: Choose 3 men from 10 men:
10
3=10!
3!(10 −3)! = 120
Step 2: Choose 2 women from 8 women:
8
2=8!
2!(8 −2)! = 28
Step 3: Multiply the number of ways to choose men and women together:
120 ×28 = 3360
Case 3: 4 men and 1 woman Step 1: Choose 4 men from 10 men:
10
4=10!
4!(10 −4)! = 210
12
Step 2: Choose 1 woman from 8 women:
8
1=8!
1!(8 −1)! = 8
Step 3: Multiply the number of ways to choose men and women together:
210 ×8 = 1680
Case 4: 5 men and 0 women Step 1: Choose 5 men from 10 men:
10
5=10!
5!(10 −5)! = 252
Step 2: Multiply the number of ways to choose men:
252 ×1 = 252
Now, we add up the number of committees from each case to get the total
number of different committees that can be formed:
2520 + 3360 + 1680 + 252 = 8520
Therefore, there are 8520 different committees that can be formed.
Question 18
Question
A committee of 5 people is to be formed from a group of 10 men and 8 women. If
2 men refuse to serve together on the committee, how many different committees
can be formed?
Solution
Step 1: Calculate the number of committees without any restrictions.
Since we are choosing 5 people from 18, this can be represented by the combi-
nation formula:
18
5=18!
5!(18 −5)! =18 ×17 ×16 ×15 ×14
5×4×3×2×1= 8568
Step 2: Calculate the number of committees with the two men who refuse
to serve together.
To find the number of committees with the two men who refuse to serve together,
we first calculate the number of committees with both of them on the committee
and then subtract this from the total number of committees. We choose one
of the 10 men to be on the committee (excluding the 2 who refuse to serve
13
together), and then choose the remaining 4 committee members from the 16
remaining people, which can be done with the combinations formula:
10
1×16
4= 10 ×16!
4!12! = 10 ×16 ×15 ×14 ×13
4×3×2×1= 3640
Step 3: Calculate the number of valid committees.
To find the number of valid committees (where the 2 men who refuse to serve
together are not on the committee together), we subtract the number of com-
mittees with the two men together from the total number of committees:
8568 −3640 = 4928
Answer: There are 4928 different committees that can be formed.
Question 19
Question
In a group of 10 friends, how many ways can we choose a committee with 3
members and a president if two particular friends, Alice and Bob, refuse to
serve on the committee together?
Solution
Step 1: First, we find the total number of ways to choose a committee with 3
members and a president from the group of 10 friends. This is represented by
the combination formula nCk=n!
k!(n−k)! , where nis the total number of friends
and kis the number of committee members.
Total ways to choose committee with 3 members =10 C3=10!
3!7! = 120
Step 2: Next, we find the number of ways in which Alice and Bob can serve
together on the committee. Since Alice and Bob must be chosen together, we
treat them as one entity.
Ways to choose Alice and Bob together =9C1= 9
Step 3: Now, within the committee, we have 9 friends remaining (Alice and
Bob are considered as one). From these 9 friends, we need to choose 1 more
member to complete the 3-member committee.
Ways to choose 1 more member =9C1= 9
Step 4: Finally, the total number of ways to choose a committee with 3
members and a president such that Alice and Bob do not serve together is the
14
total ways to choose a committee minus the ways to choose Alice and Bob
together times the ways to choose 1 more member.
Number of ways to choose committee without Alice and Bob together = 120−(9×9) = 120−81 = 39
Therefore, there are 39 ways to choose a committee with 3 members and a
president where Alice and Bob do not serve together.
Question 20
Question
A committee of 5 people is to be formed from a group of 10 men and 5 women. If
the committee must consist of at least 2 women, how many different committees
can be formed?
Solution
Step 1: Calculate the number of committees with exactly 2 women: Since there
are 5 women to choose from, and we need to choose 2 of them, we can calculate
the number of ways to choose 2 women.
5
2=5!
2!(5 −2)! = 10
Step 2: Calculate the number of committees with exactly 3 women: To find
the number of ways to choose 3 women out of 5:
5
3=5!
3!(5 −3)! = 10
Step 3: Calculate the number of committees with 4 or 5 women: Since we
need at least 2 women on the committee, we already counted the cases of exactly
2 or exactly 3 women. So, the remaining cases are the committees with 4 or 5
women. 5
4=5!
4!(5 −4)! = 5
5
5=5!
5!(5 −5)! = 1
Step 4: Calculate the number of committees with just men: The remaining
cases are when the committee consists only of men.
10
5=10!
5!(10 −5)! = 252
Step 5: Sum up all the cases to find the total number of different committees
that can be formed: Total = committees with exactly 2 women + committees
15
with exactly 3 women + committees with 4 women + committees with 5 women
+ committees with all men Total = 10 + 10 + 5 + 1 + 252 Total = 278
Therefore, there are 278 different committees that can be formed from the
group of 10 men and 5 women, where the committee must consist of at least 2
women.
Question 21
Question
In how many ways can we arrange the letters of the word ”MISSISSIPPI” such
that no two identical letters are adjacent?
Solution
Step 1: Count the total number of arrangements of the letters in ”MISSIS-
SIPPI”.
There are a total of 11 letters in ”MISSISSIPPI”, including 4 S’s, 4 I’s, 2
P’s, and 1 M. Therefore, the total number of arrangements is 11!.
Step 2: Count the number of arrangements where the identical letters appear
adjacent to each other.
Consider the group of 4 S’s as a single entity, the group of 4 I’s as a single
entity, and the group of 2 P’s as a single entity. Then we have a total of 4
entities: SSSS, IIII, P, M.
These 4 entities can be arranged in 4! ways, and within each entity, the
letters can be arranged in 4!, 4!, 2!, and 1! ways respectively. Thus, the number
of arrangements where the identical letters appear adjacent is 4!×4!×4!×2!×1!.
Step 3: Calculate the number of arrangements where no two identical letters
are adjacent.
The number of arrangements with no two identical letters adjacent is the
total number of arrangements minus the number of arrangements where the
identical letters appear adjacent. Therefore, the required number of arrange-
ments is:
11! −4! ×4! ×4! ×2! ×1!
Step 4: Calculate the final answer.
Substitute the values and calculate:
11! −4! ×4! ×4! ×2! ×1! = 39916800 −138240 = 39778560
Therefore, there are 39,778,560 ways to arrange the letters of the word ”MIS-
SISSIPPI” such that no two identical letters are adjacent.
16
Question 22
Question
How many ways are there to arrange the letters of the word ”MISSISSIPPI” if
no two S’s are adjacent?
Solution
Step 1: First, calculate the total number of ways to arrange the letters in
”MISSISSIPPI” without any restrictions. Step 2: Then, calculate the number
of ways where the S’s are adjacent. Step 3: Finally, subtract the number of
ways where the S’s are adjacent from the total number of ways to get the final
answer.
Step 1: The word ”MISSISSIPPI” has 11 letters (4 I’s, 4 S’s, 2 P’s, and 1
M). Therefore, there are a total of 11!
4!4!2!1! = 34,650 ways to arrange the letters
without any restrictions.
Step 2: To calculate the number of ways where the S’s are adjacent, treat
the pair of S’s as a single letter. So, we have 10 ”letters” to arrange (3 I’s, 2
P’s, 2 S’s, 2 single S’s, and 1 M), resulting in 10!
3!2!2!2!1! = 151,200 ways.
Step 3: Subtract the number of arrangements where the S’s are adjacent
from the total number of ways: 34,650 −151,200 = −116,550 .
Therefore, there are 116,550 ways to arrange the letters of ”MISSISSIPPI”
such that no two S’s are adjacent.
Question 23
Question
Let’s say you are planning a 4-course meal from a menu of 10 entrees, 8 soups,
6 salads, and 5 desserts. In how many ways can you select the 4-course meal if
each course must be different?
Solution
Step 1: Calculate the number of ways to choose 1 entree out of 10. Step 2:
Calculate the number of ways to choose 1 soup out of 8. Step 3: Calculate the
number of ways to choose 1 salad out of 6. Step 4: Calculate the number of
ways to choose 1 dessert out of 5. Step 5: Multiply the results from Steps 1-4
to determine the total number of ways to select the 4-course meal.
Step 1: There are 10 ways to choose 1 entree out of 10.
Step 2: There are 8 ways to choose 1 soup out of 8.
Step 3: There are 6 ways to choose 1 salad out of 6.
Step 4: There are 5 ways to choose 1 dessert out of 5.
Step 5: Multiply the results from Steps 1-4.
10 ×8×6×5 = 2400
17
Therefore, there are 2400 ways to select the 4-course meal if each course
must be different.
Question 24
Question
In how many ways can 4 men and 5 women be arranged in a row if at least 2
women must be next to each other?
Solution
Step 1: Find the total number of ways to arrange the 4 men and 5 women
without any restrictions. There are a total of 9 people to be arranged, so there
are 9! ways to arrange them.
Step 2: Find the number of ways to arrange the 4 men and 5 women such
that no 2 women are next to each other. To ensure that no 2 women are next to
each other, we can treat the women as blocks and place them among the men.
There are 4 gaps between the men where the women can be placed. Since we
have 5 women to be placed, this can be done in P(4,5) = 4!
(4−5)! = 4! ways.
Step 3: Find the total number of arrangements where at least 2 women are
next to each other. To find the total number of arrangements where at least 2
women are next to each other, we use the principle of complementary counting.
First, we find the number of arrangements where no 2 women are next to each
other, and then subtract this from the total number of arrangements.
Number of arrangements where at least 2 women are next to each other =
Total number of arrangements - Number of arrangements where no 2 women
are next to each other. Number of arrangements where at least 2 women are
next to each other = 9! −4!.
Step 4: Calculate the final answer. The total number of ways to arrange 4
men and 5 women in a row if at least 2 women must be next to each other is
9! −4!.
Question 25
Question
A committee of 5 people is to be formed from a group of 10 men and 6 women. If
the committee must consist of at least 3 men and 2 women, how many different
committees can be formed?
18
Solution
Step 1: Find the number of ways to select 3 men from 10 men.
10
3=10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
Step 2: Find the number of ways to select 2 women from 6 women.
6
2=6!
2!(6 −2)! =6×5
2×1= 15
Step 3: Calculate the total number of ways to form a committee with at
least 3 men and 2 women. Since the committee must consist of at least 3 men
and 2 women, we can have: - 3 men and 2 women - 4 men and 1 woman - 5
men and 0 women
Number of ways to form a committee with 3 men and 2 women: 120 ×15 =
1800
Number of ways to form a committee with 4 men and 1 woman:
10
4×6
1=10!
4!(10 −4)! ×6!
1!(6 −1)! = 2100
Number of ways to form a committee with 5 men and 0 women: 1 (only one
way as all 5 spots will be filled by men)
So, the total number of different committees that can be formed is 1800 +
2100 + 1 = 3901.
Question 26
Question
A company wants to form a committee of 5 people from a pool of 10 men and 8
women. If the committee must consist of at least 2 men and at least 2 women,
how many different committees can be formed?
Solution
Step 1: Determine the number of ways to select exactly 2 men and 3 women for
the committee. Step 2: Determine the number of ways to select exactly 3 men
and 2 women for the committee. Step 3: Add the results from Step 1 and Step
2 to find the total number of different committees that can be formed.
Step 1: To choose 2 men from 10, we have 10
2ways. To choose 3 women
from 8, we have 8
3ways. Therefore, the number of ways to select exactly 2
men and 3 women is 10
2×8
3.
Step 2: To choose 3 men from 10, we have 10
3ways. To choose 2 women
from 8, we have 8
2ways. Therefore, the number of ways to select exactly 3
men and 2 women is 10
3×8
2.
