MATH 201 - INTRODUCTION TO
PROBABILITY AND STATISTICS -
Combinatorial Analysis
Question Bank - Set 4
Liberty University
Question 1
Question
A group of 10 students are attending a conference where 5 different workshops
are being offered. Each student must attend exactly one workshop. In how
many ways can the students be assigned to workshops if each workshop must
have at least one student?
Solution
Step 1: Assign one student to each workshop. Since each workshop must have
at least one student, this can be done in 5! ways.
Step 2: Distribute the remaining 5 students among the 5 workshops. This
is equivalent to finding the number of ways to distribute identical objects into
distinct boxes (workshops) with no box left empty, which is a stars and bars
problem.
Step 3: Let’s represent the distribution of the remaining students with 5 stars
(representing the students) and 4 bars (dividing the students into 5 workshops).
For example,
∗∗|∗∗∗|∗∗|∗|∗
represents 2 students in the first workshop, 3 students in the second workshop, 2
students in the third workshop, 1 student in the fourth workshop, and 1 student
in the fifth workshop.
Step 4: The total number of ways to distribute the remaining students is (9
4)
using the stars and bars formula.
Step 5: Therefore, the total number of ways to assign the students to work-
shops is 5! ×(9
4)= 5! ×9!
4!5! = 5! ×9×8×7×6
4×3×2×1= 5 ×9×8×7×6 = 15120
ways.
Question 2
Question
A committee of 5 people is to be formed from a group of 10 students and 5
faculty members. If at least 2 faculty members must be on the committee, how
many different committees can be formed?
Solution
Step 1: Calculate the number of ways to choose 2 faculty members out of 5.
Step 2: Calculate the number of ways to choose the remaining 3 members
from the remaining 10 students and 3 faculty members (after choosing 2 faculty
members). Step 3: Multiply the results from Step 1 and Step 2 to find the total
number of committees.
Step 1: Choosing 2 faculty members Since we are choosing 2 faculty
members out of 5, the number of ways to do this is given by (5
2)=5!
2!(5−2)! = 10.
Step 2: Choosing the remaining 3 members After choosing 2 faculty
members, there are 3 faculty members left and 10 students remaining. We need
to choose 3 members from these 13 people. So, the number of ways to choose
the remaining 3 members is (13
3)=13!
3!(13−3)! = 286.
Step 3: Total number of committees To calculate the total number of
committees, we multiply the results from Step 1 and Step 2. Total number of
committees = 10 ×286 = 2860.
Therefore, there are 2,860 different committees that can be formed with at
least 2 faculty members on the committee.
Question 3
Question
Suppose you have a standard deck of 52 playing cards (13 cards in each of 4
suits: hearts, diamonds, clubs, and spades). How many ways can you choose a
5-card hand such that it contains exactly 3 hearts and 2 spades?
Solution
Step 1: Determine the number of ways to choose 3 hearts from the 13 available.
There are (13
3)ways to choose 3 hearts from the 13 available.
Step 2: Determine the number of ways to choose 2 spades from the 13
available. There are (13
2)ways to choose 2 spades from the 13 available.
Step 3: Multiply the results from Step 1 and Step 2 to find the total number
of ways to choose 3 hearts and 2 spades. Total ways = (13
3)×(13
2).
Step 4: Calculate the total number of ways. (13
3)=13!
3!(13−3)! =13×12×11
3×2×1=
286.
(13
2)=13!
2!(13−2)! =13×12
2×1= 78.
2
Therefore, the total number of ways to choose 3 hearts and 2 spades is
286 ×78 = 22236. Thus, there are 22,236 ways to choose a 5-card hand with
exactly 3 hearts and 2 spades.
Question 4
Question
In a group of 10 people, how many ways can we choose a committee of 3 people?
Solution
Step 1: To find the number of ways to choose a committee of 3 people from a
group of 10 people, we use the combination formula which is given by:
(n
r)=n!
r!(n−r)!
where nis the total number of people and ris the number of people we want
to choose.
Step 2: Substituting n= 10 and r= 3 into the formula, we get:
(10
3)=10!
3!(10 −3)!
Step 3: Calculating the factorials:
10! = 10 ×9×8×7×6×5×4×3×2×1 = 3628800
3! = 3 ×2×1 = 6
7! = 7 ×6×5×4×3×2×1 = 5040
Step 4: Substitute the factorials back into the formula:
(10
3)=3628800
6×5040
Step 5: Simplifying the expression:
(10
3)=3628800
30240 = 120
Therefore, there are 120 ways to choose a committee of 3 people from a
group of 10 people.
3
Question 5
Question
In a group of 10 students, how many ways can we choose a committee of 4 stu-
dents if two particular students, Alex and Beth, refuse to serve on the committee
together?
Solution
Step 1: Calculate the total number of ways to choose a committee of 4 out
of 10 students. There are 10 students to choose from, and we want to select
a committee of 4. This is a combination problem, so we use the formula for
combinations:
Total ways to choose a committee of 4 =(10
4)=10!
4!(10 −4)! =10 ×9×8×7
4×3×2×1= 210
Step 2: Calculate the number of ways to choose a committee of 4 when
both Alex and Beth are in the committee. Since Alex and Beth must be in the
committee together, we treat them as one entity. Then, we have 9 entities (Alex
and Beth as one, and the remaining 8 students) to choose the other 2 committee
members from. This is also a combination problem:
Ways to choose a committee with Alex and Beth =(9
2)=9!
2!(9 −2)! =9×8
2×1= 36
Step 3: Subtract the number of ways to choose a committee with Alex and
Beth from the total number of ways.
Number of ways to choose a committee without Alex and Beth =Total ways−Ways with Alex and Beth = 210−36 = 174
Therefore, there are 174 ways to choose a committee of 4 students from a
group of 10 students if Alex and Beth refuse to serve on the committee together.
Question 6
Question
A committee of 5 people is to be formed from a group of 8 men and 6 women. If
the committee must consist of at least 2 women, how many different committees
can be formed?
Solution
Step 1: Find the number of ways to choose a committee with exactly 2 women.
Since there must be at least 2 women in the committee, we will consider the
cases of choosing 2, 3, 4, or 5 women.
4
For choosing 2 women:
(6
2)ways to choose 2 women×(8
3)ways to choose 3 men = 15×56 = 840 ways
Step 2: Find the number of ways to choose a committee with exactly 3
women.
(6
3)ways to choose 3 women×(8
2)ways to choose 2 men = 20×28 = 560 ways
Step 3: Find the number of ways to choose a committee with exactly 4
women.
(6
4)ways to choose 4 women×(8
1)ways to choose 1 man = 15×8 = 120 ways
Step 4: Find the number of ways to choose a committee with all 5 women.
(6
5)ways to choose 5 women = 6 ways
Step 5: Calculate the total number of ways to form the committee.
Total ways = 840 + 560 + 120 + 6 = 1526
Therefore, there are 1526 different committees that can be formed with at
least 2 women.
Question 7
Question
A company has 8 different job positions available, and there are 12 applicants.
How many ways can the company hire 4 applicants for the job positions?
Solution
Step 1: To solve this problem, we will use the combination formula, which is
given by:
(n
k)=n!
k!(n−k)!
where nis the total number of items, kis the number of items to choose.
Step 2: First, we need to determine the total number of ways to select 4
applicants from the 12 that have applied. This is given by:
(12
4)=12!
4!(12 −4)! =12!
4!8!
5
Step 3: Simplifying the formula, we get:
(12
4)=12 ×11 ×10 ×9
4×3×2×1= 495
Therefore, there are 495 ways for the company to hire 4 applicants for the 8
job positions.
Question 8
Question
In a certain game, players must match 5 numbers (chosen from 1 to 50) to win
the jackpot. How many different ways can a player choose their numbers if the
order in which the numbers are selected does not matter?
Solution
Step 1: Calculate the number of ways to choose 5 numbers from 50 without
regard to order. This is a combination problem, which can be calculated using
the formula:
Number of ways =(n
k)=n!
k!(n−k)!
where nis the total number of items (50 in this case) and kis the number of
items to choose (5 in this case).
Step 2: Substituting n= 50 and k= 5 into the formula, we get:
(50
5)=50!
5!(50 −5)!
Step 3: Simplify the expression:
(50
5)=50 ×49 ×48 ×47 ×46
5×4×3×2×1
(50
5)=2118760
120
(50
5)= 176,160
Therefore, there are 176,160 different ways for a player to choose their 5
numbers for the game.
Question 9
Question
In a class of 30 students, how many ways can a committee of 5 students be
selected if Alex and Beth refuse to serve together?
6
Solution
Step 1: Find the total number of ways to select a committee of 5 students from
a class of 30 students. Step 2: Find the number of ways Alex and Beth can be
in the same committee. Step 3: Subtract the number of ways Alex and Beth
can be in the same committee from the total number of ways to get the final
answer.
Step 1: The total number of ways to select 5 students out of 30 is given by
the combination formula: C(30,5) = 30!
5!(30−5)! .
Step 2: If Alex and Beth are to be in the same committee, treat them as
one person. So the number of ways they can be in the same committee is the
number of ways to select 4 people out of 29 (treating Alex and Beth as one
person): C(29,4) = 29!
4!(29−4)! .
Step 3: The number of ways to select a committee where Alex and Beth
do not serve together is given by:
C(30,5) −C(29,4) = 30!
5!(30 −5)! −29!
4!(29 −4)!
Question 10
Question
A committee of 5 students is to be randomly selected from a group of 10 students.
How many different committees can be formed if one particular student must
be included?
Solution
Step 1: Since one particular student must be included in the committee, we
have already chosen 1 student. So, we need to choose 4 more students from the
remaining 9 students.
Step 2: To find the number of ways to select 4 students out of 9, we use the
combination formula (n
r)=n!
r!(n−r)! , where nis the total number of students
and ris the number of students we want to choose.
Step 3: Plugging in n= 9 and r= 4 into the formula, we get:
(9
4)=9!
4!(9 −4)!
Step 4: Calculating the factorials:
(9
4)=9×8×7×6×5
4×3×2×1
Step 5: Simplifying further:
(9
4)=30240
24
7
(9
4)= 1260
Step 6: Therefore, there are 1260 different committees that can be formed
if one particular student must be included.
Question 11
Question
In a class of 30 students, 15 are male and 15 are female. A committee of 5
students is to be formed randomly from the class. What is the probability that
the committee consists of 3 males and 2 females?
Solution
Step 1: Determine the total number of ways to form a committee of 5 students
from a class of 30 students. Given that there are 15 male students and 15 female
students in the class, the total number of ways to form a committee of 5 students
can be calculated using combinations:
Total ways =(30
5)=30!
5!(30 −5)! =30!
5!25! = 142506
Step 2: Determine the number of ways to form a committee consisting of 3
males and 2 females. To form a committee of 3 males and 2 females, we need to
choose 3 males out of 15 and 2 females out of 15. The number of ways to choose
3 males from 15 is given by (15
3)and the number of ways to choose 2 females
from 15 is (15
2). Therefore, the number of ways to form a committee consisting
of 3 males and 2 females is:
(15
3)×(15
2)=15!
3!12! ×15!
2!13! = 455 ×105 = 47850
Step 3: Calculate the probability of forming a committee consisting of 3
males and 2 females. The probability is given by the ratio of the number of
favorable outcomes (committees with 3 males and 2 females) to the total number
of outcomes (all possible committees):
Probability =Number of ways with 3 males and 2 females
Total number of ways =47850
142506 ≈0.3359
Therefore, the probability that the committee consists of 3 males and 2 females
is approximately 0.3359.
8
Question 12
Question
A company is assigning 5 different positions to 8 employees. How many different
ways can this be done if one of the employees, Alex, must be in a specific
position?
Solution
Step 1: Since Alex must be in a specific position, we will consider Alex already
assigned to that position. Therefore, we only need to assign the remaining 4
positions to the remaining 7 employees.
Step 2: The number of ways to assign the remaining 4 positions to the 7
employees is given by the number of permutations of 7 items taken 4 at a time.
Step 3: The number of permutations of 7 items taken 4 at a time is given
by: 7!
(7 −4)! =7!
3!
Step 4: Calculating the permutations:
7!
3! =7×6×5×4×3!
3×2×1= 7 ×6×5×4 = 840
Therefore, there are 840 different ways to assign 5 different positions to 8
employees if Alex must be in a specific position.
Question 13
Question
A committee of 6 people is to be formed from a group of 10 men and 7 women.
Find the probability that the committee consists of 3 men and 3 women.
Solution
Step 1: Find the total number of ways to form a committee of 6 people. To
form a committee of 6 people from a group of 17 (10 men and 7 women), we
can use the formula for combinations:
Total ways =(17
6)=17!
6!(17 −6)!
Step 2: Find the number of ways to choose 3 men from 10. Using com-
binations, we can find the number of ways to choose 3 men from a group of
10:
Ways to choose 3 men =(10
3)=10!
3!(10 −3)!
9
Step 3: Find the number of ways to choose 3 women from 7. Similar to the
previous step, we can calculate the number of ways to choose 3 women from 7:
Ways to choose 3 women =(7
3)=7!
3!(7 −3)!
Step 4: Find the total number of ways to form a committee with 3 men and
3 women. Since we want a committee with 3 men and 3 women, we need to
multiply the number of ways to choose 3 men by the number of ways to choose
3 women:
Total ways for 3 men and 3 women =(10
3)×(7
3)
Step 5: Calculate the probability. The probability that the committee con-
sists of 3 men and 3 women is the ratio of the total number of ways to form
a committee with 3 men and 3 women to the total number of ways to form a
committee of 6 people:
Probability =(10
3)×(7
3)
(17
6)=
10!
3!(10−3)! ×7!
3!(7−3)!
17!
6!(17−6)!
Question 14
Question
In a group of 10 people (5 men and 5 women), how many different ways can a
committee of 3 people be formed if there must be at least 1 man and 1 woman
on the committee?
Solution
Step 1: First, we will find the total number of ways to form a committee of 3
people from the group of 10 people. This can be done using the combination
formula (n
r)=n!
r!(n−r)! , where nis the total number of people and ris the
number of people we want to select. In this case, n= 10 and r= 3.
Total ways to form committee =(10
3)=10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
Step 2: Next, we will find the number of ways to form a committee with 3
people where all 3 members are men or all 3 members are women. Since there
are 5 men and 5 women, these scenarios are mutually exclusive.
• Ways to select 3 men out of 5: (5
3)=5!
3!(5−3)! = 10 ways
• Ways to select 3 women out of 5: (5
3)=5!
3!(5−3)! = 10 ways
10
Step 3: However, these scenarios do not satisfy the condition of having at
least 1 man and 1 woman on the committee. Therefore, we will find the total
number of ways to select a committee with either 2 men and 1 woman or 1 man
and 2 women.
• Ways to select 2 men out of 5 and 1 woman out of 5: (5
2)×(5
1)= 10×5 = 50
ways
• Ways to select 1 man out of 5 and 2 women out of 5: (5
1)×(5
2)= 5×10 = 50
ways
Step 4: Finally, we calculate the total number of ways to form a committee
of 3 people with at least 1 man and 1 woman by summing up the possibilities
from Step 2 and Step 3.
Total ways with at least 1 man and 1 woman = 10 + 10 + 50 + 50 = 120
Therefore, there are 120 different ways a committee of 3 people can be formed
from a group of 10 people (5 men and 5 women) with at least 1 man and 1 woman
on the committee.
Question 15
Question
In a group of 10 people, how many ways can we choose a committee of 4 people
if 2 of the members, Alice and Bob, refuse to serve together on the committee?
Solution
Step 1: Find the total number of ways to choose a committee of 4 people without
any restrictions. The total number of ways to choose 4 people from a group of
10 is given by the combination formula: (10
4)=10!
4!(10−4)! .
Step 2: Find the number of ways to choose a committee where Alice and
Bob are both in the committee. Let’s assume that Alice and Bob will work
together in the committee. Then, we need to choose 2 more people from the
remaining 8 people. This can be done in (8
2)ways.
Step 3: Subtract the number of ways with Alice and Bob from the total to get
the final result. Now, we need to subtract the number of ways where Alice and
Bob are both in the committee from the total number of ways: Total number
of ways −Number of ways with Alice and Bob in the committee =(10
4)−(8
2).
Therefore, the number of ways to choose a committee of 4 people where
Alice and Bob refuse to serve together is (10
4)−(8
2).
11
Question 16
Question
A committee of 4 people is to be formed from a group of 10 students. If 4 of
the students are math majors and 6 are engineering majors, how many different
committees can be formed if each committee must have at least 2 math majors
and at least 1 engineering major?
Solution
Step 1: Calculate the number of committees with exactly 2 math majors and 2
engineering majors.
There are (4
2)= 6 ways to choose 2 math majors from the 4 available, and
(6
2)= 15 ways to choose 2 engineering majors from the 6 available. Multiply
these together to get 6×15 = 90.
Step 2: Calculate the number of committees with exactly 3 math majors
and 1 engineering major.
There are (4
3)= 4 ways to choose 3 math majors from the 4 available, and
(6
1)= 6 ways to choose 1 engineering major from the 6 available. Multiply these
together to get 4×6 = 24.
Step 3: Calculate the number of committees with 4 math majors.
There is only 1 way to choose all 4 math majors from the 4 available.
Step 4: Add the results from Steps 1, 2, and 3 to find the total number of
valid committees.
90 + 24 + 1 = 115
Therefore, there are 115 different committees that can be formed where each
committee must have at least 2 math majors and at least 1 engineering major.
Question 17
Question
A committee of 5 people is to be formed from a group of 8 men and 7 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: Calculate the number of committees with exactly 2 men and 3 women.
To select 2 men from 8 men and 3 women from 7 women, we use the combination
formula: (8
2)·(7
3)=8!
2!(8 −2)! ·7!
3!(7 −3)! = 28 ·35 = 980
Step 2: Calculate the number of committees with exactly 3 men and 2
women. To select 3 men from 8 men and 2 women from 7 women, we use the
12
combination formula:
(8
3)·(7
2)=8!
3!(8 −3)! ·7!
2!(7 −2)! = 56 ·21 = 1176
Step 3: Calculate the number of committees with 4 men and 1 woman. To
select 4 men from 8 men and 1 woman from 7 women, we use the combination
formula: (8
4)·(7
1)=8!
4!(8 −4)! ·7!
1!(7 −1)! = 70 ·7 = 490
Step 4: Calculate the total number of committees with at least 2 men and 2
women. Add the results from Step 1, Step 2, and Step 3: Total = 980 + 1176
+ 490 = 2646
Therefore, there are 2646 different committees that can be formed with at
least 2 men and at least 2 women from the group.
Question 18
Question
In how many ways can a committee of 3 people be selected from a group of 7
people to serve as president, vice-president, and treasurer?
Solution
Step 1: To find the number of ways to select the committee, we first need to
determine the number of ways to select the president, then the vice-president,
and finally the treasurer.
Step 2: There are 7 choices for the president, since any of the 7 people can
be selected.
Step 3: Once the president is chosen, there are 6 remaining people to choose
from for the vice-president.
Step 4: After selecting the president and vice-president, there are 5 remain-
ing people to choose from for the treasurer.
Step 5: Therefore, the total number of ways to select the committee is the
product of the number of choices for each position:
7×6×5 = 210
Step 6: So, there are 210 ways to select a committee of 3 people from a
group of 7 to serve as president, vice-president, and treasurer.
Question 19
Question
In a group of 10 students, how many ways are there to select a committee of 3
students and a chairperson from the group?
13
Solution
Step 1: To find the number of ways to select a committee of 3 students from
10, we will use the combination formula (n
r)=n!
r!(n−r)! where nis the total
number of students and ris the number of students we want to select.
Step 2: Substitute n= 10 and r= 3 into the combination formula:
(10
3)=10!
3!(10 −3)!
Step 3: Simplify the factorial terms in the denominator:
(10
3)=10!
3!7!
Step 4: Calculate the factorials:
(10
3)=10 ×9×8×7!
3×2×1×7!
Step 5: Simplify further to find the number of ways to select the committee:
(10
3)=10 ×9×8
3×2×1= 120
Step 6: After selecting the 3 students for the committee, we need to choose
a chairperson from those 3 students. There are 3 ways to choose a chairperson
from the committee.
Step 7: Multiply the number of ways to select the committee by the number
of ways to choose a chairperson to find the total number of ways:
120 ×3 = 360
Therefore, there are 360 ways to select a committee of 3 students and a
chairperson from a group of 10 students.
Question 20
Question
A committee of 5 people is to be formed from a group of 10 women and 8 men.
How many different committees can be formed if the committee must have at
least 2 women and at least 2 men?
14
Solution
Step 1: Calculate the total number of ways to form a committee with 5 people:
There are 10 women and 8 men to choose from, so the total number of ways to
choose 5 people without any restrictions is (18
5)= 8,568.
Step 2: Calculate the number of ways to form a committee with at least 2
women and at least 2 men: To form a committee with at least 2 women and at
least 2 men, we can have the following cases: - 2 women and 3 men - 3 women
and 2 men - 4 women and 1 man - 5 women and 0 men
For the case of 2 women and 3 men: Choose 2 women from 10: (10
2)ways
Choose 3 men from 8: (8
3)ways Multiply the two to get the total number of
ways: (10
2)×(8
3)= 2,520 ways
For the case of 3 women and 2 men: Choose 3 women from 10: (10
3)ways
Choose 2 men from 8: (8
2)ways Multiply the two to get the total number of
ways: (10
3)×(8
2)= 2,520 ways
For the case of 4 women and 1 man: Choose 4 women from 10: (10
4)ways
Choose 1 man from 8: (8
1)ways Multiply the two to get the total number of
ways: (10
4)×(8
1)= 560 ways
For the case of 5 women and 0 men: Choose 5 women from 10: (10
5)ways
There are no men to choose, so only 1 way The total number of ways: (10
5)= 252
ways
Step 3: Add up the number of ways for each case to get the total number
of ways to form the committee with at least 2 women and at least 2 men:
2,520 + 2,520 + 560 + 252 = 5,852 ways
Therefore, there are 5,852 different committees that can be formed with at
least 2 women and at least 2 men.
Question 21
Question
In a group of 10 students, how many ways can we choose a committee of 4
students?
Solution
Step 1: To solve this problem, we will use the formula for combinations, which
is given by:
C(n, k) = n!
k!(n−k)!
where nis the total number of students and kis the number of students we
want to choose for the committee.
15
Step 2: Substituting n= 10 and k= 4 into the formula, we get:
C(10,4) = 10!
4!(10 −4)!
Step 3: Calculating the factorials:
10! = 10 ×9×8×7×6×5×4×3×2×1
4! = 4 ×3×2×1
6! = 6 ×5×4×3×2×1
Step 4: Substituting the factorials into the formula:
C(10,4) = 10 ×9×8×7×6×5×4×3×2×1
4×3×2×1×6×5×4×3×2×1
Step 5: Simplifying the expression:
C(10,4) = 10 ×9×8×7
4×3×2×1= 210
Step 6: Therefore, there are 210 ways to choose a committee of 4 students
from a group of 10.
Question 22
Question
A committee of 5 people is to be formed from a group of 8 men and 6 women.
If the committee must consist of 3 men and 2 women, how many different
committees can be formed?
Solution
Step 1: Calculate the number of ways to choose 3 men from 8. Step 2: Calculate
the number of ways to choose 2 women from 6. Step 3: Multiply the results of
Step 1 and Step 2 to find the total number of different committees that can be
formed.
Step 1: Number of ways to choose 3 men from 8.
C(8,3) = 8!
3!(8 −3)! =8×7×6
3×2×1= 56
Step 2: Number of ways to choose 2 women from 6.
C(6,2) = 6!
2!(6 −2)! =6×5
2×1= 15
Step 3: Total number of different committees that can be formed.
56 ×15 = 840
Therefore, there are 840 different committees that can be formed with 3 men
and 2 women.
16
Question 23
Question
In a group of 10 students, how many ways can we select a committee of 3
students, where one student is designated as the president, one as the vice
president, and one as the secretary?
Solution
To solve this problem, we can break it down into two steps: Step 1: Calculate
the number of ways to select a committee of 3 students. Step 2: After selecting
the committee, assign the roles of president, vice president, and secretary to the
selected students.
Step 1: To select a committee of 3 students out of 10, we can use the
combination formula:
Number of ways =(10
3)=10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
So, there are 120 ways to select a committee of 3 students.
Step 2: After selecting the committee, we need to assign the roles of pres-
ident, vice president, and secretary. For the president, there are 3 students to
choose from the committee. After selecting the president, there are 2 students
left for the vice president role. Finally, the last remaining student will be the
secretary.
Therefore, the total number of ways to select the committee with assigned
roles is:
120 ×3×2×1 = 720
Therefore, there are 720 ways to select a committee of 3 students, where one
is the president, one is the vice president, and one is the secretary.
Question 24
Question
A committee of 4 people is to be formed from a group of 10 people. How many
different committees can be formed if 2 specific people must be included in the
committee?
Solution
Step 1: Choose the 2 specific people to be included in the committee. There
are (10
2)ways to do this.
Step 2: Choose the remaining 2 people from the remaining 8 people. There
are (8
2)ways to do this.
17
Step 3: Multiply the results from Step 1 and Step 2 to find the total number
of different committees that can be formed.
Therefore, the total number of different committees that can be formed is
given by:
(10
2)×(8
2)=10!
2!8! ×8!
2!6!
=10 ×9
2×1×8×7
2×1= 45 ×28
= 1260
So, there are 1260 different committees that can be formed if 2 specific people
must be included in the committee.
Question 25
Question
In a group of 10 students, how many ways can we choose a committee of 3
students to serve on the student council if one student, Alice, refuses to serve
with Bob due to a conflict?
Solution
Step 1: First, let’s find the total number of ways to choose a committee of 3
students from a group of 10 students. This is given by the combination formula
(n
r)=n!
r!(n−r)! , where nis the total number of students and ris the number of
students we want to choose for the committee. In this case, n= 10 and r= 3.