19
Step 3: Adding the results from Step 1 and Step 2 gives the total number of
different committees:
10
2×8
3+10
3×8
2
Question 27
Question
How many ways are there to choose a committee of 4 people from a group of
8 men and 5 women, if the committee must consist of exactly 3 men and 1
woman?
Solution
To find the number of ways to choose a committee of 3 men and 1 woman from
the group, we will use the concepts of permutations and combinations.
Step 1: Find the number of ways to choose 3 men from the 8 men.
We will use combinations here because the order of selection does not matter.
Number of ways to choose 3 men from 8 men = 8
3=8!
3!(8 −3)! = 56
Step 2: Find the number of ways to choose 1 woman from the 5
women. Again, we will use combinations because the order of selection does
not matter.
Number of ways to choose 1 woman from 5 women = 5
1=5!
1!(5 −1)! = 5
Step 3: Find the total number of ways to choose the committee. To
find the total number of ways to choose the committee consisting of 3 men and
1 woman, we multiply the number of ways to choose the men and the number
of ways to choose the woman.
Total number of ways to choose the committee = Number of ways to choose 3 men×Number of ways to choose 1 woman = 56×5 = 280
Therefore, there are 280 ways to choose a committee of 4 people consisting
of exactly 3 men and 1 woman from the group.
Question 28
Question
In a group of 10 students, calculate the number of ways to choose 4 students to
form a committee with a president, vice-president, secretary, and treasurer.
20
Solution
Step 1: Calculate the number of ways to choose 4 students out of 10. Step 2: For
each group of 4 students, assign one student to each position on the committee.
Step 3: Find the total number of ways by multiplying the results from Steps 1
and 2.
Step 1: To choose 4 students out of 10, we will use the combination formula
n
k=n!
k!(n−k)! . Substitute n= 10 and k= 4:
10
4=10!
4!(10 −4)! =10 ×9×8×7
4×3×2×1= 210.
Step 2: For each group of 4 students, we need to assign one student to
each position on the committee. - There are 4 choices for the president. -
After choosing the president, there are 3 choices left for the vice-president. -
After choosing the president and vice-president, there are 2 choices left for the
secretary. - Finally, the treasurer will be the remaining student.
Step 3: Multiply the results from Step 1 and Step 2 to find the total number
of ways to form the committee:
210 ×4×3×2 = 5040.
Therefore, there are 5040 ways to choose 4 students to form a committee
with a president, vice-president, secretary, and treasurer from a group of 10
students.
Question 29
Question
A committee of 5 people is to be formed from a group of 10 students and 5
professors. If the committee must consist of at least 2 students and 1 professor,
how many different committees can be formed?
Solution
Step 1: Calculate the number of ways to form a committee with 2 students and
3 professors. There are 10
2ways to choose 2 students from the 10 available
students, and 5
3ways to choose 3 professors from the 5 available professors.
Therefore, the number of ways to form a committee with 2 students and 3
professors is 10
2·5
3.
Step 2: Calculate the number of ways to form a committee with 3 students
and 2 professors. Similarly, there are 10
3ways to choose 3 students and 5
2
ways to choose 2 professors. Therefore, the number of ways to form a committee
with 3 students and 2 professors is 10
3·5
2.
Step 3: Calculate the total number of different committees that can be
formed. The total number of different committees that can be formed is the
21
sum of the committees with 2 students and 3 professors, and the committees
with 3 students and 2 professors: 10
2·5
3+10
3·5
2.
Now we can perform the calculations to find the total number of different
committees that can be formed.
Question 30
Question
How many ways are there to form a committee of 5 people from a group of 10
people, where 3 of them are men and 7 of them are women, if the committee
must have at least 2 men?
Solution
To find the number of ways to form a committee of 5 people with at least 2
men, we will consider the different possible scenarios separately and then add
them up.
Step 1: Calculate the number of ways to choose 2 men and 3 women: Since
there are 3 men and 7 women, the number of ways to choose 2 men out of
3 is 3
2= 3. Similarly, the number of ways to choose 3 women out of 7 is
7
3= 35. Therefore, the total number of ways to choose 2 men and 3 women is
3×35 = 105.
Step 2: Calculate the number of ways to choose 3 men and 2 women: Simi-
larly, the number of ways to choose 3 men out of 3 is 3
3= 1. The number of
ways to choose 2 women out of 7 is 7
2= 21. Therefore, the total number of
ways to choose 3 men and 2 women is 1 ×21 = 21.
Step 3: Add up the two cases to get the total number of ways: The total
number of ways to form a committee of 5 people with at least 2 men is 105+21 =
126 ways.
Question 31
Question
A group of 8 friends want to take a group photo. In how many ways can they
arrange themselves in a row for the photo if:
a) There are no restrictions on the arrangement.
b) The tallest friend must be at one end of the row.
22
Solution
a) When there are no restrictions on the arrangement, the number of ways the 8
friends can line up in a row is given by the number of permutations of 8 objects,
which is 8! = 40,320.
b) If the tallest friend must be at one end of the row, we treat the tallest
friend as one entity. This way, there are 2 entities (tallest friend entity and the
rest of the friends) to arrange in a row. The number of ways to arrange these 2
entities is 2! = 2.
For each of these arrangements, the friends within the ”rest of the friends”
entity can be arranged in 7! ways.
Therefore, the total number of ways the friends can line up in a row with
the tallest friend at one end is 2 ×7! = 10,080.
Question 32
Question
In how many ways can you arrange the letters in the word ”UNIVERSITY”
such that no two vowels are adjacent?
Solution
Step 1: First, we identify the total number of ways to arrange the letters in the
word ”UNIVERSITY” without any restrictions.
The word ”UNIVERSITY” has 10 letters, but the letters ’I’ and ’U’ appear
twice.
Therefore, the total number of ways to arrange the letters without any re-
strictions is given by:
10!/(2! ·2!) = 9,072,000
Step 2: Next, we consider the restriction that no two vowels can be adjacent.
Let’s treat the letters ’U’, ’I’, and ’E’ as a single entity (a block) since
they cannot be adjacent. This reduces the problem to arranging the letters
’NVRSTYB’ plus the block ’UEI’.
The total number of ways to arrange these letters and the block is given by:
7! ×3! = 30,240
Step 3: Therefore, the number of ways to arrange the letters in the word
”UNIVERSITY” such that no two vowels are adjacent is 30,240.
23
Question 33
Question
A committee of 5 people is to be formed from a group of 10 men and 7 women. If
the committee must consist of at least 2 women, how many different committees
can be formed?
Solution
To find the number of different committees that can be formed, we need to
consider the different cases where the number of women in the committee ranges
from 2 to 5.
Step 1: Calculate the number of committees with 2 women and 3
men
We need to choose 2 women out of 7 and 3 men out of 10. Number of ways
to choose 2 women from 7: 7
2Number of ways to choose 3 men from 10: 10
3
Total number of committees with 2 women and 3 men: 7
2×10
3
Step 2: Calculate the number of committees with 3 women and 2
men
Similarly, we need to choose 3 women out of 7 and 2 men out of 10. Number
of ways to choose 3 women from 7: 7
3Number of ways to choose 2 men from
10: 10
2Total number of committees with 3 women and 2 men: 7
3×10
2
Step 3: Calculate the number of committees with 4 women and 1
man
Next, we need to choose 4 women out of 7 and 1 man out of 10. Number of
ways to choose 4 women from 7: 7
4Number of ways to choose 1 man from 10:
10
1Total number of committees with 4 women and 1 man: 7
4×10
1
Step 4: Calculate the number of committees with 5 women and 0
men
Lastly, we need to choose all 5 women out of 7 (as there are no men in this
case). Number of ways to choose 5 women from 7: 7
5
Step 5: Add up the total number of committees
Now, we can find the total number of committees by summing up the com-
mittees from each case. Total number of committees = 7
2×10
3+7
3×10
2+
7
4×10
1+7
5
Question 34
Question
A committee of 5 people is to be formed from a group of 10 students. If 3 of
the students are in math club and 7 are in science club, how many different
committees can be formed if each committee must have at least 1 math club
member and 1 science club member?
24
Solution
Let’s break this problem down into cases to account for the different possibilities.
Case 1: 1 math club member and 4 science club members Step 1:
Choose 1 math club member out of 3.
Step 2: Choose 4 science club members out of 7.
Number of ways for Case 1 = 3
1×7
4
Case 2: 2 math club members and 3 science club members Step 1:
Choose 2 math club members out of 3.
Step 2: Choose 3 science club members out of 7.
Number of ways for Case 2 = 3
2×7
3
Case 3: 3 math club members and 2 science club members Step 1:
Choose 3 math club members out of 3.
Step 2: Choose 2 science club members out of 7.
Number of ways for Case 3 = 3
3×7
2
Finally, to find the total number of different committees that can be formed,
we sum up the results of each case.
Total number of committees = 3
1×7
4+3
2×7
3+3
3×7
2
Question 35
Question
In a card game, each player is dealt 5 cards from a standard deck of 52 playing
cards. How many ways can the cards be dealt such that each player receives
exactly 2 red cards and 3 black cards?
Solution
Step 1: Determine the total number of ways to choose 2 red cards out of 26 red
cards. There are 26
2ways to choose 2 red cards.
Step 2: Determine the total number of ways to choose 3 black cards out of
26 black cards. There are 26
3ways to choose 3 black cards.
Step 3: Multiply the results from Step 1 and Step 2 to find the total number
of ways to choose 2 red cards and 3 black cards. Total ways = 26
2×26
3.
Step 4: Calculate the total number of ways. 26
2=26!
2!(26−2)! =26×25
2= 325
26
3=26!
3!(26−3)! =26×25×24
3×2= 2600
Therefore, total ways = 325 ×2600 = 845,000.
25
Question 4
Question
A committee of 5 people is to be formed from a group of 8 men and 5 women.
If the committee must have at least 3 men, how many different committees can
be formed?
Solution
Step 1: Find the number of ways to choose a committee with exactly 3 men.
Step 2: Find the number of ways to choose a committee with exactly 4 men.
Step 3: Find the number of ways to choose a committee with all 5 men. Step
4: Add the results from Step 1, Step 2, and Step 3 to get the total number of
different committees.
Step 1: Choose 3 men from 8: 8
3=8!
3!(8−3)! =8×7×6
3×2×1= 56 ways.
Step 2: Choose 4 men from 8: 8
4=8!
4!(8−4)! =8×7×6×5
4×3×2×1= 70 ways.
Step 3: Choose all 5 men from 8: 8
5=8!
5!(8−5)! =8×7×6×5×4
5×4×3×2×1= 56 ways.
Step 4: Total number of different committees = 56 + 70 + 56 = 182 commit-
tees.
Question 5
Question
A committee of 5 members is to be formed from a group of 10 students. If 3 stu-
dents refuse to serve on the committee together, how many different committees
can be formed?
Solution
Step 1: Find the total number of committees without any restrictions. There
are 10 students to choose from to form a committee of 5 members. This can be
represented by the combination formula, n
k=n!
k!(n−k)! . So, the total number
of committees without restrictions is:
10
5=10!
5!(10 −5)! =10 ×9×8×7×6
5×4×3×2×1= 252.
Step 2: Find the number of committees where the 3 students refuse to serve
together. Let’s consider those 3 students as one group. Then we have 8 students
(10 students - 2 groups of 1 and 1 group of 3) to choose the remaining 5 members
of the committee from. This can be represented by the combination formula,
n
k=n!
k!(n−k)! . So, the number of committees where the 3 students refuse to
serve together is:
8
5=8!
5!(8 −5)! =8×7×6×5×4
5×4×3×2×1= 56.
4
Step 3: Calculate the number of committees that can be formed. We subtract
the number of committees where the 3 students refuse to serve together from
the total number of committees without restrictions:
252 −56 = 196.
Therefore, there are 196 different committees that can be formed when 3
students refuse to serve on the committee together.
Question 6
Question
A group of 10 friends go out to eat at a restaurant that has 5 different appetizers,
7 different main courses, and 4 different desserts on the menu. Each person
orders one appetizer, one main course, and one dessert. In how many ways
can the group’s orders be arranged if no two people can order the exact same
combination of dishes?
Solution
Step 1: Calculate the number of ways to choose one appetizer for each person.
Since there are 5 different appetizers on the menu, the first person has 5 choices,
the second person has 4 choices (as they can’t choose the same appetizer as the
first person), the third person has 3 choices, and so on. Therefore, the total
number of ways to choose appetizers for all 10 people is 5 ×4×3×2×1.
Step 2: Calculate the number of ways to choose one main course for each
person. Following a similar logic to Step 1, there are 7 choices for the first
person, 6 choices for the second person, 5 choices for the third person, and so
on. Therefore, the total number of ways to choose main courses for all 10 people
is 7 ×6×5×4×3×2×1.
Step 3: Calculate the number of ways to choose one dessert for each person.
Similarly, there are 4 choices for the first person, 3 choices for the second person,
2 choices for the third person, and so on. Therefore, the total number of ways
to choose desserts for all 10 people is 4 ×3×2×1.
Step 4: Multiply the results from Steps 1, 2, and 3 to find the total number
of ways the group’s orders can be arranged without any two people ordering the
same combination of dishes. Therefore, the total number of ways is 5 ×4×3×
2×1×7×6×5×4×3×2×1×4×3×2×1 = 604,800. Hence, there are
604,800 ways the group’s orders can be arranged.
5
Question 7
Question
In a group of 10 people, how many ways can we choose a committee of 3 people
such that at least one person must be included?
Solution
Step 1: Calculate the total number of ways to choose a committee of 3 people
from a group of 10 people. We can do this using the combination formula
n
r=n!
r!(n−r)! .
Total ways to choose a committee of 3 from 10 = 10
3=10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
Step 2: Calculate the number of ways to choose a committee of 3 people
such that no one is included. In this case, we need to choose 3 people from the
remaining 9 people (excluding the 1 person who must be included).
Ways to choose 3 people from 9 (none from the mandatory person) = 9
3=9!
3!(9 −3)! =9×8×7
3×2×1= 84
Step 3: Subtract the number of ways we can choose a committee of 3 people
with none from the mandatory person from the total number of ways to choose
a committee of 3 people.
Number of ways to choose a committee of 3 with at least 1 person included = 120−84 = 36
Therefore, there are 36 ways to choose a committee of 3 people from a group of
10 people such that at least 1 person must be included.
Question 8
Question
How many 5-digit numbers can be formed using the digits 1, 2, 3, 4, 5, 6, 7, 8,
9, and 0 if repetition of digits is not allowed and the number cannot start with
0?
Solution
Step 1: Determine the number of ways to choose the first digit (not a zero).
There are 9 possible choices (1-9) for the first digit.
Step 2: Determine the number of ways to choose the remaining 4 digits.
Since repetition is not allowed, there are 9 choices for the second digit, 8 choices
for the third digit, 7 choices for the fourth digit, and 6 choices for the fifth digit.
6
Step 3: Calculate the total number of 5-digit numbers. Multiply the number
of choices at each step:
9×9×8×7×6 = 27216
So, there are 27,216 different 5-digit numbers that can be formed using the
digits 1-9 without repetition and not starting with 0.
Question 9
Question
A committee of 5 people is to be formed from a group of 10 mathematicians
and 6 physicists. If the committee must consist of 3 mathematicians and 2
physicists, how many different committees can be formed?
Solution
Step 1: Calculate the number of ways to choose 3 mathematicians and 2 physi-
cists from their respective groups. Step 2: Multiply the results from Step 1 to
find the total number of different committees that can be formed.
Step 1: The number of ways to choose 3 mathematicians from 10 is given
by the combination formula 10C3:
10C3=10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
The number of ways to choose 2 physicists from 6 is given by the combination
formula 6C2:
6C2=6!
2!(6 −2)! =6×5
2×1= 15
Step 2: Multiply the results from Step 1 to find the total number of different
committees that can be formed:
120 ×15 = 1800
Therefore, there are 1800 different committees that can be formed with 3
mathematicians and 2 physicists.
Question 10
Question
In how many ways can we form a committee of 5 people from a group of 10 men
and 8 women if the committee must have at least 2 men and at least 2 women?
7
Solution
Step 1: Calculate the number of ways to choose 2 men and 3 women. Since
there are 10 men and 8 women, the number of ways to choose 2 men from 10 is
10
2and the number of ways to choose 3 women from 8 is 8
3. Therefore, the
total number of ways to choose 2 men and 3 women is 10
2×8
3.
Step 2: Calculate the number of ways to choose 3 men and 2 women. Sim-
ilarly, the number of ways to choose 3 men from 10 is 10
3and the number of
ways to choose 2 women from 8 is 8
2. So, the total number of ways to choose
3 men and 2 women is 10
3×8
2.
Step 3: Add the results from Step 1 and Step 2 to find the total number of
valid committee formations. The total number of ways to form the committee
is the sum of the results from Step 1 and Step 2: Total = 10
2×8
3+10
3×8
2.
Question 11
Question
A committee of 5 students is to be formed from a group of 10 students. If 4
of the students are math majors and 6 are computer science majors, how many
different committees can be formed if each committee must have at least 2 math
majors and at least 1 computer science major?
Solution
Step 1: Calculate the number of committees with 2 math majors and 3 computer
science majors. Step 2: Calculate the number of committees with 3 math majors
and 2 computer science majors. Step 3: Add the results from Step 1 and Step
2 to find the total number of valid committees.
Step 1: To select 2 math majors from 4, we use combinations: 4
2= 6 ways.
To select 3 computer science majors from 6, we use combinations: 6
3= 20 ways.
Multiplying these choices together, we get 6 ×20 = 120 committees.
Step 2: To select 3 math majors from 4, we use combinations: 4
3= 4 ways.
To select 2 computer science majors from 6, we use combinations: 6
2= 15 ways.
Multiplying these choices together, we get 4 ×15 = 60 committees.
Step 3: Adding the results from Step 1 and Step 2, we get a total of
120 + 60 = 180 valid committees that can be formed.
Question 12
Question
A committee of 5 people is to be formed from a group of 8 women and 6 men.
In how many ways can the committee be formed if it must contain at least 2
women and 2 men?
8
Solution
Step 1: Calculate the total number of ways to form a committee with 5 people.
This can be calculated using combinations. The total number of ways to choose
5 people from 14 is 14
5.
Step 2: Calculate the number of ways to form a committee with only 2
women and 3 men. This can be calculated by choosing 2 women out of 8 and
3 men out of 6. The number of ways to choose 2 women from 8 is 8
2, and the
number of ways to choose 3 men from 6 is 6
3. Therefore, the total number of
ways to form a committee with 2 women and 3 men is 8
2×6
3.
Step 3: Calculate the number of ways to form a committee with more than
2 women and 2 men. Since the committee must have at least 2 women and
2 men, we can calculate the number of ways by subtracting the cases with
exactly 2 women and 3 men (calculated in Step 2) from the total number of
ways (calculated in Step 1). Hence, the number of ways to form a committee
with more than 2 women and 2 men is 14
5−(8
2×6
3).
Question 13
Question
In a group of 12 people, how many ways can we choose a committee of 5 people
if 2 specific people refuse to be on the committee together?
Solution
Step 1: Consider the total number of ways to choose a committee of 5 people
from 12. There are 12
5ways to choose a committee of 5 people from a group
of 12 without any restrictions.
Step 2: Consider the number of ways to choose a committee when the 2
specific people refuse to be on the committee together. Let’s first find the
number of ways to choose a committee when both of the specific people are on
the committee. This can be done in 10
3ways, choosing the remaining 3 people
from the remaining 10 people.
Step 3: Next, consider the number of ways to choose a committee when both
of the specific people are not on the committee. This can be done in 10
5ways,
since we are choosing all 5 people from the remaining 10 people. However, we
need to subtract the cases where both specific people are together. This occurs
in 10
3ways. Thus, the number of ways to choose a committee of 5 people
without the 2 specific people being together is 10
5−10
3.
Step 4: Compute the final answer. The total number of ways to choose a
committee when the 2 specific people refuse to be on the committee together is:
12
5−10
5−10
3 = 792 −(252 −120) = 660. Therefore, there are 660 ways
to choose a committee of 5 people from a group of 12 when 2 specific people
refuse to be on the committee together.
9
Question 14
Question
In how many ways can a committee of 4 people be formed from a group of 7
men and 5 women if the committee must consist of at least 2 women?
Solution
Step 1: Calculate the number of ways to choose a committee with exactly 2
women: There are 5
2ways to choose 2 women from the 5 available women,
and 7
2ways to choose the remaining 2 people from the 7 men. So, the number
of ways to choose a committee with exactly 2 women is 5
2×7
2.
Step 2: Calculate the number of ways to choose a committee with exactly
3 women: There are 5
3ways to choose 3 women from the 5 available women,
and 7
1ways to choose the remaining person from the 7 men. So, the number
of ways to choose a committee with exactly 3 women is 5
3×7
1.
Step 3: Calculate the number of ways to choose a committee with all 4
women: There is 5
4ways to choose 4 women from the 5 available women. So,
the number of ways to choose a committee with all 4 women is 5
4.
Step 4: Add up the number of ways from each case to get the total number of
ways to form the committee: Total number of ways = 5
2×7
2+5
3×7
1+5
4.
Calculating the values, we get: 5
2= 10, 7
2= 21, 5
3= 10, 7
1= 7,
5
4= 5.
Therefore, the total number of ways to form the committee is 10 ×21 + 10 ×
7 + 5 = 210 + 70 + 5 = 285.
Question 15
Question
A committee of 5 people is to be selected from a group of 10 students and
8 professors. If the committee must contain at least 3 students, how many
different committees can be formed?
Solution
Step 1: Calculate the number of committees with exactly 3, 4, or 5 students.
Choose 3 students from 10: 10
3
Choose 2 other members from the remaining 15 (5 students and 8 profes-
sors): 15
2
Multiply these to get the number of committees with exactly 3 students:
10
3·15
2
10
Choose 4 students from 10: 10
4
Choose 1 other member from the remaining 14: 14
1
Multiply these to get the number of committees with exactly 4 students:
10
4·14
1
Choose 5 students from 10: 10
5
Since we can’t have all professors in the committee, only 3 professors are
left to choose from. So, choose 3 professors from 8: 8
3
Multiply these to get the number of committees with all 5 students: 10
5·
8
3
Step 2: Add up the number of committees with exactly 3, 4, or 5 students
to find the total number of committees.
10
3·15
2+10
4·14
1+10
5·8
3
= 120 ·105 + 210 ·14 + 252 ·56
= 12600 + 2940 + 14112
= 29652
Answer: There are 29,652 different committees that can be formed.
Question 16
Question
A committee of 5 people is to be formed from a group of 8 students and 4
professors. If the committee must consist of 2 students and 3 professors, how
many different committees can be formed?
Solution
Step 1: Calculate the number of ways to choose 2 students out of 8. There are
8
2ways to choose 2 students from a group of 8.
8
2=8!
2!(8 −2)! =8×7
2×1= 28.
Step 2: Calculate the number of ways to choose 3 professors out of 4. There
are 4
3ways to choose 3 professors from a group of 4.
4
3=4!
3!(4 −3)! =4
1= 4.