Thus, the total number of ways to choose a committee of 3 students from 10 is:
(10
3)=10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
Step 2: Next, let’s find the number of ways to choose a committee of 3
students from the 9 remaining students after Alice refuses to serve with Bob.
This is given by the combination formula. So, the number of ways to choose a
committee of 3 students from 9 is:
(9
3)=9!
3!(9 −3)! =9×8×7
3×2×1= 84
Step 3: Now, to find the number of ways to choose a committee of 3 stu-
dents that does not include Alice and Bob together, we need to subtract the
number of committees where Alice and Bob are together from the total number
of committees. There are 84 ways to choose a committee of 3 students from the
9 remaining students, and Alice and Bob can be together in (2
1)×(7
1)= 14 ways
(either Alice and Bob or Bob and Alice). Therefore, the number of ways to
choose a committee of 3 students that does not include Alice and Bob together
is 84 −14 = 70 ways.
18
Question 2
Question
A committee of 5 people is to be formed from a group of 10 students and 5
faculty members. If at least 2 faculty members must be on the committee, how
many different committees can be formed?
Solution
Step 1: Calculate the number of ways to choose 2 faculty members out of 5.
Step 2: Calculate the number of ways to choose the remaining 3 members
from the remaining 10 students and 3 faculty members (after choosing 2 faculty
members). Step 3: Multiply the results from Step 1 and Step 2 to find the total
number of committees.
Step 1: Choosing 2 faculty members Since we are choosing 2 faculty
members out of 5, the number of ways to do this is given by (5
2)=5!
2!(5−2)! = 10.
Step 2: Choosing the remaining 3 members After choosing 2 faculty
members, there are 3 faculty members left and 10 students remaining. We need
to choose 3 members from these 13 people. So, the number of ways to choose
the remaining 3 members is (13
3)=13!
3!(13−3)! = 286.
Step 3: Total number of committees To calculate the total number of
committees, we multiply the results from Step 1 and Step 2. Total number of
committees = 10 ×286 = 2860.
Therefore, there are 2,860 different committees that can be formed with at
least 2 faculty members on the committee.
Question 3
Question
Suppose you have a standard deck of 52 playing cards (13 cards in each of 4
suits: hearts, diamonds, clubs, and spades). How many ways can you choose a
5-card hand such that it contains exactly 3 hearts and 2 spades?
Solution
Step 1: Determine the number of ways to choose 3 hearts from the 13 available.
There are (13
3)ways to choose 3 hearts from the 13 available.
Step 2: Determine the number of ways to choose 2 spades from the 13
available. There are (13
2)ways to choose 2 spades from the 13 available.
Step 3: Multiply the results from Step 1 and Step 2 to find the total number
of ways to choose 3 hearts and 2 spades. Total ways = (13
3)×(13
2).
Step 4: Calculate the total number of ways. (13
3)=13!
3!(13−3)! =13×12×11
3×2×1=
286.
(13
2)=13!
2!(13−2)! =13×12
2×1= 78.
2
Therefore, the total number of ways to choose 3 hearts and 2 spades is
286 ×78 = 22236. Thus, there are 22,236 ways to choose a 5-card hand with
exactly 3 hearts and 2 spades.
Question 4
Question
In a group of 10 people, how many ways can we choose a committee of 3 people?
Solution
Step 1: To find the number of ways to choose a committee of 3 people from a
group of 10 people, we use the combination formula which is given by:
(n
r)=n!
r!(n−r)!
where nis the total number of people and ris the number of people we want
to choose.
Step 2: Substituting n= 10 and r= 3 into the formula, we get:
(10
3)=10!
3!(10 −3)!
Step 3: Calculating the factorials:
10! = 10 ×9×8×7×6×5×4×3×2×1 = 3628800
3! = 3 ×2×1 = 6
7! = 7 ×6×5×4×3×2×1 = 5040
Step 4: Substitute the factorials back into the formula:
(10
3)=3628800
6×5040
Step 5: Simplifying the expression:
(10
3)=3628800
30240 = 120
Therefore, there are 120 ways to choose a committee of 3 people from a
group of 10 people.
3
Question 5
Question
In a group of 10 students, how many ways can we choose a committee of 4 stu-
dents if two particular students, Alex and Beth, refuse to serve on the committee
together?
Solution
Step 1: Calculate the total number of ways to choose a committee of 4 out
of 10 students. There are 10 students to choose from, and we want to select
a committee of 4. This is a combination problem, so we use the formula for
combinations:
Total ways to choose a committee of 4 =(10
4)=10!
4!(10 −4)! =10 ×9×8×7
4×3×2×1= 210
Step 2: Calculate the number of ways to choose a committee of 4 when
both Alex and Beth are in the committee. Since Alex and Beth must be in the
committee together, we treat them as one entity. Then, we have 9 entities (Alex
and Beth as one, and the remaining 8 students) to choose the other 2 committee
members from. This is also a combination problem:
Ways to choose a committee with Alex and Beth =(9
2)=9!
2!(9 −2)! =9×8
2×1= 36
Step 3: Subtract the number of ways to choose a committee with Alex and
Beth from the total number of ways.
Number of ways to choose a committee without Alex and Beth =Total ways−Ways with Alex and Beth = 210−36 = 174
Therefore, there are 174 ways to choose a committee of 4 students from a
group of 10 students if Alex and Beth refuse to serve on the committee together.
Question 6
Question
A committee of 5 people is to be formed from a group of 8 men and 6 women. If
the committee must consist of at least 2 women, how many different committees
can be formed?
Solution
Step 1: Find the number of ways to choose a committee with exactly 2 women.
Since there must be at least 2 women in the committee, we will consider the
cases of choosing 2, 3, 4, or 5 women.
4
For choosing 2 women:
(6
2)ways to choose 2 women×(8
3)ways to choose 3 men = 15×56 = 840 ways
Step 2: Find the number of ways to choose a committee with exactly 3
women.
(6
3)ways to choose 3 women×(8
2)ways to choose 2 men = 20×28 = 560 ways
Step 3: Find the number of ways to choose a committee with exactly 4
women.
(6
4)ways to choose 4 women×(8
1)ways to choose 1 man = 15×8 = 120 ways
Step 4: Find the number of ways to choose a committee with all 5 women.
(6
5)ways to choose 5 women = 6 ways
Step 5: Calculate the total number of ways to form the committee.
Total ways = 840 + 560 + 120 + 6 = 1526
Therefore, there are 1526 different committees that can be formed with at
least 2 women.
Question 7
Question
A company has 8 different job positions available, and there are 12 applicants.
How many ways can the company hire 4 applicants for the job positions?
Solution
Step 1: To solve this problem, we will use the combination formula, which is
given by:
(n
k)=n!
k!(n−k)!
where nis the total number of items, kis the number of items to choose.
Step 2: First, we need to determine the total number of ways to select 4
applicants from the 12 that have applied. This is given by:
(12
4)=12!
4!(12 −4)! =12!
4!8!
5
Step 3: Simplifying the formula, we get:
(12
4)=12 ×11 ×10 ×9
4×3×2×1= 495
Therefore, there are 495 ways for the company to hire 4 applicants for the 8
job positions.
Question 8
Question
In a certain game, players must match 5 numbers (chosen from 1 to 50) to win
the jackpot. How many different ways can a player choose their numbers if the
order in which the numbers are selected does not matter?
Solution
Step 1: Calculate the number of ways to choose 5 numbers from 50 without
regard to order. This is a combination problem, which can be calculated using
the formula:
Number of ways =(n
k)=n!
k!(n−k)!
where nis the total number of items (50 in this case) and kis the number of
items to choose (5 in this case).
Step 2: Substituting n= 50 and k= 5 into the formula, we get:
(50
5)=50!
5!(50 −5)!
Step 3: Simplify the expression:
(50
5)=50 ×49 ×48 ×47 ×46
5×4×3×2×1
(50
5)=2118760
120
(50
5)= 176,160
Therefore, there are 176,160 different ways for a player to choose their 5
numbers for the game.
Question 9
Question
In a class of 30 students, how many ways can a committee of 5 students be
selected if Alex and Beth refuse to serve together?
6
Solution
Step 1: Find the total number of ways to select a committee of 5 students from
a class of 30 students. Step 2: Find the number of ways Alex and Beth can be
in the same committee. Step 3: Subtract the number of ways Alex and Beth
can be in the same committee from the total number of ways to get the final
answer.
Step 1: The total number of ways to select 5 students out of 30 is given by
the combination formula: C(30,5) = 30!
5!(30−5)! .
Step 2: If Alex and Beth are to be in the same committee, treat them as
one person. So the number of ways they can be in the same committee is the
number of ways to select 4 people out of 29 (treating Alex and Beth as one
person): C(29,4) = 29!
4!(29−4)! .
Step 3: The number of ways to select a committee where Alex and Beth
do not serve together is given by:
C(30,5) −C(29,4) = 30!
5!(30 −5)! −29!
4!(29 −4)!
Question 10
Question
A committee of 5 students is to be randomly selected from a group of 10 students.
How many different committees can be formed if one particular student must
be included?
Solution
Step 1: Since one particular student must be included in the committee, we
have already chosen 1 student. So, we need to choose 4 more students from the
remaining 9 students.
Step 2: To find the number of ways to select 4 students out of 9, we use the
combination formula (n
r)=n!
r!(n−r)! , where nis the total number of students
and ris the number of students we want to choose.
Step 3: Plugging in n= 9 and r= 4 into the formula, we get:
(9
4)=9!
4!(9 −4)!
Step 4: Calculating the factorials:
(9
4)=9×8×7×6×5
4×3×2×1
Step 5: Simplifying further:
(9
4)=30240
24
7
(9
4)= 1260
Step 6: Therefore, there are 1260 different committees that can be formed
if one particular student must be included.
Question 11
Question
In a class of 30 students, 15 are male and 15 are female. A committee of 5
students is to be formed randomly from the class. What is the probability that
the committee consists of 3 males and 2 females?
Solution
Step 1: Determine the total number of ways to form a committee of 5 students
from a class of 30 students. Given that there are 15 male students and 15 female
students in the class, the total number of ways to form a committee of 5 students
can be calculated using combinations:
Total ways =(30
5)=30!
5!(30 −5)! =30!
5!25! = 142506
Step 2: Determine the number of ways to form a committee consisting of 3
males and 2 females. To form a committee of 3 males and 2 females, we need to
choose 3 males out of 15 and 2 females out of 15. The number of ways to choose
3 males from 15 is given by (15
3)and the number of ways to choose 2 females
from 15 is (15
2). Therefore, the number of ways to form a committee consisting
of 3 males and 2 females is:
(15
3)×(15
2)=15!
3!12! ×15!
2!13! = 455 ×105 = 47850
Step 3: Calculate the probability of forming a committee consisting of 3
males and 2 females. The probability is given by the ratio of the number of
favorable outcomes (committees with 3 males and 2 females) to the total number
of outcomes (all possible committees):
Probability =Number of ways with 3 males and 2 females
Total number of ways =47850
142506 ≈0.3359
Therefore, the probability that the committee consists of 3 males and 2 females
is approximately 0.3359.
8
Question 12
Question
A company is assigning 5 different positions to 8 employees. How many different
ways can this be done if one of the employees, Alex, must be in a specific
position?
Solution
Step 1: Since Alex must be in a specific position, we will consider Alex already
assigned to that position. Therefore, we only need to assign the remaining 4
positions to the remaining 7 employees.
Step 2: The number of ways to assign the remaining 4 positions to the 7
employees is given by the number of permutations of 7 items taken 4 at a time.
Step 3: The number of permutations of 7 items taken 4 at a time is given
by: 7!
(7 −4)! =7!
3!
Step 4: Calculating the permutations:
7!
3! =7×6×5×4×3!
3×2×1= 7 ×6×5×4 = 840
Therefore, there are 840 different ways to assign 5 different positions to 8
employees if Alex must be in a specific position.
Question 13
Question
A committee of 6 people is to be formed from a group of 10 men and 7 women.
Find the probability that the committee consists of 3 men and 3 women.
Solution
Step 1: Find the total number of ways to form a committee of 6 people. To
form a committee of 6 people from a group of 17 (10 men and 7 women), we
can use the formula for combinations:
Total ways =(17
6)=17!
6!(17 −6)!
Step 2: Find the number of ways to choose 3 men from 10. Using com-
binations, we can find the number of ways to choose 3 men from a group of
10:
Ways to choose 3 men =(10
3)=10!
3!(10 −3)!
9
Step 3: Find the number of ways to choose 3 women from 7. Similar to the
previous step, we can calculate the number of ways to choose 3 women from 7:
Ways to choose 3 women =(7
3)=7!
3!(7 −3)!