11
Step 3: Find the total number of different committees that can be formed
by multiplying the number of ways to choose students and professors. Total
number of committees = 8
2×4
3= 28 ×4 = 112.
Therefore, there are 112 different committees that can be formed with 2
students and 3 professors.
Question 17
Question
A committee of 5 people is to be formed from a group of 10 men and 8 women. If
the committee must consist of at least 2 men and 2 women, how many different
committees can be formed?
Solution
Let’s break this problem down into cases: Case 1: 2 men and 3 women Case 2:
3 men and 2 women Case 3: 4 men and 1 woman Case 4: 5 men and 0 women
Case 1: 2 men and 3 women Step 1: Choose 2 men from 10 men:
10
2=10!
2!(10 −2)! = 45
Step 2: Choose 3 women from 8 women:
8
3=8!
3!(8 −3)! = 56
Step 3: Multiply the number of ways to choose men and women together:
45 ×56 = 2520
Case 2: 3 men and 2 women Step 1: Choose 3 men from 10 men:
10
3=10!
3!(10 −3)! = 120
Step 2: Choose 2 women from 8 women:
8
2=8!
2!(8 −2)! = 28
Step 3: Multiply the number of ways to choose men and women together:
120 ×28 = 3360
Case 3: 4 men and 1 woman Step 1: Choose 4 men from 10 men:
10
4=10!
4!(10 −4)! = 210
12
Step 2: Choose 1 woman from 8 women:
8
1=8!
1!(8 −1)! = 8
Step 3: Multiply the number of ways to choose men and women together:
210 ×8 = 1680
Case 4: 5 men and 0 women Step 1: Choose 5 men from 10 men:
10
5=10!
5!(10 −5)! = 252
Step 2: Multiply the number of ways to choose men:
252 ×1 = 252
Now, we add up the number of committees from each case to get the total
number of different committees that can be formed:
2520 + 3360 + 1680 + 252 = 8520
Therefore, there are 8520 different committees that can be formed.
Question 18
Question
A committee of 5 people is to be formed from a group of 10 men and 8 women. If
2 men refuse to serve together on the committee, how many different committees
can be formed?
Solution
Step 1: Calculate the number of committees without any restrictions.
Since we are choosing 5 people from 18, this can be represented by the combi-
nation formula:
18
5=18!
5!(18 −5)! =18 ×17 ×16 ×15 ×14
5×4×3×2×1= 8568
Step 2: Calculate the number of committees with the two men who refuse
to serve together.
To find the number of committees with the two men who refuse to serve together,
we first calculate the number of committees with both of them on the committee
and then subtract this from the total number of committees. We choose one
of the 10 men to be on the committee (excluding the 2 who refuse to serve
13
together), and then choose the remaining 4 committee members from the 16
remaining people, which can be done with the combinations formula:
10
1×16
4= 10 ×16!
4!12! = 10 ×16 ×15 ×14 ×13
4×3×2×1= 3640
Step 3: Calculate the number of valid committees.
To find the number of valid committees (where the 2 men who refuse to serve
together are not on the committee together), we subtract the number of com-
mittees with the two men together from the total number of committees:
8568 −3640 = 4928
Answer: There are 4928 different committees that can be formed.
Question 19
Question
In a group of 10 friends, how many ways can we choose a committee with 3
members and a president if two particular friends, Alice and Bob, refuse to
serve on the committee together?
Solution
Step 1: First, we find the total number of ways to choose a committee with 3
members and a president from the group of 10 friends. This is represented by
the combination formula nCk=n!
k!(n−k)! , where nis the total number of friends
and kis the number of committee members.
Total ways to choose committee with 3 members =10 C3=10!
3!7! = 120
Step 2: Next, we find the number of ways in which Alice and Bob can serve
together on the committee. Since Alice and Bob must be chosen together, we
treat them as one entity.
Ways to choose Alice and Bob together =9C1= 9
Step 3: Now, within the committee, we have 9 friends remaining (Alice and
Bob are considered as one). From these 9 friends, we need to choose 1 more
member to complete the 3-member committee.
Ways to choose 1 more member =9C1= 9
Step 4: Finally, the total number of ways to choose a committee with 3
members and a president such that Alice and Bob do not serve together is the
14
total ways to choose a committee minus the ways to choose Alice and Bob
together times the ways to choose 1 more member.
Number of ways to choose committee without Alice and Bob together = 120−(9×9) = 120−81 = 39
Therefore, there are 39 ways to choose a committee with 3 members and a
president where Alice and Bob do not serve together.
Question 20
Question
A committee of 5 people is to be formed from a group of 10 men and 5 women. If
the committee must consist of at least 2 women, how many different committees
can be formed?
Solution
Step 1: Calculate the number of committees with exactly 2 women: Since there
are 5 women to choose from, and we need to choose 2 of them, we can calculate
the number of ways to choose 2 women.
5
2=5!
2!(5 −2)! = 10
Step 2: Calculate the number of committees with exactly 3 women: To find
the number of ways to choose 3 women out of 5:
5
3=5!
3!(5 −3)! = 10
Step 3: Calculate the number of committees with 4 or 5 women: Since we
need at least 2 women on the committee, we already counted the cases of exactly
2 or exactly 3 women. So, the remaining cases are the committees with 4 or 5
women. 5
4=5!
4!(5 −4)! = 5
5
5=5!
5!(5 −5)! = 1
Step 4: Calculate the number of committees with just men: The remaining
cases are when the committee consists only of men.
10
5=10!
5!(10 −5)! = 252
Step 5: Sum up all the cases to find the total number of different committees
that can be formed: Total = committees with exactly 2 women + committees
15
with exactly 3 women + committees with 4 women + committees with 5 women
+ committees with all men Total = 10 + 10 + 5 + 1 + 252 Total = 278
Therefore, there are 278 different committees that can be formed from the
group of 10 men and 5 women, where the committee must consist of at least 2
women.
Question 21
Question
In how many ways can we arrange the letters of the word ”MISSISSIPPI” such
that no two identical letters are adjacent?
Solution
Step 1: Count the total number of arrangements of the letters in ”MISSIS-
SIPPI”.
There are a total of 11 letters in ”MISSISSIPPI”, including 4 S’s, 4 I’s, 2
P’s, and 1 M. Therefore, the total number of arrangements is 11!.
Step 2: Count the number of arrangements where the identical letters appear
adjacent to each other.
Consider the group of 4 S’s as a single entity, the group of 4 I’s as a single
entity, and the group of 2 P’s as a single entity. Then we have a total of 4
entities: SSSS, IIII, P, M.
These 4 entities can be arranged in 4! ways, and within each entity, the
letters can be arranged in 4!, 4!, 2!, and 1! ways respectively. Thus, the number
of arrangements where the identical letters appear adjacent is 4!×4!×4!×2!×1!.
Step 3: Calculate the number of arrangements where no two identical letters
are adjacent.
The number of arrangements with no two identical letters adjacent is the
total number of arrangements minus the number of arrangements where the
identical letters appear adjacent. Therefore, the required number of arrange-
ments is:
11! −4! ×4! ×4! ×2! ×1!
Step 4: Calculate the final answer.
Substitute the values and calculate:
11! −4! ×4! ×4! ×2! ×1! = 39916800 −138240 = 39778560
Therefore, there are 39,778,560 ways to arrange the letters of the word ”MIS-
SISSIPPI” such that no two identical letters are adjacent.
16
Question 22
Question
How many ways are there to arrange the letters of the word ”MISSISSIPPI” if
no two S’s are adjacent?
Solution
Step 1: First, calculate the total number of ways to arrange the letters in
”MISSISSIPPI” without any restrictions. Step 2: Then, calculate the number
of ways where the S’s are adjacent. Step 3: Finally, subtract the number of
ways where the S’s are adjacent from the total number of ways to get the final
answer.
Step 1: The word ”MISSISSIPPI” has 11 letters (4 I’s, 4 S’s, 2 P’s, and 1
M). Therefore, there are a total of 11!
4!4!2!1! = 34,650 ways to arrange the letters
without any restrictions.
Step 2: To calculate the number of ways where the S’s are adjacent, treat
the pair of S’s as a single letter. So, we have 10 ”letters” to arrange (3 I’s, 2
P’s, 2 S’s, 2 single S’s, and 1 M), resulting in 10!
3!2!2!2!1! = 151,200 ways.
Step 3: Subtract the number of arrangements where the S’s are adjacent
from the total number of ways: 34,650 −151,200 = −116,550 .
Therefore, there are 116,550 ways to arrange the letters of ”MISSISSIPPI”
such that no two S’s are adjacent.
Question 23
Question
Let’s say you are planning a 4-course meal from a menu of 10 entrees, 8 soups,
6 salads, and 5 desserts. In how many ways can you select the 4-course meal if
each course must be different?
Solution
Step 1: Calculate the number of ways to choose 1 entree out of 10. Step 2:
Calculate the number of ways to choose 1 soup out of 8. Step 3: Calculate the
number of ways to choose 1 salad out of 6. Step 4: Calculate the number of
ways to choose 1 dessert out of 5. Step 5: Multiply the results from Steps 1-4
to determine the total number of ways to select the 4-course meal.
Step 1: There are 10 ways to choose 1 entree out of 10.
Step 2: There are 8 ways to choose 1 soup out of 8.
Step 3: There are 6 ways to choose 1 salad out of 6.
Step 4: There are 5 ways to choose 1 dessert out of 5.
Step 5: Multiply the results from Steps 1-4.
10 ×8×6×5 = 2400
17
Therefore, there are 2400 ways to select the 4-course meal if each course
must be different.
Question 24
Question
In how many ways can 4 men and 5 women be arranged in a row if at least 2
women must be next to each other?
Solution
Step 1: Find the total number of ways to arrange the 4 men and 5 women
without any restrictions. There are a total of 9 people to be arranged, so there
are 9! ways to arrange them.
Step 2: Find the number of ways to arrange the 4 men and 5 women such
that no 2 women are next to each other. To ensure that no 2 women are next to
each other, we can treat the women as blocks and place them among the men.
There are 4 gaps between the men where the women can be placed. Since we
have 5 women to be placed, this can be done in P(4,5) = 4!
(4−5)! = 4! ways.
Step 3: Find the total number of arrangements where at least 2 women are
next to each other. To find the total number of arrangements where at least 2
women are next to each other, we use the principle of complementary counting.
First, we find the number of arrangements where no 2 women are next to each
other, and then subtract this from the total number of arrangements.
Number of arrangements where at least 2 women are next to each other =
Total number of arrangements - Number of arrangements where no 2 women
are next to each other. Number of arrangements where at least 2 women are
next to each other = 9! −4!.
Step 4: Calculate the final answer. The total number of ways to arrange 4
men and 5 women in a row if at least 2 women must be next to each other is
9! −4!.
Question 25
Question
A committee of 5 people is to be formed from a group of 10 men and 6 women. If
the committee must consist of at least 3 men and 2 women, how many different
committees can be formed?
18
Solution
Step 1: Find the number of ways to select 3 men from 10 men.
10
3=10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
Step 2: Find the number of ways to select 2 women from 6 women.
6
2=6!
2!(6 −2)! =6×5
2×1= 15
Step 3: Calculate the total number of ways to form a committee with at
least 3 men and 2 women. Since the committee must consist of at least 3 men
and 2 women, we can have: - 3 men and 2 women - 4 men and 1 woman - 5
men and 0 women
Number of ways to form a committee with 3 men and 2 women: 120 ×15 =
1800
Number of ways to form a committee with 4 men and 1 woman:
10
4×6
1=10!
4!(10 −4)! ×6!
1!(6 −1)! = 2100
Number of ways to form a committee with 5 men and 0 women: 1 (only one
way as all 5 spots will be filled by men)
So, the total number of different committees that can be formed is 1800 +
2100 + 1 = 3901.
Question 26
Question
A company wants to form a committee of 5 people from a pool of 10 men and 8
women. If the committee must consist of at least 2 men and at least 2 women,
how many different committees can be formed?
Solution
Step 1: Determine the number of ways to select exactly 2 men and 3 women for
the committee. Step 2: Determine the number of ways to select exactly 3 men
and 2 women for the committee. Step 3: Add the results from Step 1 and Step
2 to find the total number of different committees that can be formed.
Step 1: To choose 2 men from 10, we have 10
2ways. To choose 3 women
from 8, we have 8
3ways. Therefore, the number of ways to select exactly 2
men and 3 women is 10
2×8
3.
Step 2: To choose 3 men from 10, we have 10
3ways. To choose 2 women
from 8, we have 8
2ways. Therefore, the number of ways to select exactly 3
men and 2 women is 10
3×8
2.
19
Step 3: Adding the results from Step 1 and Step 2 gives the total number of
different committees:
10
2×8
3+10
3×8
2
Question 27
Question
How many ways are there to choose a committee of 4 people from a group of
8 men and 5 women, if the committee must consist of exactly 3 men and 1
woman?
Solution
To find the number of ways to choose a committee of 3 men and 1 woman from
the group, we will use the concepts of permutations and combinations.
Step 1: Find the number of ways to choose 3 men from the 8 men.
We will use combinations here because the order of selection does not matter.
Number of ways to choose 3 men from 8 men = 8
3=8!
3!(8 −3)! = 56
Step 2: Find the number of ways to choose 1 woman from the 5
women. Again, we will use combinations because the order of selection does
not matter.
Number of ways to choose 1 woman from 5 women = 5
1=5!
1!(5 −1)! = 5
Step 3: Find the total number of ways to choose the committee. To
find the total number of ways to choose the committee consisting of 3 men and
1 woman, we multiply the number of ways to choose the men and the number
of ways to choose the woman.
Total number of ways to choose the committee = Number of ways to choose 3 men×Number of ways to choose 1 woman = 56×5 = 280
Therefore, there are 280 ways to choose a committee of 4 people consisting
of exactly 3 men and 1 woman from the group.
Question 28
Question
In a group of 10 students, calculate the number of ways to choose 4 students to
form a committee with a president, vice-president, secretary, and treasurer.
20
Solution
Step 1: Calculate the number of ways to choose 4 students out of 10. Step 2: For
each group of 4 students, assign one student to each position on the committee.
Step 3: Find the total number of ways by multiplying the results from Steps 1
and 2.
Step 1: To choose 4 students out of 10, we will use the combination formula
n
k=n!
k!(n−k)! . Substitute n= 10 and k= 4:
10
4=10!
4!(10 −4)! =10 ×9×8×7
4×3×2×1= 210.
Step 2: For each group of 4 students, we need to assign one student to
each position on the committee. - There are 4 choices for the president. -
After choosing the president, there are 3 choices left for the vice-president. -
After choosing the president and vice-president, there are 2 choices left for the
secretary. - Finally, the treasurer will be the remaining student.
Step 3: Multiply the results from Step 1 and Step 2 to find the total number
of ways to form the committee:
210 ×4×3×2 = 5040.
Therefore, there are 5040 ways to choose 4 students to form a committee
with a president, vice-president, secretary, and treasurer from a group of 10
students.
Question 29
Question
A committee of 5 people is to be formed from a group of 10 students and 5
professors. If the committee must consist of at least 2 students and 1 professor,
how many different committees can be formed?
Solution
Step 1: Calculate the number of ways to form a committee with 2 students and
3 professors. There are 10
2ways to choose 2 students from the 10 available
students, and 5
3ways to choose 3 professors from the 5 available professors.
Therefore, the number of ways to form a committee with 2 students and 3
professors is 10
2·5
3.
Step 2: Calculate the number of ways to form a committee with 3 students
and 2 professors. Similarly, there are 10
3ways to choose 3 students and 5
2
ways to choose 2 professors. Therefore, the number of ways to form a committee
with 3 students and 2 professors is 10
3·5
2.
Step 3: Calculate the total number of different committees that can be
formed. The total number of different committees that can be formed is the
21
sum of the committees with 2 students and 3 professors, and the committees
with 3 students and 2 professors: 10
2·5
3+10
3·5
2.
Now we can perform the calculations to find the total number of different
committees that can be formed.
Question 30
Question
How many ways are there to form a committee of 5 people from a group of 10
people, where 3 of them are men and 7 of them are women, if the committee
must have at least 2 men?
Solution
To find the number of ways to form a committee of 5 people with at least 2
men, we will consider the different possible scenarios separately and then add
them up.
Step 1: Calculate the number of ways to choose 2 men and 3 women: Since
there are 3 men and 7 women, the number of ways to choose 2 men out of
3 is 3
2= 3. Similarly, the number of ways to choose 3 women out of 7 is
7
3= 35. Therefore, the total number of ways to choose 2 men and 3 women is
3×35 = 105.
Step 2: Calculate the number of ways to choose 3 men and 2 women: Simi-
larly, the number of ways to choose 3 men out of 3 is 3
3= 1. The number of
ways to choose 2 women out of 7 is 7
2= 21. Therefore, the total number of
ways to choose 3 men and 2 women is 1 ×21 = 21.
Step 3: Add up the two cases to get the total number of ways: The total
number of ways to form a committee of 5 people with at least 2 men is 105+21 =
126 ways.
Question 31
Question
A group of 8 friends want to take a group photo. In how many ways can they
arrange themselves in a row for the photo if:
a) There are no restrictions on the arrangement.
b) The tallest friend must be at one end of the row.
22
Solution
a) When there are no restrictions on the arrangement, the number of ways the 8
friends can line up in a row is given by the number of permutations of 8 objects,
which is 8! = 40,320.
b) If the tallest friend must be at one end of the row, we treat the tallest
friend as one entity. This way, there are 2 entities (tallest friend entity and the
rest of the friends) to arrange in a row. The number of ways to arrange these 2
entities is 2! = 2.
For each of these arrangements, the friends within the ”rest of the friends”
entity can be arranged in 7! ways.
Therefore, the total number of ways the friends can line up in a row with
the tallest friend at one end is 2 ×7! = 10,080.
Question 32
Question
In how many ways can you arrange the letters in the word ”UNIVERSITY”
such that no two vowels are adjacent?
Solution
Step 1: First, we identify the total number of ways to arrange the letters in the
word ”UNIVERSITY” without any restrictions.
The word ”UNIVERSITY” has 10 letters, but the letters ’I’ and ’U’ appear
twice.
Therefore, the total number of ways to arrange the letters without any re-
strictions is given by:
10!/(2! ·2!) = 9,072,000
Step 2: Next, we consider the restriction that no two vowels can be adjacent.
Let’s treat the letters ’U’, ’I’, and ’E’ as a single entity (a block) since
they cannot be adjacent. This reduces the problem to arranging the letters
’NVRSTYB’ plus the block ’UEI’.
The total number of ways to arrange these letters and the block is given by:
7! ×3! = 30,240
Step 3: Therefore, the number of ways to arrange the letters in the word
”UNIVERSITY” such that no two vowels are adjacent is 30,240.
23
Question 33
Question
A committee of 5 people is to be formed from a group of 10 men and 7 women. If
the committee must consist of at least 2 women, how many different committees
can be formed?
Solution
To find the number of different committees that can be formed, we need to
consider the different cases where the number of women in the committee ranges
from 2 to 5.
Step 1: Calculate the number of committees with 2 women and 3
men
We need to choose 2 women out of 7 and 3 men out of 10. Number of ways
to choose 2 women from 7: 7
2Number of ways to choose 3 men from 10: 10
3
Total number of committees with 2 women and 3 men: 7
2×10
3
Step 2: Calculate the number of committees with 3 women and 2
men
Similarly, we need to choose 3 women out of 7 and 2 men out of 10. Number
of ways to choose 3 women from 7: 7
3Number of ways to choose 2 men from
10: 10
2Total number of committees with 3 women and 2 men: 7
3×10
2
Step 3: Calculate the number of committees with 4 women and 1
man
Next, we need to choose 4 women out of 7 and 1 man out of 10. Number of
ways to choose 4 women from 7: 7
4Number of ways to choose 1 man from 10:
10
1Total number of committees with 4 women and 1 man: 7
4×10
1
Step 4: Calculate the number of committees with 5 women and 0
men
Lastly, we need to choose all 5 women out of 7 (as there are no men in this
case). Number of ways to choose 5 women from 7: 7
5
Step 5: Add up the total number of committees
Now, we can find the total number of committees by summing up the com-
mittees from each case. Total number of committees = 7
2×10
3+7
3×10
2+
7
4×10
1+7
5
Question 34
Question
A committee of 5 people is to be formed from a group of 10 students. If 3 of
the students are in math club and 7 are in science club, how many different
committees can be formed if each committee must have at least 1 math club
member and 1 science club member?
24
Solution
Let’s break this problem down into cases to account for the different possibilities.
Case 1: 1 math club member and 4 science club members Step 1:
Choose 1 math club member out of 3.
Step 2: Choose 4 science club members out of 7.
Number of ways for Case 1 = 3
1×7
4
Case 2: 2 math club members and 3 science club members Step 1:
Choose 2 math club members out of 3.
Step 2: Choose 3 science club members out of 7.
Number of ways for Case 2 = 3
2×7
3
Case 3: 3 math club members and 2 science club members Step 1:
Choose 3 math club members out of 3.
Step 2: Choose 2 science club members out of 7.
Number of ways for Case 3 = 3
3×7
2
Finally, to find the total number of different committees that can be formed,
we sum up the results of each case.
Total number of committees = 3
1×7
4+3
2×7
3+3
3×7
2
Question 35
Question
In a card game, each player is dealt 5 cards from a standard deck of 52 playing
cards. How many ways can the cards be dealt such that each player receives
exactly 2 red cards and 3 black cards?
Solution
Step 1: Determine the total number of ways to choose 2 red cards out of 26 red
cards. There are 26
2ways to choose 2 red cards.
Step 2: Determine the total number of ways to choose 3 black cards out of
26 black cards. There are 26
3ways to choose 3 black cards.
Step 3: Multiply the results from Step 1 and Step 2 to find the total number
of ways to choose 2 red cards and 3 black cards. Total ways = 26
2×26
3.
Step 4: Calculate the total number of ways. 26
2=26!
2!(26−2)! =26×25
2= 325
26
3=26!
3!(26−3)! =26×25×24
3×2= 2600
Therefore, total ways = 325 ×2600 = 845,000.
25
Question 4
Question
A committee of 5 people is to be formed from a group of 8 men and 5 women.
If the committee must have at least 3 men, how many different committees can
be formed?
Solution
Step 1: Find the number of ways to choose a committee with exactly 3 men.
Step 2: Find the number of ways to choose a committee with exactly 4 men.
Step 3: Find the number of ways to choose a committee with all 5 men. Step
4: Add the results from Step 1, Step 2, and Step 3 to get the total number of
different committees.
Step 1: Choose 3 men from 8: 8
3=8!
3!(8−3)! =8×7×6
3×2×1= 56 ways.
Step 2: Choose 4 men from 8: 8
4=8!
4!(8−4)! =8×7×6×5
4×3×2×1= 70 ways.
Step 3: Choose all 5 men from 8: 8
5=8!
5!(8−5)! =8×7×6×5×4
5×4×3×2×1= 56 ways.
Step 4: Total number of different committees = 56 + 70 + 56 = 182 commit-
tees.
Question 5
Question
A committee of 5 members is to be formed from a group of 10 students. If 3 stu-
dents refuse to serve on the committee together, how many different committees
can be formed?