Step 4: Find the total number of ways to form a committee with 3 men and
3 women. Since we want a committee with 3 men and 3 women, we need to
multiply the number of ways to choose 3 men by the number of ways to choose
3 women:
Total ways for 3 men and 3 women =(10
3)×(7
3)
Step 5: Calculate the probability. The probability that the committee con-
sists of 3 men and 3 women is the ratio of the total number of ways to form
a committee with 3 men and 3 women to the total number of ways to form a
committee of 6 people:
Probability =(10
3)×(7
3)
(17
6)=
10!
3!(10−3)! ×7!
3!(7−3)!
17!
6!(17−6)!
Question 14
Question
In a group of 10 people (5 men and 5 women), how many different ways can a
committee of 3 people be formed if there must be at least 1 man and 1 woman
on the committee?
Solution
Step 1: First, we will find the total number of ways to form a committee of 3
people from the group of 10 people. This can be done using the combination
formula (n
r)=n!
r!(n−r)! , where nis the total number of people and ris the
number of people we want to select. In this case, n= 10 and r= 3.
Total ways to form committee =(10
3)=10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
Step 2: Next, we will find the number of ways to form a committee with 3
people where all 3 members are men or all 3 members are women. Since there
are 5 men and 5 women, these scenarios are mutually exclusive.
• Ways to select 3 men out of 5: (5
3)=5!
3!(5−3)! = 10 ways
• Ways to select 3 women out of 5: (5
3)=5!
3!(5−3)! = 10 ways
10
Step 3: However, these scenarios do not satisfy the condition of having at
least 1 man and 1 woman on the committee. Therefore, we will find the total
number of ways to select a committee with either 2 men and 1 woman or 1 man
and 2 women.
• Ways to select 2 men out of 5 and 1 woman out of 5: (5
2)×(5
1)= 10×5 = 50
ways
• Ways to select 1 man out of 5 and 2 women out of 5: (5
1)×(5
2)= 5×10 = 50
ways
Step 4: Finally, we calculate the total number of ways to form a committee
of 3 people with at least 1 man and 1 woman by summing up the possibilities
from Step 2 and Step 3.
Total ways with at least 1 man and 1 woman = 10 + 10 + 50 + 50 = 120
Therefore, there are 120 different ways a committee of 3 people can be formed
from a group of 10 people (5 men and 5 women) with at least 1 man and 1 woman
on the committee.
Question 15
Question
In a group of 10 people, how many ways can we choose a committee of 4 people
if 2 of the members, Alice and Bob, refuse to serve together on the committee?
Solution
Step 1: Find the total number of ways to choose a committee of 4 people without
any restrictions. The total number of ways to choose 4 people from a group of
10 is given by the combination formula: (10
4)=10!
4!(10−4)! .
Step 2: Find the number of ways to choose a committee where Alice and
Bob are both in the committee. Let’s assume that Alice and Bob will work
together in the committee. Then, we need to choose 2 more people from the
remaining 8 people. This can be done in (8
2)ways.
Step 3: Subtract the number of ways with Alice and Bob from the total to get
the final result. Now, we need to subtract the number of ways where Alice and
Bob are both in the committee from the total number of ways: Total number
of ways −Number of ways with Alice and Bob in the committee =(10
4)−(8
2).
Therefore, the number of ways to choose a committee of 4 people where
Alice and Bob refuse to serve together is (10
4)−(8
2).
11
Question 16
Question
A committee of 4 people is to be formed from a group of 10 students. If 4 of
the students are math majors and 6 are engineering majors, how many different
committees can be formed if each committee must have at least 2 math majors
and at least 1 engineering major?
Solution
Step 1: Calculate the number of committees with exactly 2 math majors and 2
engineering majors.
There are (4
2)= 6 ways to choose 2 math majors from the 4 available, and
(6
2)= 15 ways to choose 2 engineering majors from the 6 available. Multiply
these together to get 6×15 = 90.
Step 2: Calculate the number of committees with exactly 3 math majors
and 1 engineering major.
There are (4
3)= 4 ways to choose 3 math majors from the 4 available, and
(6
1)= 6 ways to choose 1 engineering major from the 6 available. Multiply these
together to get 4×6 = 24.
Step 3: Calculate the number of committees with 4 math majors.
There is only 1 way to choose all 4 math majors from the 4 available.
Step 4: Add the results from Steps 1, 2, and 3 to find the total number of
valid committees.
90 + 24 + 1 = 115
Therefore, there are 115 different committees that can be formed where each
committee must have at least 2 math majors and at least 1 engineering major.
Question 17
Question
A committee of 5 people is to be formed from a group of 8 men and 7 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: Calculate the number of committees with exactly 2 men and 3 women.
To select 2 men from 8 men and 3 women from 7 women, we use the combination
formula: (8
2)·(7
3)=8!
2!(8 −2)! ·7!
3!(7 −3)! = 28 ·35 = 980
Step 2: Calculate the number of committees with exactly 3 men and 2
women. To select 3 men from 8 men and 2 women from 7 women, we use the
12
combination formula:
(8
3)·(7
2)=8!
3!(8 −3)! ·7!
2!(7 −2)! = 56 ·21 = 1176
Step 3: Calculate the number of committees with 4 men and 1 woman. To
select 4 men from 8 men and 1 woman from 7 women, we use the combination
formula: (8
4)·(7
1)=8!
4!(8 −4)! ·7!
1!(7 −1)! = 70 ·7 = 490
Step 4: Calculate the total number of committees with at least 2 men and 2
women. Add the results from Step 1, Step 2, and Step 3: Total = 980 + 1176
+ 490 = 2646
Therefore, there are 2646 different committees that can be formed with at
least 2 men and at least 2 women from the group.
Question 18
Question
In how many ways can a committee of 3 people be selected from a group of 7
people to serve as president, vice-president, and treasurer?
Solution
Step 1: To find the number of ways to select the committee, we first need to
determine the number of ways to select the president, then the vice-president,
and finally the treasurer.
Step 2: There are 7 choices for the president, since any of the 7 people can
be selected.
Step 3: Once the president is chosen, there are 6 remaining people to choose
from for the vice-president.
Step 4: After selecting the president and vice-president, there are 5 remain-
ing people to choose from for the treasurer.
Step 5: Therefore, the total number of ways to select the committee is the
product of the number of choices for each position:
7×6×5 = 210
Step 6: So, there are 210 ways to select a committee of 3 people from a
group of 7 to serve as president, vice-president, and treasurer.
Question 19
Question
In a group of 10 students, how many ways are there to select a committee of 3
students and a chairperson from the group?
13
Solution
Step 1: To find the number of ways to select a committee of 3 students from
10, we will use the combination formula (n
r)=n!
r!(n−r)! where nis the total
number of students and ris the number of students we want to select.
Step 2: Substitute n= 10 and r= 3 into the combination formula:
(10
3)=10!
3!(10 −3)!
Step 3: Simplify the factorial terms in the denominator:
(10
3)=10!
3!7!
Step 4: Calculate the factorials:
(10
3)=10 ×9×8×7!
3×2×1×7!
Step 5: Simplify further to find the number of ways to select the committee:
(10
3)=10 ×9×8
3×2×1= 120
Step 6: After selecting the 3 students for the committee, we need to choose
a chairperson from those 3 students. There are 3 ways to choose a chairperson
from the committee.
Step 7: Multiply the number of ways to select the committee by the number
of ways to choose a chairperson to find the total number of ways:
120 ×3 = 360
Therefore, there are 360 ways to select a committee of 3 students and a
chairperson from a group of 10 students.
Question 20
Question
A committee of 5 people is to be formed from a group of 10 women and 8 men.
How many different committees can be formed if the committee must have at
least 2 women and at least 2 men?
14
Solution
Step 1: Calculate the total number of ways to form a committee with 5 people:
There are 10 women and 8 men to choose from, so the total number of ways to
choose 5 people without any restrictions is (18
5)= 8,568.
Step 2: Calculate the number of ways to form a committee with at least 2
women and at least 2 men: To form a committee with at least 2 women and at
least 2 men, we can have the following cases: - 2 women and 3 men - 3 women
and 2 men - 4 women and 1 man - 5 women and 0 men
For the case of 2 women and 3 men: Choose 2 women from 10: (10
2)ways
Choose 3 men from 8: (8
3)ways Multiply the two to get the total number of
ways: (10
2)×(8
3)= 2,520 ways
For the case of 3 women and 2 men: Choose 3 women from 10: (10
3)ways
Choose 2 men from 8: (8
2)ways Multiply the two to get the total number of
ways: (10
3)×(8
2)= 2,520 ways
For the case of 4 women and 1 man: Choose 4 women from 10: (10
4)ways
Choose 1 man from 8: (8
1)ways Multiply the two to get the total number of
ways: (10
4)×(8
1)= 560 ways
For the case of 5 women and 0 men: Choose 5 women from 10: (10
5)ways
There are no men to choose, so only 1 way The total number of ways: (10
5)= 252
ways
Step 3: Add up the number of ways for each case to get the total number
of ways to form the committee with at least 2 women and at least 2 men:
2,520 + 2,520 + 560 + 252 = 5,852 ways
Therefore, there are 5,852 different committees that can be formed with at
least 2 women and at least 2 men.
Question 21
Question
In a group of 10 students, how many ways can we choose a committee of 4
students?
Solution
Step 1: To solve this problem, we will use the formula for combinations, which
is given by:
C(n, k) = n!
k!(n−k)!
where nis the total number of students and kis the number of students we
want to choose for the committee.
15
Step 2: Substituting n= 10 and k= 4 into the formula, we get:
C(10,4) = 10!
4!(10 −4)!
Step 3: Calculating the factorials:
10! = 10 ×9×8×7×6×5×4×3×2×1
4! = 4 ×3×2×1
6! = 6 ×5×4×3×2×1
Step 4: Substituting the factorials into the formula:
C(10,4) = 10 ×9×8×7×6×5×4×3×2×1
4×3×2×1×6×5×4×3×2×1
Step 5: Simplifying the expression:
C(10,4) = 10 ×9×8×7
4×3×2×1= 210
Step 6: Therefore, there are 210 ways to choose a committee of 4 students
from a group of 10.
Question 22
Question
A committee of 5 people is to be formed from a group of 8 men and 6 women.
If the committee must consist of 3 men and 2 women, how many different
committees can be formed?
Solution
Step 1: Calculate the number of ways to choose 3 men from 8. Step 2: Calculate
the number of ways to choose 2 women from 6. Step 3: Multiply the results of
Step 1 and Step 2 to find the total number of different committees that can be
formed.
Step 1: Number of ways to choose 3 men from 8.
C(8,3) = 8!
3!(8 −3)! =8×7×6
3×2×1= 56
Step 2: Number of ways to choose 2 women from 6.
C(6,2) = 6!
2!(6 −2)! =6×5
2×1= 15
Step 3: Total number of different committees that can be formed.
56 ×15 = 840
Therefore, there are 840 different committees that can be formed with 3 men
and 2 women.
16
Question 23
Question
In a group of 10 students, how many ways can we select a committee of 3
students, where one student is designated as the president, one as the vice
president, and one as the secretary?
Solution
To solve this problem, we can break it down into two steps: Step 1: Calculate
the number of ways to select a committee of 3 students. Step 2: After selecting
the committee, assign the roles of president, vice president, and secretary to the
selected students.
Step 1: To select a committee of 3 students out of 10, we can use the
combination formula:
Number of ways =(10
3)=10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
So, there are 120 ways to select a committee of 3 students.
Step 2: After selecting the committee, we need to assign the roles of pres-
ident, vice president, and secretary. For the president, there are 3 students to
choose from the committee. After selecting the president, there are 2 students
left for the vice president role. Finally, the last remaining student will be the
secretary.
Therefore, the total number of ways to select the committee with assigned
roles is:
120 ×3×2×1 = 720
Therefore, there are 720 ways to select a committee of 3 students, where one
is the president, one is the vice president, and one is the secretary.
Question 24
Question
A committee of 4 people is to be formed from a group of 10 people. How many
different committees can be formed if 2 specific people must be included in the
committee?
Solution
Step 1: Choose the 2 specific people to be included in the committee. There
are (10
2)ways to do this.
Step 2: Choose the remaining 2 people from the remaining 8 people. There
are (8
2)ways to do this.
17
Step 3: Multiply the results from Step 1 and Step 2 to find the total number
of different committees that can be formed.
Therefore, the total number of different committees that can be formed is
given by:
(10
2)×(8
2)=10!
2!8! ×8!
2!6!
=10 ×9
2×1×8×7
2×1= 45 ×28
= 1260
So, there are 1260 different committees that can be formed if 2 specific people
must be included in the committee.
Question 25
Question
In a group of 10 students, how many ways can we choose a committee of 3
students to serve on the student council if one student, Alice, refuses to serve
with Bob due to a conflict?
Solution
Step 1: First, let’s find the total number of ways to choose a committee of 3
students from a group of 10 students. This is given by the combination formula
(n
r)=n!
r!(n−r)! , where nis the total number of students and ris the number of
students we want to choose for the committee. In this case, n= 10 and r= 3.
Thus, the total number of ways to choose a committee of 3 students from 10 is:
(10
3)=10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
Step 2: Next, let’s find the number of ways to choose a committee of 3
students from the 9 remaining students after Alice refuses to serve with Bob.