Solution
Step 1: Find the total number of committees without any restrictions. There
are 10 students to choose from to form a committee of 5 members. This can be
represented by the combination formula, n
k=n!
k!(n−k)! . So, the total number
of committees without restrictions is:
10
5=10!
5!(10 −5)! =10 ×9×8×7×6
5×4×3×2×1= 252.
Step 2: Find the number of committees where the 3 students refuse to serve
together. Let’s consider those 3 students as one group. Then we have 8 students
(10 students - 2 groups of 1 and 1 group of 3) to choose the remaining 5 members
of the committee from. This can be represented by the combination formula,
n
k=n!
k!(n−k)! . So, the number of committees where the 3 students refuse to
serve together is:
8
5=8!
5!(8 −5)! =8×7×6×5×4
5×4×3×2×1= 56.
4
Step 3: Calculate the number of committees that can be formed. We subtract
the number of committees where the 3 students refuse to serve together from
the total number of committees without restrictions:
252 −56 = 196.
Therefore, there are 196 different committees that can be formed when 3
students refuse to serve on the committee together.
Question 6
Question
A group of 10 friends go out to eat at a restaurant that has 5 different appetizers,
7 different main courses, and 4 different desserts on the menu. Each person
orders one appetizer, one main course, and one dessert. In how many ways
can the group’s orders be arranged if no two people can order the exact same
combination of dishes?
Solution
Step 1: Calculate the number of ways to choose one appetizer for each person.
Since there are 5 different appetizers on the menu, the first person has 5 choices,
the second person has 4 choices (as they can’t choose the same appetizer as the
first person), the third person has 3 choices, and so on. Therefore, the total
number of ways to choose appetizers for all 10 people is 5 ×4×3×2×1.
Step 2: Calculate the number of ways to choose one main course for each
person. Following a similar logic to Step 1, there are 7 choices for the first
person, 6 choices for the second person, 5 choices for the third person, and so
on. Therefore, the total number of ways to choose main courses for all 10 people
is 7 ×6×5×4×3×2×1.
Step 3: Calculate the number of ways to choose one dessert for each person.
Similarly, there are 4 choices for the first person, 3 choices for the second person,
2 choices for the third person, and so on. Therefore, the total number of ways
to choose desserts for all 10 people is 4 ×3×2×1.
Step 4: Multiply the results from Steps 1, 2, and 3 to find the total number
of ways the group’s orders can be arranged without any two people ordering the
same combination of dishes. Therefore, the total number of ways is 5 ×4×3×
2×1×7×6×5×4×3×2×1×4×3×2×1 = 604,800. Hence, there are
604,800 ways the group’s orders can be arranged.
5
Question 7
Question
In a group of 10 people, how many ways can we choose a committee of 3 people
such that at least one person must be included?
Solution
Step 1: Calculate the total number of ways to choose a committee of 3 people
from a group of 10 people. We can do this using the combination formula
n
r=n!
r!(n−r)! .
Total ways to choose a committee of 3 from 10 = 10
3=10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
Step 2: Calculate the number of ways to choose a committee of 3 people
such that no one is included. In this case, we need to choose 3 people from the
remaining 9 people (excluding the 1 person who must be included).
Ways to choose 3 people from 9 (none from the mandatory person) = 9
3=9!
3!(9 −3)! =9×8×7
3×2×1= 84
Step 3: Subtract the number of ways we can choose a committee of 3 people
with none from the mandatory person from the total number of ways to choose
a committee of 3 people.
Number of ways to choose a committee of 3 with at least 1 person included = 120−84 = 36
Therefore, there are 36 ways to choose a committee of 3 people from a group of
10 people such that at least 1 person must be included.
Question 8
Question
How many 5-digit numbers can be formed using the digits 1, 2, 3, 4, 5, 6, 7, 8,
9, and 0 if repetition of digits is not allowed and the number cannot start with
0?
Solution
Step 1: Determine the number of ways to choose the first digit (not a zero).
There are 9 possible choices (1-9) for the first digit.
Step 2: Determine the number of ways to choose the remaining 4 digits.
Since repetition is not allowed, there are 9 choices for the second digit, 8 choices
for the third digit, 7 choices for the fourth digit, and 6 choices for the fifth digit.
6
Step 3: Calculate the total number of 5-digit numbers. Multiply the number
of choices at each step:
9×9×8×7×6 = 27216
So, there are 27,216 different 5-digit numbers that can be formed using the
digits 1-9 without repetition and not starting with 0.
Question 9
Question
A committee of 5 people is to be formed from a group of 10 mathematicians
and 6 physicists. If the committee must consist of 3 mathematicians and 2
physicists, how many different committees can be formed?
Solution
Step 1: Calculate the number of ways to choose 3 mathematicians and 2 physi-
cists from their respective groups. Step 2: Multiply the results from Step 1 to
find the total number of different committees that can be formed.
Step 1: The number of ways to choose 3 mathematicians from 10 is given
by the combination formula 10C3:
10C3=10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
The number of ways to choose 2 physicists from 6 is given by the combination
formula 6C2:
6C2=6!
2!(6 −2)! =6×5
2×1= 15
Step 2: Multiply the results from Step 1 to find the total number of different
committees that can be formed:
120 ×15 = 1800
Therefore, there are 1800 different committees that can be formed with 3
mathematicians and 2 physicists.
Question 10
Question
In how many ways can we form a committee of 5 people from a group of 10 men
and 8 women if the committee must have at least 2 men and at least 2 women?
7
Solution
Step 1: Calculate the number of ways to choose 2 men and 3 women. Since
there are 10 men and 8 women, the number of ways to choose 2 men from 10 is
10
2and the number of ways to choose 3 women from 8 is 8
3. Therefore, the
total number of ways to choose 2 men and 3 women is 10
2×8
3.
Step 2: Calculate the number of ways to choose 3 men and 2 women. Sim-
ilarly, the number of ways to choose 3 men from 10 is 10
3and the number of
ways to choose 2 women from 8 is 8
2. So, the total number of ways to choose
3 men and 2 women is 10
3×8
2.
Step 3: Add the results from Step 1 and Step 2 to find the total number of
valid committee formations. The total number of ways to form the committee
is the sum of the results from Step 1 and Step 2: Total = 10
2×8
3+10
3×8
2.
Question 11
Question
A committee of 5 students is to be formed from a group of 10 students. If 4
of the students are math majors and 6 are computer science majors, how many
different committees can be formed if each committee must have at least 2 math
majors and at least 1 computer science major?
Solution
Step 1: Calculate the number of committees with 2 math majors and 3 computer
science majors. Step 2: Calculate the number of committees with 3 math majors
and 2 computer science majors. Step 3: Add the results from Step 1 and Step
2 to find the total number of valid committees.
Step 1: To select 2 math majors from 4, we use combinations: 4
2= 6 ways.
To select 3 computer science majors from 6, we use combinations: 6
3= 20 ways.
Multiplying these choices together, we get 6 ×20 = 120 committees.
Step 2: To select 3 math majors from 4, we use combinations: 4
3= 4 ways.
To select 2 computer science majors from 6, we use combinations: 6
2= 15 ways.
Multiplying these choices together, we get 4 ×15 = 60 committees.
Step 3: Adding the results from Step 1 and Step 2, we get a total of
120 + 60 = 180 valid committees that can be formed.
Question 12
Question
A committee of 5 people is to be formed from a group of 8 women and 6 men.
In how many ways can the committee be formed if it must contain at least 2
women and 2 men?
8
Solution
Step 1: Calculate the total number of ways to form a committee with 5 people.
This can be calculated using combinations. The total number of ways to choose
5 people from 14 is 14
5.
Step 2: Calculate the number of ways to form a committee with only 2
women and 3 men. This can be calculated by choosing 2 women out of 8 and
3 men out of 6. The number of ways to choose 2 women from 8 is 8
2, and the
number of ways to choose 3 men from 6 is 6
3. Therefore, the total number of
ways to form a committee with 2 women and 3 men is 8
2×6
3.
Step 3: Calculate the number of ways to form a committee with more than
2 women and 2 men. Since the committee must have at least 2 women and
2 men, we can calculate the number of ways by subtracting the cases with
exactly 2 women and 3 men (calculated in Step 2) from the total number of
ways (calculated in Step 1). Hence, the number of ways to form a committee
with more than 2 women and 2 men is 14
5−(8
2×6
3).
Question 13
Question
In a group of 12 people, how many ways can we choose a committee of 5 people
if 2 specific people refuse to be on the committee together?
Solution
Step 1: Consider the total number of ways to choose a committee of 5 people
from 12. There are 12
5ways to choose a committee of 5 people from a group
of 12 without any restrictions.
Step 2: Consider the number of ways to choose a committee when the 2
specific people refuse to be on the committee together. Let’s first find the
number of ways to choose a committee when both of the specific people are on
the committee. This can be done in 10
3ways, choosing the remaining 3 people
from the remaining 10 people.
Step 3: Next, consider the number of ways to choose a committee when both
of the specific people are not on the committee. This can be done in 10
5ways,
since we are choosing all 5 people from the remaining 10 people. However, we
need to subtract the cases where both specific people are together. This occurs
in 10
3ways. Thus, the number of ways to choose a committee of 5 people
without the 2 specific people being together is 10
5−10
3.
Step 4: Compute the final answer. The total number of ways to choose a
committee when the 2 specific people refuse to be on the committee together is:
12
5−10
5−10
3 = 792 −(252 −120) = 660. Therefore, there are 660 ways
to choose a committee of 5 people from a group of 12 when 2 specific people
refuse to be on the committee together.
9
Question 14
Question
In how many ways can a committee of 4 people be formed from a group of 7
men and 5 women if the committee must consist of at least 2 women?
Solution
Step 1: Calculate the number of ways to choose a committee with exactly 2
women: There are 5
2ways to choose 2 women from the 5 available women,
and 7
2ways to choose the remaining 2 people from the 7 men. So, the number
of ways to choose a committee with exactly 2 women is 5
2×7
2.
Step 2: Calculate the number of ways to choose a committee with exactly
3 women: There are 5
3ways to choose 3 women from the 5 available women,
and 7
1ways to choose the remaining person from the 7 men. So, the number
of ways to choose a committee with exactly 3 women is 5
3×7
1.
Step 3: Calculate the number of ways to choose a committee with all 4
women: There is 5
4ways to choose 4 women from the 5 available women. So,
the number of ways to choose a committee with all 4 women is 5
4.
Step 4: Add up the number of ways from each case to get the total number of
ways to form the committee: Total number of ways = 5
2×7
2+5
3×7
1+5
4.
Calculating the values, we get: 5
2= 10, 7
2= 21, 5
3= 10, 7
1= 7,
5
4= 5.
Therefore, the total number of ways to form the committee is 10 ×21 + 10 ×
7 + 5 = 210 + 70 + 5 = 285.
Question 15
Question
A committee of 5 people is to be selected from a group of 10 students and
8 professors. If the committee must contain at least 3 students, how many
different committees can be formed?
Solution
Step 1: Calculate the number of committees with exactly 3, 4, or 5 students.
Choose 3 students from 10: 10
3
Choose 2 other members from the remaining 15 (5 students and 8 profes-
sors): 15
2
Multiply these to get the number of committees with exactly 3 students:
10
3·15
2
10
Choose 4 students from 10: 10
4
Choose 1 other member from the remaining 14: 14
1
Multiply these to get the number of committees with exactly 4 students:
10
4·14
1
Choose 5 students from 10: 10
5
Since we can’t have all professors in the committee, only 3 professors are
left to choose from. So, choose 3 professors from 8: 8
3
Multiply these to get the number of committees with all 5 students: 10
5·
8
3
Step 2: Add up the number of committees with exactly 3, 4, or 5 students
to find the total number of committees.
10
3·15
2+10
4·14
1+10
5·8
3
= 120 ·105 + 210 ·14 + 252 ·56
= 12600 + 2940 + 14112
= 29652
Answer: There are 29,652 different committees that can be formed.
Question 16
Question
A committee of 5 people is to be formed from a group of 8 students and 4
professors. If the committee must consist of 2 students and 3 professors, how
many different committees can be formed?
Solution
Step 1: Calculate the number of ways to choose 2 students out of 8. There are
8
2ways to choose 2 students from a group of 8.
8
2=8!
2!(8 −2)! =8×7
2×1= 28.
Step 2: Calculate the number of ways to choose 3 professors out of 4. There
are 4
3ways to choose 3 professors from a group of 4.
4
3=4!
3!(4 −3)! =4
1= 4.
11
Step 3: Find the total number of different committees that can be formed
by multiplying the number of ways to choose students and professors. Total
number of committees = 8
2×4
3= 28 ×4 = 112.
Therefore, there are 112 different committees that can be formed with 2
students and 3 professors.
Question 17
Question
A committee of 5 people is to be formed from a group of 10 men and 8 women. If
the committee must consist of at least 2 men and 2 women, how many different
committees can be formed?
Solution
Let’s break this problem down into cases: Case 1: 2 men and 3 women Case 2:
3 men and 2 women Case 3: 4 men and 1 woman Case 4: 5 men and 0 women
Case 1: 2 men and 3 women Step 1: Choose 2 men from 10 men:
10
2=10!
2!(10 −2)! = 45
Step 2: Choose 3 women from 8 women:
8
3=8!
3!(8 −3)! = 56
Step 3: Multiply the number of ways to choose men and women together:
45 ×56 = 2520
Case 2: 3 men and 2 women Step 1: Choose 3 men from 10 men:
10
3=10!
3!(10 −3)! = 120
Step 2: Choose 2 women from 8 women:
8
2=8!
2!(8 −2)! = 28
Step 3: Multiply the number of ways to choose men and women together:
120 ×28 = 3360
Case 3: 4 men and 1 woman Step 1: Choose 4 men from 10 men:
10
4=10!
4!(10 −4)! = 210
12
Step 2: Choose 1 woman from 8 women:
8
1=8!
1!(8 −1)! = 8
Step 3: Multiply the number of ways to choose men and women together:
210 ×8 = 1680
Case 4: 5 men and 0 women Step 1: Choose 5 men from 10 men:
10
5=10!
5!(10 −5)! = 252
Step 2: Multiply the number of ways to choose men:
252 ×1 = 252
Now, we add up the number of committees from each case to get the total
number of different committees that can be formed:
2520 + 3360 + 1680 + 252 = 8520
Therefore, there are 8520 different committees that can be formed.
Question 18
Question
A committee of 5 people is to be formed from a group of 10 men and 8 women. If
2 men refuse to serve together on the committee, how many different committees
can be formed?
Solution
Step 1: Calculate the number of committees without any restrictions.
Since we are choosing 5 people from 18, this can be represented by the combi-
nation formula:
18
5=18!
5!(18 −5)! =18 ×17 ×16 ×15 ×14
5×4×3×2×1= 8568
Step 2: Calculate the number of committees with the two men who refuse
to serve together.
To find the number of committees with the two men who refuse to serve together,
we first calculate the number of committees with both of them on the committee
and then subtract this from the total number of committees. We choose one
of the 10 men to be on the committee (excluding the 2 who refuse to serve
13
together), and then choose the remaining 4 committee members from the 16
remaining people, which can be done with the combinations formula:
10
1×16
4= 10 ×16!
4!12! = 10 ×16 ×15 ×14 ×13
4×3×2×1= 3640
Step 3: Calculate the number of valid committees.
To find the number of valid committees (where the 2 men who refuse to serve
together are not on the committee together), we subtract the number of com-
mittees with the two men together from the total number of committees:
8568 −3640 = 4928
Answer: There are 4928 different committees that can be formed.
Question 19
Question
In a group of 10 friends, how many ways can we choose a committee with 3
members and a president if two particular friends, Alice and Bob, refuse to
serve on the committee together?
Solution
Step 1: First, we find the total number of ways to choose a committee with 3
members and a president from the group of 10 friends. This is represented by
the combination formula nCk=n!
k!(n−k)! , where nis the total number of friends
and kis the number of committee members.
Total ways to choose committee with 3 members =10 C3=10!
3!7! = 120
Step 2: Next, we find the number of ways in which Alice and Bob can serve
together on the committee. Since Alice and Bob must be chosen together, we
treat them as one entity.
Ways to choose Alice and Bob together =9C1= 9
Step 3: Now, within the committee, we have 9 friends remaining (Alice and
Bob are considered as one). From these 9 friends, we need to choose 1 more
member to complete the 3-member committee.
Ways to choose 1 more member =9C1= 9
Step 4: Finally, the total number of ways to choose a committee with 3
members and a president such that Alice and Bob do not serve together is the
14
total ways to choose a committee minus the ways to choose Alice and Bob
together times the ways to choose 1 more member.
Number of ways to choose committee without Alice and Bob together = 120−(9×9) = 120−81 = 39
Therefore, there are 39 ways to choose a committee with 3 members and a
president where Alice and Bob do not serve together.
Question 20
Question
A committee of 5 people is to be formed from a group of 10 men and 5 women. If
the committee must consist of at least 2 women, how many different committees
can be formed?
Solution
Step 1: Calculate the number of committees with exactly 2 women: Since there
are 5 women to choose from, and we need to choose 2 of them, we can calculate
the number of ways to choose 2 women.
5
2=5!
2!(5 −2)! = 10
Step 2: Calculate the number of committees with exactly 3 women: To find
the number of ways to choose 3 women out of 5:
5
3=5!
3!(5 −3)! = 10
Step 3: Calculate the number of committees with 4 or 5 women: Since we
need at least 2 women on the committee, we already counted the cases of exactly
2 or exactly 3 women. So, the remaining cases are the committees with 4 or 5
women. 5
4=5!
4!(5 −4)! = 5
5
5=5!
5!(5 −5)! = 1
Step 4: Calculate the number of committees with just men: The remaining
cases are when the committee consists only of men.
10
5=10!
5!(10 −5)! = 252
Step 5: Sum up all the cases to find the total number of different committees
that can be formed: Total = committees with exactly 2 women + committees
15
with exactly 3 women + committees with 4 women + committees with 5 women
+ committees with all men Total = 10 + 10 + 5 + 1 + 252 Total = 278
Therefore, there are 278 different committees that can be formed from the
group of 10 men and 5 women, where the committee must consist of at least 2
women.
Question 21
Question
In how many ways can we arrange the letters of the word ”MISSISSIPPI” such
that no two identical letters are adjacent?
Solution
Step 1: Count the total number of arrangements of the letters in ”MISSIS-
SIPPI”.
There are a total of 11 letters in ”MISSISSIPPI”, including 4 S’s, 4 I’s, 2
P’s, and 1 M. Therefore, the total number of arrangements is 11!.
Step 2: Count the number of arrangements where the identical letters appear
adjacent to each other.
Consider the group of 4 S’s as a single entity, the group of 4 I’s as a single
entity, and the group of 2 P’s as a single entity. Then we have a total of 4
entities: SSSS, IIII, P, M.
These 4 entities can be arranged in 4! ways, and within each entity, the
letters can be arranged in 4!, 4!, 2!, and 1! ways respectively. Thus, the number
of arrangements where the identical letters appear adjacent is 4!×4!×4!×2!×1!.
Step 3: Calculate the number of arrangements where no two identical letters
are adjacent.
The number of arrangements with no two identical letters adjacent is the
total number of arrangements minus the number of arrangements where the
identical letters appear adjacent. Therefore, the required number of arrange-
ments is:
11! −4! ×4! ×4! ×2! ×1!
Step 4: Calculate the final answer.
Substitute the values and calculate:
11! −4! ×4! ×4! ×2! ×1! = 39916800 −138240 = 39778560
Therefore, there are 39,778,560 ways to arrange the letters of the word ”MIS-
SISSIPPI” such that no two identical letters are adjacent.
16
Question 22
Question
How many ways are there to arrange the letters of the word ”MISSISSIPPI” if
no two S’s are adjacent?
Solution
Step 1: First, calculate the total number of ways to arrange the letters in
”MISSISSIPPI” without any restrictions. Step 2: Then, calculate the number
of ways where the S’s are adjacent. Step 3: Finally, subtract the number of
ways where the S’s are adjacent from the total number of ways to get the final
answer.
Step 1: The word ”MISSISSIPPI” has 11 letters (4 I’s, 4 S’s, 2 P’s, and 1
M). Therefore, there are a total of 11!
4!4!2!1! = 34,650 ways to arrange the letters
without any restrictions.
Step 2: To calculate the number of ways where the S’s are adjacent, treat
the pair of S’s as a single letter. So, we have 10 ”letters” to arrange (3 I’s, 2
P’s, 2 S’s, 2 single S’s, and 1 M), resulting in 10!
3!2!2!2!1! = 151,200 ways.
Step 3: Subtract the number of arrangements where the S’s are adjacent
from the total number of ways: 34,650 −151,200 = −116,550 .
Therefore, there are 116,550 ways to arrange the letters of ”MISSISSIPPI”
such that no two S’s are adjacent.
Question 23
Question
Let’s say you are planning a 4-course meal from a menu of 10 entrees, 8 soups,
6 salads, and 5 desserts. In how many ways can you select the 4-course meal if
each course must be different?
Solution
Step 1: Calculate the number of ways to choose 1 entree out of 10. Step 2:
Calculate the number of ways to choose 1 soup out of 8. Step 3: Calculate the
number of ways to choose 1 salad out of 6. Step 4: Calculate the number of
ways to choose 1 dessert out of 5. Step 5: Multiply the results from Steps 1-4
to determine the total number of ways to select the 4-course meal.
Step 1: There are 10 ways to choose 1 entree out of 10.
Step 2: There are 8 ways to choose 1 soup out of 8.
Step 3: There are 6 ways to choose 1 salad out of 6.
Step 4: There are 5 ways to choose 1 dessert out of 5.
Step 5: Multiply the results from Steps 1-4.
10 ×8×6×5 = 2400
17
Therefore, there are 2400 ways to select the 4-course meal if each course
must be different.
Question 24
Question
In how many ways can 4 men and 5 women be arranged in a row if at least 2
women must be next to each other?
Solution
Step 1: Find the total number of ways to arrange the 4 men and 5 women
without any restrictions. There are a total of 9 people to be arranged, so there
are 9! ways to arrange them.
Step 2: Find the number of ways to arrange the 4 men and 5 women such
that no 2 women are next to each other. To ensure that no 2 women are next to
each other, we can treat the women as blocks and place them among the men.
There are 4 gaps between the men where the women can be placed. Since we
have 5 women to be placed, this can be done in P(4,5) = 4!
(4−5)! = 4! ways.
Step 3: Find the total number of arrangements where at least 2 women are
next to each other. To find the total number of arrangements where at least 2
women are next to each other, we use the principle of complementary counting.
First, we find the number of arrangements where no 2 women are next to each
other, and then subtract this from the total number of arrangements.
Number of arrangements where at least 2 women are next to each other =
Total number of arrangements - Number of arrangements where no 2 women
are next to each other. Number of arrangements where at least 2 women are
next to each other = 9! −4!.
Step 4: Calculate the final answer. The total number of ways to arrange 4
men and 5 women in a row if at least 2 women must be next to each other is
9! −4!.