This is given by the combination formula. So, the number of ways to choose a
committee of 3 students from 9 is:
(9
3)=9!
3!(9 −3)! =9×8×7
3×2×1= 84
Step 3: Now, to find the number of ways to choose a committee of 3 stu-
dents that does not include Alice and Bob together, we need to subtract the
number of committees where Alice and Bob are together from the total number
of committees. There are 84 ways to choose a committee of 3 students from the
9 remaining students, and Alice and Bob can be together in (2
1)×(7
1)= 14 ways
(either Alice and Bob or Bob and Alice). Therefore, the number of ways to
choose a committee of 3 students that does not include Alice and Bob together
is 84 −14 = 70 ways.
18
Question 2
Question
A committee of 5 people is to be formed from a group of 10 students and 5
faculty members. If at least 2 faculty members must be on the committee, how
many different committees can be formed?
Solution
Step 1: Calculate the number of ways to choose 2 faculty members out of 5.
Step 2: Calculate the number of ways to choose the remaining 3 members
from the remaining 10 students and 3 faculty members (after choosing 2 faculty
members). Step 3: Multiply the results from Step 1 and Step 2 to find the total
number of committees.
Step 1: Choosing 2 faculty members Since we are choosing 2 faculty
members out of 5, the number of ways to do this is given by (5
2)=5!
2!(5−2)! = 10.
Step 2: Choosing the remaining 3 members After choosing 2 faculty
members, there are 3 faculty members left and 10 students remaining. We need
to choose 3 members from these 13 people. So, the number of ways to choose
the remaining 3 members is (13
3)=13!
3!(13−3)! = 286.
Step 3: Total number of committees To calculate the total number of
committees, we multiply the results from Step 1 and Step 2. Total number of
committees = 10 ×286 = 2860.
Therefore, there are 2,860 different committees that can be formed with at
least 2 faculty members on the committee.
Question 3
Question
Suppose you have a standard deck of 52 playing cards (13 cards in each of 4
suits: hearts, diamonds, clubs, and spades). How many ways can you choose a
5-card hand such that it contains exactly 3 hearts and 2 spades?
Solution
Step 1: Determine the number of ways to choose 3 hearts from the 13 available.
There are (13
3)ways to choose 3 hearts from the 13 available.
Step 2: Determine the number of ways to choose 2 spades from the 13
available. There are (13
2)ways to choose 2 spades from the 13 available.
Step 3: Multiply the results from Step 1 and Step 2 to find the total number
of ways to choose 3 hearts and 2 spades. Total ways = (13
3)×(13
2).
Step 4: Calculate the total number of ways. (13
3)=13!
3!(13−3)! =13×12×11
3×2×1=
286.
(13
2)=13!
2!(13−2)! =13×12
2×1= 78.
2
Therefore, the total number of ways to choose 3 hearts and 2 spades is
286 ×78 = 22236. Thus, there are 22,236 ways to choose a 5-card hand with
exactly 3 hearts and 2 spades.
Question 4
Question
In a group of 10 people, how many ways can we choose a committee of 3 people?
Solution
Step 1: To find the number of ways to choose a committee of 3 people from a
group of 10 people, we use the combination formula which is given by:
(n
r)=n!
r!(n−r)!
where nis the total number of people and ris the number of people we want
to choose.
Step 2: Substituting n= 10 and r= 3 into the formula, we get:
(10
3)=10!
3!(10 −3)!
Step 3: Calculating the factorials:
10! = 10 ×9×8×7×6×5×4×3×2×1 = 3628800
3! = 3 ×2×1 = 6
7! = 7 ×6×5×4×3×2×1 = 5040
Step 4: Substitute the factorials back into the formula:
(10
3)=3628800
6×5040
Step 5: Simplifying the expression:
(10
3)=3628800
30240 = 120
Therefore, there are 120 ways to choose a committee of 3 people from a
group of 10 people.
3
Question 5
Question
In a group of 10 students, how many ways can we choose a committee of 4 stu-
dents if two particular students, Alex and Beth, refuse to serve on the committee
together?
Solution
Step 1: Calculate the total number of ways to choose a committee of 4 out
of 10 students. There are 10 students to choose from, and we want to select
a committee of 4. This is a combination problem, so we use the formula for
combinations:
Total ways to choose a committee of 4 =(10
4)=10!
4!(10 −4)! =10 ×9×8×7
4×3×2×1= 210
Step 2: Calculate the number of ways to choose a committee of 4 when
both Alex and Beth are in the committee. Since Alex and Beth must be in the
committee together, we treat them as one entity. Then, we have 9 entities (Alex
and Beth as one, and the remaining 8 students) to choose the other 2 committee
members from. This is also a combination problem:
Ways to choose a committee with Alex and Beth =(9
2)=9!
2!(9 −2)! =9×8
2×1= 36
Step 3: Subtract the number of ways to choose a committee with Alex and
Beth from the total number of ways.
Number of ways to choose a committee without Alex and Beth =Total ways−Ways with Alex and Beth = 210−36 = 174
Therefore, there are 174 ways to choose a committee of 4 students from a
group of 10 students if Alex and Beth refuse to serve on the committee together.
Question 6
Question
A committee of 5 people is to be formed from a group of 8 men and 6 women. If
the committee must consist of at least 2 women, how many different committees
can be formed?
Solution
Step 1: Find the number of ways to choose a committee with exactly 2 women.
Since there must be at least 2 women in the committee, we will consider the
cases of choosing 2, 3, 4, or 5 women.
4
For choosing 2 women:
(6
2)ways to choose 2 women×(8
3)ways to choose 3 men = 15×56 = 840 ways
Step 2: Find the number of ways to choose a committee with exactly 3
women.
(6
3)ways to choose 3 women×(8
2)ways to choose 2 men = 20×28 = 560 ways
Step 3: Find the number of ways to choose a committee with exactly 4
women.
(6
4)ways to choose 4 women×(8
1)ways to choose 1 man = 15×8 = 120 ways
Step 4: Find the number of ways to choose a committee with all 5 women.
(6
5)ways to choose 5 women = 6 ways
Step 5: Calculate the total number of ways to form the committee.
Total ways = 840 + 560 + 120 + 6 = 1526
Therefore, there are 1526 different committees that can be formed with at
least 2 women.
Question 7
Question
A company has 8 different job positions available, and there are 12 applicants.
How many ways can the company hire 4 applicants for the job positions?
Solution
Step 1: To solve this problem, we will use the combination formula, which is
given by:
(n
k)=n!
k!(n−k)!
where nis the total number of items, kis the number of items to choose.
Step 2: First, we need to determine the total number of ways to select 4
applicants from the 12 that have applied. This is given by:
(12
4)=12!
4!(12 −4)! =12!
4!8!
5
Step 3: Simplifying the formula, we get:
(12
4)=12 ×11 ×10 ×9
4×3×2×1= 495
Therefore, there are 495 ways for the company to hire 4 applicants for the 8
job positions.
Question 8
Question
In a certain game, players must match 5 numbers (chosen from 1 to 50) to win
the jackpot. How many different ways can a player choose their numbers if the
order in which the numbers are selected does not matter?
Solution
Step 1: Calculate the number of ways to choose 5 numbers from 50 without
regard to order. This is a combination problem, which can be calculated using
the formula:
Number of ways =(n
k)=n!
k!(n−k)!
where nis the total number of items (50 in this case) and kis the number of
items to choose (5 in this case).
Step 2: Substituting n= 50 and k= 5 into the formula, we get:
(50
5)=50!
5!(50 −5)!
Step 3: Simplify the expression:
(50
5)=50 ×49 ×48 ×47 ×46
5×4×3×2×1
(50
5)=2118760
120
(50
5)= 176,160
Therefore, there are 176,160 different ways for a player to choose their 5
numbers for the game.
Question 9
Question
In a class of 30 students, how many ways can a committee of 5 students be
selected if Alex and Beth refuse to serve together?
6
Solution
Step 1: Find the total number of ways to select a committee of 5 students from
a class of 30 students. Step 2: Find the number of ways Alex and Beth can be
in the same committee. Step 3: Subtract the number of ways Alex and Beth
can be in the same committee from the total number of ways to get the final
answer.
Step 1: The total number of ways to select 5 students out of 30 is given by
the combination formula: C(30,5) = 30!
5!(30−5)! .
Step 2: If Alex and Beth are to be in the same committee, treat them as
one person. So the number of ways they can be in the same committee is the
number of ways to select 4 people out of 29 (treating Alex and Beth as one
person): C(29,4) = 29!
4!(29−4)! .
Step 3: The number of ways to select a committee where Alex and Beth
do not serve together is given by:
C(30,5) −C(29,4) = 30!
5!(30 −5)! −29!
4!(29 −4)!
Question 10
Question
A committee of 5 students is to be randomly selected from a group of 10 students.
How many different committees can be formed if one particular student must
be included?
Solution
Step 1: Since one particular student must be included in the committee, we
have already chosen 1 student. So, we need to choose 4 more students from the
remaining 9 students.
Step 2: To find the number of ways to select 4 students out of 9, we use the
combination formula (n
r)=n!
r!(n−r)! , where nis the total number of students
and ris the number of students we want to choose.
Step 3: Plugging in n= 9 and r= 4 into the formula, we get:
(9
4)=9!
4!(9 −4)!
Step 4: Calculating the factorials:
(9
4)=9×8×7×6×5
4×3×2×1
Step 5: Simplifying further:
(9
4)=30240
24
7
(9
4)= 1260
Step 6: Therefore, there are 1260 different committees that can be formed
if one particular student must be included.
Question 11
Question
In a class of 30 students, 15 are male and 15 are female. A committee of 5
students is to be formed randomly from the class. What is the probability that
the committee consists of 3 males and 2 females?
Solution
Step 1: Determine the total number of ways to form a committee of 5 students
from a class of 30 students. Given that there are 15 male students and 15 female
students in the class, the total number of ways to form a committee of 5 students
can be calculated using combinations:
Total ways =(30
5)=30!
5!(30 −5)! =30!
5!25! = 142506
Step 2: Determine the number of ways to form a committee consisting of 3
males and 2 females. To form a committee of 3 males and 2 females, we need to
choose 3 males out of 15 and 2 females out of 15. The number of ways to choose
3 males from 15 is given by (15
3)and the number of ways to choose 2 females
from 15 is (15
2). Therefore, the number of ways to form a committee consisting
of 3 males and 2 females is:
(15
3)×(15
2)=15!
3!12! ×15!
2!13! = 455 ×105 = 47850
Step 3: Calculate the probability of forming a committee consisting of 3
males and 2 females. The probability is given by the ratio of the number of
favorable outcomes (committees with 3 males and 2 females) to the total number
of outcomes (all possible committees):
Probability =Number of ways with 3 males and 2 females
Total number of ways =47850
142506 ≈0.3359
Therefore, the probability that the committee consists of 3 males and 2 females
is approximately 0.3359.
8
Question 12
Question
A company is assigning 5 different positions to 8 employees. How many different
ways can this be done if one of the employees, Alex, must be in a specific
position?
Solution
Step 1: Since Alex must be in a specific position, we will consider Alex already
assigned to that position. Therefore, we only need to assign the remaining 4
positions to the remaining 7 employees.
Step 2: The number of ways to assign the remaining 4 positions to the 7
employees is given by the number of permutations of 7 items taken 4 at a time.
Step 3: The number of permutations of 7 items taken 4 at a time is given
by: 7!
(7 −4)! =7!
3!
Step 4: Calculating the permutations:
7!
3! =7×6×5×4×3!
3×2×1= 7 ×6×5×4 = 840
Therefore, there are 840 different ways to assign 5 different positions to 8
employees if Alex must be in a specific position.
Question 13
Question
A committee of 6 people is to be formed from a group of 10 men and 7 women.
Find the probability that the committee consists of 3 men and 3 women.
Solution
Step 1: Find the total number of ways to form a committee of 6 people. To
form a committee of 6 people from a group of 17 (10 men and 7 women), we
can use the formula for combinations:
Total ways =(17
6)=17!
6!(17 −6)!
Step 2: Find the number of ways to choose 3 men from 10. Using com-
binations, we can find the number of ways to choose 3 men from a group of
10:
Ways to choose 3 men =(10
3)=10!
3!(10 −3)!
9
Step 3: Find the number of ways to choose 3 women from 7. Similar to the
previous step, we can calculate the number of ways to choose 3 women from 7:
Ways to choose 3 women =(7
3)=7!
3!(7 −3)!
Step 4: Find the total number of ways to form a committee with 3 men and
3 women. Since we want a committee with 3 men and 3 women, we need to
multiply the number of ways to choose 3 men by the number of ways to choose
3 women:
Total ways for 3 men and 3 women =(10
3)×(7
3)
Step 5: Calculate the probability. The probability that the committee con-
sists of 3 men and 3 women is the ratio of the total number of ways to form
a committee with 3 men and 3 women to the total number of ways to form a
committee of 6 people:
Probability =(10
3)×(7
3)
(17
6)=
10!
3!(10−3)! ×7!
3!(7−3)!
17!
6!(17−6)!
Question 14
Question
In a group of 10 people (5 men and 5 women), how many different ways can a
committee of 3 people be formed if there must be at least 1 man and 1 woman
on the committee?
Solution
Step 1: First, we will find the total number of ways to form a committee of 3
people from the group of 10 people. This can be done using the combination
formula (n
r)=n!
r!(n−r)! , where nis the total number of people and ris the
number of people we want to select. In this case, n= 10 and r= 3.
Total ways to form committee =(10
3)=10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
Step 2: Next, we will find the number of ways to form a committee with 3
people where all 3 members are men or all 3 members are women. Since there
are 5 men and 5 women, these scenarios are mutually exclusive.
• Ways to select 3 men out of 5: (5
3)=5!
3!(5−3)! = 10 ways
• Ways to select 3 women out of 5: (5
3)=5!
3!(5−3)! = 10 ways
10
Step 3: However, these scenarios do not satisfy the condition of having at
least 1 man and 1 woman on the committee. Therefore, we will find the total
number of ways to select a committee with either 2 men and 1 woman or 1 man
and 2 women.
• Ways to select 2 men out of 5 and 1 woman out of 5: (5
2)×(5
1)= 10×5 = 50
ways
• Ways to select 1 man out of 5 and 2 women out of 5: (5
1)×(5
2)= 5×10 = 50
ways
Step 4: Finally, we calculate the total number of ways to form a committee
of 3 people with at least 1 man and 1 woman by summing up the possibilities
from Step 2 and Step 3.
Total ways with at least 1 man and 1 woman = 10 + 10 + 50 + 50 = 120
Therefore, there are 120 different ways a committee of 3 people can be formed
from a group of 10 people (5 men and 5 women) with at least 1 man and 1 woman
on the committee.
Question 15
Question
In a group of 10 people, how many ways can we choose a committee of 4 people
if 2 of the members, Alice and Bob, refuse to serve together on the committee?
Solution
Step 1: Find the total number of ways to choose a committee of 4 people without
any restrictions. The total number of ways to choose 4 people from a group of
10 is given by the combination formula: (10
4)=10!
4!(10−4)! .
Step 2: Find the number of ways to choose a committee where Alice and
Bob are both in the committee. Let’s assume that Alice and Bob will work
together in the committee. Then, we need to choose 2 more people from the
remaining 8 people. This can be done in (8
2)ways.
Step 3: Subtract the number of ways with Alice and Bob from the total to get
the final result. Now, we need to subtract the number of ways where Alice and
Bob are both in the committee from the total number of ways: Total number
of ways −Number of ways with Alice and Bob in the committee =(10
4)−(8
2).
Therefore, the number of ways to choose a committee of 4 people where
Alice and Bob refuse to serve together is (10
4)−(8
2).
11
Question 16
Question
A committee of 4 people is to be formed from a group of 10 students. If 4 of
the students are math majors and 6 are engineering majors, how many different
committees can be formed if each committee must have at least 2 math majors
and at least 1 engineering major?
Solution
Step 1: Calculate the number of committees with exactly 2 math majors and 2
engineering majors.
There are (4
2)= 6 ways to choose 2 math majors from the 4 available, and
(6
2)= 15 ways to choose 2 engineering majors from the 6 available. Multiply
these together to get 6×15 = 90.
Step 2: Calculate the number of committees with exactly 3 math majors
and 1 engineering major.
There are (4
3)= 4 ways to choose 3 math majors from the 4 available, and
(6
1)= 6 ways to choose 1 engineering major from the 6 available. Multiply these
together to get 4×6 = 24.
Step 3: Calculate the number of committees with 4 math majors.
There is only 1 way to choose all 4 math majors from the 4 available.
Step 4: Add the results from Steps 1, 2, and 3 to find the total number of
valid committees.
90 + 24 + 1 = 115
Therefore, there are 115 different committees that can be formed where each
committee must have at least 2 math majors and at least 1 engineering major.
Question 17
Question
A committee of 5 people is to be formed from a group of 8 men and 7 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: Calculate the number of committees with exactly 2 men and 3 women.
To select 2 men from 8 men and 3 women from 7 women, we use the combination
formula: (8
2)·(7
3)=8!
2!(8 −2)! ·7!
3!(7 −3)! = 28 ·35 = 980
Step 2: Calculate the number of committees with exactly 3 men and 2
women. To select 3 men from 8 men and 2 women from 7 women, we use the
12
combination formula:
(8
3)·(7
2)=8!
3!(8 −3)! ·7!
2!(7 −2)! = 56 ·21 = 1176
Step 3: Calculate the number of committees with 4 men and 1 woman. To
select 4 men from 8 men and 1 woman from 7 women, we use the combination
formula: (8
4)·(7
1)=8!
4!(8 −4)! ·7!
1!(7 −1)! = 70 ·7 = 490
Step 4: Calculate the total number of committees with at least 2 men and 2
women. Add the results from Step 1, Step 2, and Step 3: Total = 980 + 1176
+ 490 = 2646
Therefore, there are 2646 different committees that can be formed with at
least 2 men and at least 2 women from the group.
Question 18
Question
In how many ways can a committee of 3 people be selected from a group of 7
people to serve as president, vice-president, and treasurer?
Solution
Step 1: To find the number of ways to select the committee, we first need to
determine the number of ways to select the president, then the vice-president,
and finally the treasurer.
Step 2: There are 7 choices for the president, since any of the 7 people can
be selected.
Step 3: Once the president is chosen, there are 6 remaining people to choose
from for the vice-president.
Step 4: After selecting the president and vice-president, there are 5 remain-
ing people to choose from for the treasurer.
Step 5: Therefore, the total number of ways to select the committee is the
product of the number of choices for each position:
7×6×5 = 210
Step 6: So, there are 210 ways to select a committee of 3 people from a
group of 7 to serve as president, vice-president, and treasurer.
Question 19
Question
In a group of 10 students, how many ways are there to select a committee of 3
students and a chairperson from the group?
13
Solution
Step 1: To find the number of ways to select a committee of 3 students from
10, we will use the combination formula (n
r)=n!
r!(n−r)! where nis the total
number of students and ris the number of students we want to select.
Step 2: Substitute n= 10 and r= 3 into the combination formula:
(10
3)=10!
3!(10 −3)!
Step 3: Simplify the factorial terms in the denominator:
(10
3)=10!
3!7!
Step 4: Calculate the factorials:
(10
3)=10 ×9×8×7!
3×2×1×7!
Step 5: Simplify further to find the number of ways to select the committee:
(10
3)=10 ×9×8
3×2×1= 120
Step 6: After selecting the 3 students for the committee, we need to choose
a chairperson from those 3 students. There are 3 ways to choose a chairperson
from the committee.
Step 7: Multiply the number of ways to select the committee by the number
of ways to choose a chairperson to find the total number of ways:
120 ×3 = 360
Therefore, there are 360 ways to select a committee of 3 students and a
chairperson from a group of 10 students.
Question 20
Question
A committee of 5 people is to be formed from a group of 10 women and 8 men.
How many different committees can be formed if the committee must have at
least 2 women and at least 2 men?
14
Solution
Step 1: Calculate the total number of ways to form a committee with 5 people:
There are 10 women and 8 men to choose from, so the total number of ways to
choose 5 people without any restrictions is (18
5)= 8,568.
Step 2: Calculate the number of ways to form a committee with at least 2
women and at least 2 men: To form a committee with at least 2 women and at
least 2 men, we can have the following cases: - 2 women and 3 men - 3 women
and 2 men - 4 women and 1 man - 5 women and 0 men
For the case of 2 women and 3 men: Choose 2 women from 10: (10
2)ways
Choose 3 men from 8: (8
3)ways Multiply the two to get the total number of
ways: (10
2)×(8
3)= 2,520 ways
For the case of 3 women and 2 men: Choose 3 women from 10: (10
3)ways
Choose 2 men from 8: (8
2)ways Multiply the two to get the total number of
ways: (10
3)×(8
2)= 2,520 ways
For the case of 4 women and 1 man: Choose 4 women from 10: (10
4)ways
Choose 1 man from 8: (8
1)ways Multiply the two to get the total number of
ways: (10
4)×(8
1)= 560 ways
For the case of 5 women and 0 men: Choose 5 women from 10: (10
5)ways
There are no men to choose, so only 1 way The total number of ways: (10
5)= 252
ways
Step 3: Add up the number of ways for each case to get the total number
of ways to form the committee with at least 2 women and at least 2 men:
2,520 + 2,520 + 560 + 252 = 5,852 ways
Therefore, there are 5,852 different committees that can be formed with at
least 2 women and at least 2 men.
Question 21
Question
In a group of 10 students, how many ways can we choose a committee of 4
students?
Solution
Step 1: To solve this problem, we will use the formula for combinations, which
is given by:
C(n, k) = n!
k!(n−k)!
where nis the total number of students and kis the number of students we
want to choose for the committee.
15
Step 2: Substituting n= 10 and k= 4 into the formula, we get:
C(10,4) = 10!
4!(10 −4)!
Step 3: Calculating the factorials:
10! = 10 ×9×8×7×6×5×4×3×2×1
4! = 4 ×3×2×1
6! = 6 ×5×4×3×2×1
Step 4: Substituting the factorials into the formula:
C(10,4) = 10 ×9×8×7×6×5×4×3×2×1
4×3×2×1×6×5×4×3×2×1
Step 5: Simplifying the expression:
C(10,4) = 10 ×9×8×7
4×3×2×1= 210
Step 6: Therefore, there are 210 ways to choose a committee of 4 students
from a group of 10.
Question 22
Question
A committee of 5 people is to be formed from a group of 8 men and 6 women.
If the committee must consist of 3 men and 2 women, how many different
committees can be formed?
Solution
Step 1: Calculate the number of ways to choose 3 men from 8. Step 2: Calculate
the number of ways to choose 2 women from 6. Step 3: Multiply the results of
Step 1 and Step 2 to find the total number of different committees that can be
formed.
Step 1: Number of ways to choose 3 men from 8.
C(8,3) = 8!
3!(8 −3)! =8×7×6
3×2×1= 56
Step 2: Number of ways to choose 2 women from 6.
C(6,2) = 6!
2!(6 −2)! =6×5
2×1= 15
Step 3: Total number of different committees that can be formed.
56 ×15 = 840
Therefore, there are 840 different committees that can be formed with 3 men
and 2 women.
16
Question 23
Question
In a group of 10 students, how many ways can we select a committee of 3
students, where one student is designated as the president, one as the vice
president, and one as the secretary?
Solution
To solve this problem, we can break it down into two steps: Step 1: Calculate
the number of ways to select a committee of 3 students. Step 2: After selecting
the committee, assign the roles of president, vice president, and secretary to the
selected students.
Step 1: To select a committee of 3 students out of 10, we can use the
combination formula:
Number of ways =(10
3)=10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
So, there are 120 ways to select a committee of 3 students.
Step 2: After selecting the committee, we need to assign the roles of pres-
ident, vice president, and secretary. For the president, there are 3 students to
choose from the committee. After selecting the president, there are 2 students
left for the vice president role. Finally, the last remaining student will be the
secretary.
Therefore, the total number of ways to select the committee with assigned
roles is:
120 ×3×2×1 = 720
Therefore, there are 720 ways to select a committee of 3 students, where one
is the president, one is the vice president, and one is the secretary.
Question 24
Question
A committee of 4 people is to be formed from a group of 10 people. How many
different committees can be formed if 2 specific people must be included in the
committee?
Solution
Step 1: Choose the 2 specific people to be included in the committee. There
are (10
2)ways to do this.
Step 2: Choose the remaining 2 people from the remaining 8 people. There
are (8
2)ways to do this.
17
Step 3: Multiply the results from Step 1 and Step 2 to find the total number
of different committees that can be formed.
Therefore, the total number of different committees that can be formed is
given by:
(10
2)×(8
2)=10!
2!8! ×8!
2!6!
=10 ×9
2×1×8×7
2×1= 45 ×28
= 1260
So, there are 1260 different committees that can be formed if 2 specific people
must be included in the committee.
Question 25
Question
In a group of 10 students, how many ways can we choose a committee of 3
students to serve on the student council if one student, Alice, refuses to serve
with Bob due to a conflict?
Solution
Step 1: First, let’s find the total number of ways to choose a committee of 3
students from a group of 10 students. This is given by the combination formula
(n
r)=n!
r!(n−r)! , where nis the total number of students and ris the number of
students we want to choose for the committee. In this case, n= 10 and r= 3.
Thus, the total number of ways to choose a committee of 3 students from 10 is:
(10
3)=10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
Step 2: Next, let’s find the number of ways to choose a committee of 3
students from the 9 remaining students after Alice refuses to serve with Bob.
This is given by the combination formula. So, the number of ways to choose a
committee of 3 students from 9 is:
(9
3)=9!