Question 25
Question
A committee of 5 people is to be formed from a group of 10 men and 6 women. If
the committee must consist of at least 3 men and 2 women, how many different
committees can be formed?
18
Solution
Step 1: Find the number of ways to select 3 men from 10 men.
10
3=10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
Step 2: Find the number of ways to select 2 women from 6 women.
6
2=6!
2!(6 −2)! =6×5
2×1= 15
Step 3: Calculate the total number of ways to form a committee with at
least 3 men and 2 women. Since the committee must consist of at least 3 men
and 2 women, we can have: - 3 men and 2 women - 4 men and 1 woman - 5
men and 0 women
Number of ways to form a committee with 3 men and 2 women: 120 ×15 =
1800
Number of ways to form a committee with 4 men and 1 woman:
10
4×6
1=10!
4!(10 −4)! ×6!
1!(6 −1)! = 2100
Number of ways to form a committee with 5 men and 0 women: 1 (only one
way as all 5 spots will be filled by men)
So, the total number of different committees that can be formed is 1800 +
2100 + 1 = 3901.
Question 26
Question
A company wants to form a committee of 5 people from a pool of 10 men and 8
women. If the committee must consist of at least 2 men and at least 2 women,
how many different committees can be formed?
Solution
Step 1: Determine the number of ways to select exactly 2 men and 3 women for
the committee. Step 2: Determine the number of ways to select exactly 3 men
and 2 women for the committee. Step 3: Add the results from Step 1 and Step
2 to find the total number of different committees that can be formed.
Step 1: To choose 2 men from 10, we have 10
2ways. To choose 3 women
from 8, we have 8
3ways. Therefore, the number of ways to select exactly 2
men and 3 women is 10
2×8
3.
Step 2: To choose 3 men from 10, we have 10
3ways. To choose 2 women
from 8, we have 8
2ways. Therefore, the number of ways to select exactly 3
men and 2 women is 10
3×8
2.
19
Step 3: Adding the results from Step 1 and Step 2 gives the total number of
different committees:
10
2×8
3+10
3×8
2
Question 27
Question
How many ways are there to choose a committee of 4 people from a group of
8 men and 5 women, if the committee must consist of exactly 3 men and 1
woman?
Solution
To find the number of ways to choose a committee of 3 men and 1 woman from
the group, we will use the concepts of permutations and combinations.
Step 1: Find the number of ways to choose 3 men from the 8 men.
We will use combinations here because the order of selection does not matter.
Number of ways to choose 3 men from 8 men = 8
3=8!
3!(8 −3)! = 56
Step 2: Find the number of ways to choose 1 woman from the 5
women. Again, we will use combinations because the order of selection does
not matter.
Number of ways to choose 1 woman from 5 women = 5
1=5!
1!(5 −1)! = 5
Step 3: Find the total number of ways to choose the committee. To
find the total number of ways to choose the committee consisting of 3 men and
1 woman, we multiply the number of ways to choose the men and the number
of ways to choose the woman.
Total number of ways to choose the committee = Number of ways to choose 3 men×Number of ways to choose 1 woman = 56×5 = 280
Therefore, there are 280 ways to choose a committee of 4 people consisting
of exactly 3 men and 1 woman from the group.
Question 28
Question
In a group of 10 students, calculate the number of ways to choose 4 students to
form a committee with a president, vice-president, secretary, and treasurer.
20
Solution
Step 1: Calculate the number of ways to choose 4 students out of 10. Step 2: For
each group of 4 students, assign one student to each position on the committee.
Step 3: Find the total number of ways by multiplying the results from Steps 1
and 2.
Step 1: To choose 4 students out of 10, we will use the combination formula
n
k=n!
k!(n−k)! . Substitute n= 10 and k= 4:
10
4=10!
4!(10 −4)! =10 ×9×8×7
4×3×2×1= 210.
Step 2: For each group of 4 students, we need to assign one student to
each position on the committee. - There are 4 choices for the president. -
After choosing the president, there are 3 choices left for the vice-president. -
After choosing the president and vice-president, there are 2 choices left for the
secretary. - Finally, the treasurer will be the remaining student.
Step 3: Multiply the results from Step 1 and Step 2 to find the total number
of ways to form the committee:
210 ×4×3×2 = 5040.
Therefore, there are 5040 ways to choose 4 students to form a committee
with a president, vice-president, secretary, and treasurer from a group of 10
students.
Question 29
Question
A committee of 5 people is to be formed from a group of 10 students and 5
professors. If the committee must consist of at least 2 students and 1 professor,
how many different committees can be formed?
Solution
Step 1: Calculate the number of ways to form a committee with 2 students and
3 professors. There are 10
2ways to choose 2 students from the 10 available
students, and 5
3ways to choose 3 professors from the 5 available professors.
Therefore, the number of ways to form a committee with 2 students and 3
professors is 10
2·5
3.
Step 2: Calculate the number of ways to form a committee with 3 students
and 2 professors. Similarly, there are 10
3ways to choose 3 students and 5
2
ways to choose 2 professors. Therefore, the number of ways to form a committee
with 3 students and 2 professors is 10
3·5
2.
Step 3: Calculate the total number of different committees that can be
formed. The total number of different committees that can be formed is the
21
sum of the committees with 2 students and 3 professors, and the committees
with 3 students and 2 professors: 10
2·5
3+10
3·5
2.
Now we can perform the calculations to find the total number of different
committees that can be formed.
Question 30
Question
How many ways are there to form a committee of 5 people from a group of 10
people, where 3 of them are men and 7 of them are women, if the committee
must have at least 2 men?
Solution
To find the number of ways to form a committee of 5 people with at least 2
men, we will consider the different possible scenarios separately and then add
them up.
Step 1: Calculate the number of ways to choose 2 men and 3 women: Since
there are 3 men and 7 women, the number of ways to choose 2 men out of
3 is 3
2= 3. Similarly, the number of ways to choose 3 women out of 7 is
7
3= 35. Therefore, the total number of ways to choose 2 men and 3 women is
3×35 = 105.
Step 2: Calculate the number of ways to choose 3 men and 2 women: Simi-
larly, the number of ways to choose 3 men out of 3 is 3
3= 1. The number of
ways to choose 2 women out of 7 is 7
2= 21. Therefore, the total number of
ways to choose 3 men and 2 women is 1 ×21 = 21.
Step 3: Add up the two cases to get the total number of ways: The total
number of ways to form a committee of 5 people with at least 2 men is 105+21 =
126 ways.
Question 31
Question
A group of 8 friends want to take a group photo. In how many ways can they
arrange themselves in a row for the photo if:
a) There are no restrictions on the arrangement.
b) The tallest friend must be at one end of the row.
22
Solution
a) When there are no restrictions on the arrangement, the number of ways the 8
friends can line up in a row is given by the number of permutations of 8 objects,
which is 8! = 40,320.
b) If the tallest friend must be at one end of the row, we treat the tallest
friend as one entity. This way, there are 2 entities (tallest friend entity and the
rest of the friends) to arrange in a row. The number of ways to arrange these 2
entities is 2! = 2.
For each of these arrangements, the friends within the ”rest of the friends”
entity can be arranged in 7! ways.
Therefore, the total number of ways the friends can line up in a row with
the tallest friend at one end is 2 ×7! = 10,080.
Question 32
Question
In how many ways can you arrange the letters in the word ”UNIVERSITY”
such that no two vowels are adjacent?
Solution
Step 1: First, we identify the total number of ways to arrange the letters in the
word ”UNIVERSITY” without any restrictions.
The word ”UNIVERSITY” has 10 letters, but the letters ’I’ and ’U’ appear
twice.
Therefore, the total number of ways to arrange the letters without any re-
strictions is given by:
10!/(2! ·2!) = 9,072,000
Step 2: Next, we consider the restriction that no two vowels can be adjacent.
Let’s treat the letters ’U’, ’I’, and ’E’ as a single entity (a block) since
they cannot be adjacent. This reduces the problem to arranging the letters
’NVRSTYB’ plus the block ’UEI’.
The total number of ways to arrange these letters and the block is given by:
7! ×3! = 30,240
Step 3: Therefore, the number of ways to arrange the letters in the word
”UNIVERSITY” such that no two vowels are adjacent is 30,240.
23
Question 33
Question
A committee of 5 people is to be formed from a group of 10 men and 7 women. If
the committee must consist of at least 2 women, how many different committees
can be formed?
Solution
To find the number of different committees that can be formed, we need to
consider the different cases where the number of women in the committee ranges
from 2 to 5.
Step 1: Calculate the number of committees with 2 women and 3
men
We need to choose 2 women out of 7 and 3 men out of 10. Number of ways
to choose 2 women from 7: 7
2Number of ways to choose 3 men from 10: 10
3
Total number of committees with 2 women and 3 men: 7
2×10
3
Step 2: Calculate the number of committees with 3 women and 2
men
Similarly, we need to choose 3 women out of 7 and 2 men out of 10. Number
of ways to choose 3 women from 7: 7
3Number of ways to choose 2 men from
10: 10
2Total number of committees with 3 women and 2 men: 7
3×10
2
Step 3: Calculate the number of committees with 4 women and 1
man
Next, we need to choose 4 women out of 7 and 1 man out of 10. Number of
ways to choose 4 women from 7: 7
4Number of ways to choose 1 man from 10:
10
1Total number of committees with 4 women and 1 man: 7
4×10
1
Step 4: Calculate the number of committees with 5 women and 0
men
Lastly, we need to choose all 5 women out of 7 (as there are no men in this
case). Number of ways to choose 5 women from 7: 7
5
Step 5: Add up the total number of committees
Now, we can find the total number of committees by summing up the com-
mittees from each case. Total number of committees = 7
2×10
3+7
3×10
2+
7
4×10
1+7
5
Question 34
Question
A committee of 5 people is to be formed from a group of 10 students. If 3 of
the students are in math club and 7 are in science club, how many different
committees can be formed if each committee must have at least 1 math club
member and 1 science club member?
24
Solution
Let’s break this problem down into cases to account for the different possibilities.
Case 1: 1 math club member and 4 science club members Step 1:
Choose 1 math club member out of 3.
Step 2: Choose 4 science club members out of 7.
Number of ways for Case 1 = 3
1×7
4
Case 2: 2 math club members and 3 science club members Step 1:
Choose 2 math club members out of 3.
Step 2: Choose 3 science club members out of 7.
Number of ways for Case 2 = 3
2×7
3
Case 3: 3 math club members and 2 science club members Step 1:
Choose 3 math club members out of 3.
Step 2: Choose 2 science club members out of 7.
Number of ways for Case 3 = 3
3×7
2
Finally, to find the total number of different committees that can be formed,
we sum up the results of each case.
Total number of committees = 3
1×7
4+3
2×7
3+3
3×7
2
Question 35
Question
In a card game, each player is dealt 5 cards from a standard deck of 52 playing
cards. How many ways can the cards be dealt such that each player receives
exactly 2 red cards and 3 black cards?
Solution
Step 1: Determine the total number of ways to choose 2 red cards out of 26 red
cards. There are 26
2ways to choose 2 red cards.
Step 2: Determine the total number of ways to choose 3 black cards out of
26 black cards. There are 26
3ways to choose 3 black cards.
Step 3: Multiply the results from Step 1 and Step 2 to find the total number
of ways to choose 2 red cards and 3 black cards. Total ways = 26
2×26
3.
Step 4: Calculate the total number of ways. 26
2=26!
2!(26−2)! =26×25
2= 325
26
3=26!
3!(26−3)! =26×25×24
3×2= 2600
Therefore, total ways = 325 ×2600 = 845,000.
25
So, there are 845,000 ways to deal 5 cards such that each player receives
exactly 2 red cards and 3 black cards.
26