3!(9 −3)! =9×8×7
3×2×1= 84
Step 3: Now, to find the number of ways to choose a committee of 3 stu-
dents that does not include Alice and Bob together, we need to subtract the
number of committees where Alice and Bob are together from the total number
of committees. There are 84 ways to choose a committee of 3 students from the
9 remaining students, and Alice and Bob can be together in (2
1)×(7
1)= 14 ways
(either Alice and Bob or Bob and Alice). Therefore, the number of ways to
choose a committee of 3 students that does not include Alice and Bob together
is 84 −14 = 70 ways.
18
Question 2
Question
A committee of 5 people is to be formed from a group of 10 students and 5
faculty members. If at least 2 faculty members must be on the committee, how
many different committees can be formed?
Solution
Step 1: Calculate the number of ways to choose 2 faculty members out of 5.
Step 2: Calculate the number of ways to choose the remaining 3 members
from the remaining 10 students and 3 faculty members (after choosing 2 faculty
members). Step 3: Multiply the results from Step 1 and Step 2 to find the total
number of committees.
Step 1: Choosing 2 faculty members Since we are choosing 2 faculty
members out of 5, the number of ways to do this is given by (5
2)=5!
2!(5−2)! = 10.
Step 2: Choosing the remaining 3 members After choosing 2 faculty
members, there are 3 faculty members left and 10 students remaining. We need
to choose 3 members from these 13 people. So, the number of ways to choose
the remaining 3 members is (13
3)=13!
3!(13−3)! = 286.
Step 3: Total number of committees To calculate the total number of
committees, we multiply the results from Step 1 and Step 2. Total number of
committees = 10 ×286 = 2860.
Therefore, there are 2,860 different committees that can be formed with at
least 2 faculty members on the committee.
Question 3
Question
Suppose you have a standard deck of 52 playing cards (13 cards in each of 4
suits: hearts, diamonds, clubs, and spades). How many ways can you choose a
5-card hand such that it contains exactly 3 hearts and 2 spades?
Solution
Step 1: Determine the number of ways to choose 3 hearts from the 13 available.
There are (13
3)ways to choose 3 hearts from the 13 available.
Step 2: Determine the number of ways to choose 2 spades from the 13
available. There are (13
2)ways to choose 2 spades from the 13 available.
Step 3: Multiply the results from Step 1 and Step 2 to find the total number
of ways to choose 3 hearts and 2 spades. Total ways = (13
3)×(13
2).
Step 4: Calculate the total number of ways. (13
3)=13!
3!(13−3)! =13×12×11
3×2×1=
286.
(13
2)=13!
2!(13−2)! =13×12
2×1= 78.
2
Therefore, the total number of ways to choose 3 hearts and 2 spades is
286 ×78 = 22236. Thus, there are 22,236 ways to choose a 5-card hand with
exactly 3 hearts and 2 spades.
Question 4
Question
In a group of 10 people, how many ways can we choose a committee of 3 people?
Solution
Step 1: To find the number of ways to choose a committee of 3 people from a
group of 10 people, we use the combination formula which is given by:
(n
r)=n!
r!(n−r)!
where nis the total number of people and ris the number of people we want
to choose.
Step 2: Substituting n= 10 and r= 3 into the formula, we get:
(10
3)=10!
3!(10 −3)!
Step 3: Calculating the factorials:
10! = 10 ×9×8×7×6×5×4×3×2×1 = 3628800
3! = 3 ×2×1 = 6
7! = 7 ×6×5×4×3×2×1 = 5040
Step 4: Substitute the factorials back into the formula:
(10
3)=3628800
6×5040
Step 5: Simplifying the expression:
(10
3)=3628800
30240 = 120
Therefore, there are 120 ways to choose a committee of 3 people from a
group of 10 people.
3
Question 5
Question
In a group of 10 students, how many ways can we choose a committee of 4 stu-
dents if two particular students, Alex and Beth, refuse to serve on the committee
together?
Solution
Step 1: Calculate the total number of ways to choose a committee of 4 out
of 10 students. There are 10 students to choose from, and we want to select
a committee of 4. This is a combination problem, so we use the formula for
combinations:
Total ways to choose a committee of 4 =(10
4)=10!
4!(10 −4)! =10 ×9×8×7
4×3×2×1= 210
Step 2: Calculate the number of ways to choose a committee of 4 when
both Alex and Beth are in the committee. Since Alex and Beth must be in the
committee together, we treat them as one entity. Then, we have 9 entities (Alex
and Beth as one, and the remaining 8 students) to choose the other 2 committee
members from. This is also a combination problem:
Ways to choose a committee with Alex and Beth =(9
2)=9!
2!(9 −2)! =9×8
2×1= 36
Step 3: Subtract the number of ways to choose a committee with Alex and
Beth from the total number of ways.
Number of ways to choose a committee without Alex and Beth =Total ways−Ways with Alex and Beth = 210−36 = 174
Therefore, there are 174 ways to choose a committee of 4 students from a
group of 10 students if Alex and Beth refuse to serve on the committee together.
Question 6
Question
A committee of 5 people is to be formed from a group of 8 men and 6 women. If
the committee must consist of at least 2 women, how many different committees
can be formed?
Solution
Step 1: Find the number of ways to choose a committee with exactly 2 women.
Since there must be at least 2 women in the committee, we will consider the
cases of choosing 2, 3, 4, or 5 women.
4
For choosing 2 women:
(6
2)ways to choose 2 women×(8
3)ways to choose 3 men = 15×56 = 840 ways
Step 2: Find the number of ways to choose a committee with exactly 3
women.
(6
3)ways to choose 3 women×(8
2)ways to choose 2 men = 20×28 = 560 ways
Step 3: Find the number of ways to choose a committee with exactly 4
women.
(6
4)ways to choose 4 women×(8
1)ways to choose 1 man = 15×8 = 120 ways
Step 4: Find the number of ways to choose a committee with all 5 women.
(6
5)ways to choose 5 women = 6 ways
Step 5: Calculate the total number of ways to form the committee.
Total ways = 840 + 560 + 120 + 6 = 1526
Therefore, there are 1526 different committees that can be formed with at
least 2 women.
Question 7
Question
A company has 8 different job positions available, and there are 12 applicants.
How many ways can the company hire 4 applicants for the job positions?
Solution
Step 1: To solve this problem, we will use the combination formula, which is
given by:
(n
k)=n!
k!(n−k)!
where nis the total number of items, kis the number of items to choose.
Step 2: First, we need to determine the total number of ways to select 4
applicants from the 12 that have applied. This is given by:
(12
4)=12!
4!(12 −4)! =12!
4!8!
5
Step 3: Simplifying the formula, we get:
(12
4)=12 ×11 ×10 ×9
4×3×2×1= 495
Therefore, there are 495 ways for the company to hire 4 applicants for the 8
job positions.
Question 8
Question
In a certain game, players must match 5 numbers (chosen from 1 to 50) to win
the jackpot. How many different ways can a player choose their numbers if the
order in which the numbers are selected does not matter?
Solution
Step 1: Calculate the number of ways to choose 5 numbers from 50 without
regard to order. This is a combination problem, which can be calculated using
the formula:
Number of ways =(n
k)=n!
k!(n−k)!
where nis the total number of items (50 in this case) and kis the number of
items to choose (5 in this case).
Step 2: Substituting n= 50 and k= 5 into the formula, we get:
(50
5)=50!
5!(50 −5)!
Step 3: Simplify the expression:
(50
5)=50 ×49 ×48 ×47 ×46
5×4×3×2×1
(50
5)=2118760
120
(50
5)= 176,160
Therefore, there are 176,160 different ways for a player to choose their 5
numbers for the game.
Question 9
Question
In a class of 30 students, how many ways can a committee of 5 students be
selected if Alex and Beth refuse to serve together?
6
Solution
Step 1: Find the total number of ways to select a committee of 5 students from
a class of 30 students. Step 2: Find the number of ways Alex and Beth can be
in the same committee. Step 3: Subtract the number of ways Alex and Beth
can be in the same committee from the total number of ways to get the final
answer.
Step 1: The total number of ways to select 5 students out of 30 is given by
the combination formula: C(30,5) = 30!
5!(30−5)! .
Step 2: If Alex and Beth are to be in the same committee, treat them as
one person. So the number of ways they can be in the same committee is the
number of ways to select 4 people out of 29 (treating Alex and Beth as one
person): C(29,4) = 29!
4!(29−4)! .
Step 3: The number of ways to select a committee where Alex and Beth
do not serve together is given by:
C(30,5) −C(29,4) = 30!
5!(30 −5)! −29!
4!(29 −4)!
Question 10
Question
A committee of 5 students is to be randomly selected from a group of 10 students.
How many different committees can be formed if one particular student must
be included?
Solution
Step 1: Since one particular student must be included in the committee, we
have already chosen 1 student. So, we need to choose 4 more students from the
remaining 9 students.
Step 2: To find the number of ways to select 4 students out of 9, we use the
combination formula (n
r)=n!
r!(n−r)! , where nis the total number of students
and ris the number of students we want to choose.
Step 3: Plugging in n= 9 and r= 4 into the formula, we get:
(9
4)=9!
4!(9 −4)!
Step 4: Calculating the factorials:
(9
4)=9×8×7×6×5
4×3×2×1
Step 5: Simplifying further:
(9
4)=30240
24
7
(9
4)= 1260
Step 6: Therefore, there are 1260 different committees that can be formed
if one particular student must be included.
Question 11
Question
In a class of 30 students, 15 are male and 15 are female. A committee of 5
students is to be formed randomly from the class. What is the probability that
the committee consists of 3 males and 2 females?
Solution
Step 1: Determine the total number of ways to form a committee of 5 students
from a class of 30 students. Given that there are 15 male students and 15 female
students in the class, the total number of ways to form a committee of 5 students
can be calculated using combinations:
Total ways =(30
5)=30!
5!(30 −5)! =30!
5!25! = 142506
Step 2: Determine the number of ways to form a committee consisting of 3
males and 2 females. To form a committee of 3 males and 2 females, we need to
choose 3 males out of 15 and 2 females out of 15. The number of ways to choose
3 males from 15 is given by (15
3)and the number of ways to choose 2 females
from 15 is (15
2). Therefore, the number of ways to form a committee consisting
of 3 males and 2 females is:
(15
3)×(15
2)=15!
3!12! ×15!
2!13! = 455 ×105 = 47850
Step 3: Calculate the probability of forming a committee consisting of 3
males and 2 females. The probability is given by the ratio of the number of
favorable outcomes (committees with 3 males and 2 females) to the total number
of outcomes (all possible committees):
Probability =Number of ways with 3 males and 2 females
Total number of ways =47850
142506 ≈0.3359
Therefore, the probability that the committee consists of 3 males and 2 females
is approximately 0.3359.
8
Question 12
Question
A company is assigning 5 different positions to 8 employees. How many different
ways can this be done if one of the employees, Alex, must be in a specific
position?
Solution
Step 1: Since Alex must be in a specific position, we will consider Alex already
assigned to that position. Therefore, we only need to assign the remaining 4
positions to the remaining 7 employees.
Step 2: The number of ways to assign the remaining 4 positions to the 7
employees is given by the number of permutations of 7 items taken 4 at a time.
Step 3: The number of permutations of 7 items taken 4 at a time is given
by: 7!
(7 −4)! =7!
3!
Step 4: Calculating the permutations:
7!
3! =7×6×5×4×3!
3×2×1= 7 ×6×5×4 = 840
Therefore, there are 840 different ways to assign 5 different positions to 8
employees if Alex must be in a specific position.
Question 13
Question
A committee of 6 people is to be formed from a group of 10 men and 7 women.
Find the probability that the committee consists of 3 men and 3 women.
Solution
Step 1: Find the total number of ways to form a committee of 6 people. To
form a committee of 6 people from a group of 17 (10 men and 7 women), we
can use the formula for combinations:
Total ways =(17
6)=17!
6!(17 −6)!
Step 2: Find the number of ways to choose 3 men from 10. Using com-
binations, we can find the number of ways to choose 3 men from a group of
10:
Ways to choose 3 men =(10
3)=10!
3!(10 −3)!
9
Step 3: Find the number of ways to choose 3 women from 7. Similar to the
previous step, we can calculate the number of ways to choose 3 women from 7:
Ways to choose 3 women =(7
3)=7!
3!(7 −3)!
Step 4: Find the total number of ways to form a committee with 3 men and
3 women. Since we want a committee with 3 men and 3 women, we need to
multiply the number of ways to choose 3 men by the number of ways to choose
3 women:
Total ways for 3 men and 3 women =(10
3)×(7
3)
Step 5: Calculate the probability. The probability that the committee con-
sists of 3 men and 3 women is the ratio of the total number of ways to form
a committee with 3 men and 3 women to the total number of ways to form a
committee of 6 people:
Probability =(10
3)×(7
3)
(17
6)=
10!
3!(10−3)! ×7!
3!(7−3)!
17!
6!(17−6)!
Question 14
Question
In a group of 10 people (5 men and 5 women), how many different ways can a
committee of 3 people be formed if there must be at least 1 man and 1 woman
on the committee?
Solution
Step 1: First, we will find the total number of ways to form a committee of 3
people from the group of 10 people. This can be done using the combination
formula (n
r)=n!
r!(n−r)! , where nis the total number of people and ris the
number of people we want to select. In this case, n= 10 and r= 3.
Total ways to form committee =(10
3)=10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
Step 2: Next, we will find the number of ways to form a committee with 3
people where all 3 members are men or all 3 members are women. Since there
are 5 men and 5 women, these scenarios are mutually exclusive.
• Ways to select 3 men out of 5: (5
3)=5!
3!(5−3)! = 10 ways
• Ways to select 3 women out of 5: (5
3)=5!
3!(5−3)! = 10 ways
10
Step 3: However, these scenarios do not satisfy the condition of having at
least 1 man and 1 woman on the committee. Therefore, we will find the total
number of ways to select a committee with either 2 men and 1 woman or 1 man
and 2 women.
• Ways to select 2 men out of 5 and 1 woman out of 5: (5
2)×(5
1)= 10×5 = 50
ways
• Ways to select 1 man out of 5 and 2 women out of 5: (5
1)×(5
2)= 5×10 = 50
ways
Step 4: Finally, we calculate the total number of ways to form a committee
of 3 people with at least 1 man and 1 woman by summing up the possibilities
from Step 2 and Step 3.
Total ways with at least 1 man and 1 woman = 10 + 10 + 50 + 50 = 120
Therefore, there are 120 different ways a committee of 3 people can be formed
from a group of 10 people (5 men and 5 women) with at least 1 man and 1 woman
on the committee.
Question 15
Question
In a group of 10 people, how many ways can we choose a committee of 4 people
if 2 of the members, Alice and Bob, refuse to serve together on the committee?
Solution
Step 1: Find the total number of ways to choose a committee of 4 people without
any restrictions. The total number of ways to choose 4 people from a group of
10 is given by the combination formula: (10
4)=10!
4!(10−4)! .
Step 2: Find the number of ways to choose a committee where Alice and
Bob are both in the committee. Let’s assume that Alice and Bob will work
together in the committee. Then, we need to choose 2 more people from the
remaining 8 people. This can be done in (8
2)ways.
Step 3: Subtract the number of ways with Alice and Bob from the total to get
the final result. Now, we need to subtract the number of ways where Alice and
Bob are both in the committee from the total number of ways: Total number
of ways −Number of ways with Alice and Bob in the committee =(10
4)−(8
2).
Therefore, the number of ways to choose a committee of 4 people where
Alice and Bob refuse to serve together is (10
4)−(8
2).
11
Question 16
Question
A committee of 4 people is to be formed from a group of 10 students. If 4 of
the students are math majors and 6 are engineering majors, how many different
committees can be formed if each committee must have at least 2 math majors
and at least 1 engineering major?
Solution
Step 1: Calculate the number of committees with exactly 2 math majors and 2
engineering majors.
There are (4
2)= 6 ways to choose 2 math majors from the 4 available, and
(6
2)= 15 ways to choose 2 engineering majors from the 6 available. Multiply
these together to get 6×15 = 90.
Step 2: Calculate the number of committees with exactly 3 math majors
and 1 engineering major.
There are (4
3)= 4 ways to choose 3 math majors from the 4 available, and
(6
1)= 6 ways to choose 1 engineering major from the 6 available. Multiply these
together to get 4×6 = 24.
Step 3: Calculate the number of committees with 4 math majors.
There is only 1 way to choose all 4 math majors from the 4 available.
Step 4: Add the results from Steps 1, 2, and 3 to find the total number of
valid committees.
90 + 24 + 1 = 115
Therefore, there are 115 different committees that can be formed where each
committee must have at least 2 math majors and at least 1 engineering major.
Question 17
Question
A committee of 5 people is to be formed from a group of 8 men and 7 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: Calculate the number of committees with exactly 2 men and 3 women.
To select 2 men from 8 men and 3 women from 7 women, we use the combination
formula: (8
2)·(7
3)=8!
2!(8 −2)! ·7!
3!(7 −3)! = 28 ·35 = 980
Step 2: Calculate the number of committees with exactly 3 men and 2
women. To select 3 men from 8 men and 2 women from 7 women, we use the
12
combination formula:
(8
3)·(7
2)=8!
3!(8 −3)! ·7!
2!(7 −2)! = 56 ·21 = 1176
Step 3: Calculate the number of committees with 4 men and 1 woman. To
select 4 men from 8 men and 1 woman from 7 women, we use the combination
formula: (8
4)·(7
1)=8!
4!(8 −4)! ·7!
1!(7 −1)! = 70 ·7 = 490
Step 4: Calculate the total number of committees with at least 2 men and 2
women. Add the results from Step 1, Step 2, and Step 3: Total = 980 + 1176
+ 490 = 2646
Therefore, there are 2646 different committees that can be formed with at
least 2 men and at least 2 women from the group.
Question 18
Question
In how many ways can a committee of 3 people be selected from a group of 7
people to serve as president, vice-president, and treasurer?
Solution
Step 1: To find the number of ways to select the committee, we first need to
determine the number of ways to select the president, then the vice-president,
and finally the treasurer.
Step 2: There are 7 choices for the president, since any of the 7 people can
be selected.
Step 3: Once the president is chosen, there are 6 remaining people to choose
from for the vice-president.
Step 4: After selecting the president and vice-president, there are 5 remain-
ing people to choose from for the treasurer.
Step 5: Therefore, the total number of ways to select the committee is the
product of the number of choices for each position:
7×6×5 = 210
Step 6: So, there are 210 ways to select a committee of 3 people from a
group of 7 to serve as president, vice-president, and treasurer.
Question 19
Question
In a group of 10 students, how many ways are there to select a committee of 3
students and a chairperson from the group?
13
Solution
Step 1: To find the number of ways to select a committee of 3 students from
10, we will use the combination formula (n
r)=n!
r!(n−r)! where nis the total
number of students and ris the number of students we want to select.
Step 2: Substitute n= 10 and r= 3 into the combination formula:
(10
3)=10!
3!(10 −3)!
Step 3: Simplify the factorial terms in the denominator:
(10
3)=10!
3!7!
Step 4: Calculate the factorials:
(10
3)=10 ×9×8×7!
3×2×1×7!
Step 5: Simplify further to find the number of ways to select the committee:
(10
3)=10 ×9×8
3×2×1= 120
Step 6: After selecting the 3 students for the committee, we need to choose
a chairperson from those 3 students. There are 3 ways to choose a chairperson
from the committee.
Step 7: Multiply the number of ways to select the committee by the number
of ways to choose a chairperson to find the total number of ways:
120 ×3 = 360
Therefore, there are 360 ways to select a committee of 3 students and a
chairperson from a group of 10 students.
Question 20
Question
A committee of 5 people is to be formed from a group of 10 women and 8 men.
How many different committees can be formed if the committee must have at
least 2 women and at least 2 men?
14
Solution
Step 1: Calculate the total number of ways to form a committee with 5 people:
There are 10 women and 8 men to choose from, so the total number of ways to
choose 5 people without any restrictions is (18
5)= 8,568.
Step 2: Calculate the number of ways to form a committee with at least 2
women and at least 2 men: To form a committee with at least 2 women and at
least 2 men, we can have the following cases: - 2 women and 3 men - 3 women
and 2 men - 4 women and 1 man - 5 women and 0 men
For the case of 2 women and 3 men: Choose 2 women from 10: (10
2)ways
Choose 3 men from 8: (8
3)ways Multiply the two to get the total number of
ways: (10
2)×(8
3)= 2,520 ways
For the case of 3 women and 2 men: Choose 3 women from 10: (10
3)ways
Choose 2 men from 8: (8
2)ways Multiply the two to get the total number of
ways: (10
3)×(8
2)= 2,520 ways
For the case of 4 women and 1 man: Choose 4 women from 10: (10
4)ways
Choose 1 man from 8: (8
1)ways Multiply the two to get the total number of
ways: (10
4)×(8
1)= 560 ways
For the case of 5 women and 0 men: Choose 5 women from 10: (10
5)ways
There are no men to choose, so only 1 way The total number of ways: (10
5)= 252
ways
Step 3: Add up the number of ways for each case to get the total number
of ways to form the committee with at least 2 women and at least 2 men:
2,520 + 2,520 + 560 + 252 = 5,852 ways
Therefore, there are 5,852 different committees that can be formed with at
least 2 women and at least 2 men.
Question 21
Question
In a group of 10 students, how many ways can we choose a committee of 4
students?
Solution
Step 1: To solve this problem, we will use the formula for combinations, which
is given by:
C(n, k) = n!
k!(n−k)!
where nis the total number of students and kis the number of students we
want to choose for the committee.
15
Step 2: Substituting n= 10 and k= 4 into the formula, we get:
C(10,4) = 10!
4!(10 −4)!
Step 3: Calculating the factorials:
10! = 10 ×9×8×7×6×5×4×3×2×1
4! = 4 ×3×2×1
6! = 6 ×5×4×3×2×1
Step 4: Substituting the factorials into the formula:
C(10,4) = 10 ×9×8×7×6×5×4×3×2×1
4×3×2×1×6×5×4×3×2×1
Step 5: Simplifying the expression:
C(10,4) = 10 ×9×8×7
4×3×2×1= 210
Step 6: Therefore, there are 210 ways to choose a committee of 4 students
from a group of 10.
Question 22
Question
A committee of 5 people is to be formed from a group of 8 men and 6 women.
If the committee must consist of 3 men and 2 women, how many different
committees can be formed?
Solution
Step 1: Calculate the number of ways to choose 3 men from 8. Step 2: Calculate
the number of ways to choose 2 women from 6. Step 3: Multiply the results of
Step 1 and Step 2 to find the total number of different committees that can be
formed.
Step 1: Number of ways to choose 3 men from 8.
C(8,3) = 8!
3!(8 −3)! =8×7×6
3×2×1= 56
Step 2: Number of ways to choose 2 women from 6.
C(6,2) = 6!
2!(6 −2)! =6×5
2×1= 15
Step 3: Total number of different committees that can be formed.
56 ×15 = 840
Therefore, there are 840 different committees that can be formed with 3 men
and 2 women.
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Question 23
Question
In a group of 10 students, how many ways can we select a committee of 3
students, where one student is designated as the president, one as the vice
president, and one as the secretary?
Solution
To solve this problem, we can break it down into two steps: Step 1: Calculate
the number of ways to select a committee of 3 students. Step 2: After selecting
the committee, assign the roles of president, vice president, and secretary to the
selected students.
Step 1: To select a committee of 3 students out of 10, we can use the
combination formula:
Number of ways =(10
3)=10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
So, there are 120 ways to select a committee of 3 students.
Step 2: After selecting the committee, we need to assign the roles of pres-
ident, vice president, and secretary. For the president, there are 3 students to
choose from the committee. After selecting the president, there are 2 students
left for the vice president role. Finally, the last remaining student will be the
secretary.
Therefore, the total number of ways to select the committee with assigned
roles is:
120 ×3×2×1 = 720
Therefore, there are 720 ways to select a committee of 3 students, where one
is the president, one is the vice president, and one is the secretary.
Question 24
Question
A committee of 4 people is to be formed from a group of 10 people. How many
different committees can be formed if 2 specific people must be included in the
committee?
Solution
Step 1: Choose the 2 specific people to be included in the committee. There
are (10
2)ways to do this.
Step 2: Choose the remaining 2 people from the remaining 8 people. There
are (8
2)ways to do this.
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Step 3: Multiply the results from Step 1 and Step 2 to find the total number
of different committees that can be formed.
Therefore, the total number of different committees that can be formed is
given by:
(10
2)×(8
2)=10!
2!8! ×8!
2!6!
=10 ×9
2×1×8×7
2×1= 45 ×28
= 1260
So, there are 1260 different committees that can be formed if 2 specific people
must be included in the committee.
Question 25
Question
In a group of 10 students, how many ways can we choose a committee of 3
students to serve on the student council if one student, Alice, refuses to serve
with Bob due to a conflict?
Solution
Step 1: First, let’s find the total number of ways to choose a committee of 3
students from a group of 10 students. This is given by the combination formula
(n
r)=n!
r!(n−r)! , where nis the total number of students and ris the number of
students we want to choose for the committee. In this case, n= 10 and r= 3.
Thus, the total number of ways to choose a committee of 3 students from 10 is:
(10
3)=10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
Step 2: Next, let’s find the number of ways to choose a committee of 3
students from the 9 remaining students after Alice refuses to serve with Bob.
This is given by the combination formula. So, the number of ways to choose a
committee of 3 students from 9 is:
(9
3)=9!
3!(9 −3)! =9×8×7
3×2×1= 84
Step 3: Now, to find the number of ways to choose a committee of 3 stu-
dents that does not include Alice and Bob together, we need to subtract the
number of committees where Alice and Bob are together from the total number
of committees. There are 84 ways to choose a committee of 3 students from the
9 remaining students, and Alice and Bob can be together in (2
1)×(7
1)= 14 ways
(either Alice and Bob or Bob and Alice). Therefore, the number of ways to
choose a committee of 3 students that does not include Alice and Bob together
is 84 −14 = 70 ways.
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