MATH 350 - DISCRETE
MATHEMATICS - Permutations and
combinations
Question Bank - Set 4
Liberty University
Question 1
Question
In how many ways can 6 people be seated around a circular table if two particular
people refuse to sit next to each other?
Solution
Step 1: First, we find the total number of ways to seat 6 people around a circular
table without any restrictions. There are 6! ways to arrange 6 people in a line.
However, since the table is circular, we divide by 6 to account for the different
starting points, giving us 6!
6= 5! ways to seat the people.
Step 2: Next, we find the number of ways the two particular people can sit
next to each other. Consider the two people as a single entity. Now, we have 5
entities to arrange in a line: 1 entity from the two particular people, 4 entities
from the remaining people. There are 5! ways to arrange these entities.
Step 3: Now, we subtract the number of ways the two particular people can
sit next to each other from the total number of ways to seat 6 people. Number
of ways the two particular people can sit next to each other: 5! Total number
of ways to seat 6 people: 5! Number of ways the two particular people refuse to
sit next to each other: 5! −5! = 0
Therefore, there are zero ways to seat 6 people around a circular table if
the two particular people refuse to sit next to each other.
Question 2
Question
A committee of 4 people is to be formed from a group of 7 women and 5 men.
If at least one man must be included in the committee, how many different
committees can be formed?
Solution
Step 1: Calculate the total number of committees without any restrictions.
First, we calculate the total number of committees that can be formed with-
out any restrictions. This can be done using combinations formula:
Total number of committees = 12
4=12!
4!(12 −4)! =12!
4!8! = 495
So, there are 495 different committees that can be formed without any re-
strictions.
Step 2: Calculate the number of committees with only women.
Next, we calculate the number of committees that consist only of women.
This can be done using combinations formula:
Number of committees with only women = 7
4=7!
4!(7 −4)! =7!
4!3! = 35
Step 3: Calculate the number of committees with no men.
Now, we calculate the number of committees with no men. This can be
done by subtracting the number of committees with only women from the total
number of committees:
Number of committees with no men = Total number of committees−Number of committees with only women = 495−35 = 460
Step 4: Calculate the number of committees with at least one man.
Since we want to find the number of committees with at least one man, we
can subtract the number of committees with no men from the total number of
committees:
Number of committees with at least one man = Total number of committees−Number of committees with no men = 495−460 = 35
Therefore, there are 35 different committees that can be formed with at least
one man included.
2
Question 3
Question
In a group of 12 students, how many ways can we choose a committee of 5
students if 2 particular students must be included?
Solution
Step 1: First, we select the 2 particular students who must be included in the
committee. This can be done in 12
2ways.
Step 2: After selecting the 2 particular students, we need to select 3 more
students to complete the committee. Since we have already selected 2 students,
we are left with 10 students to choose from. We can select 3 more students in
10
3ways.
Step 3: To find the total number of ways to choose the committee of 5
students, we multiply the number of ways from Step 1 and Step 2. Therefore,
the total number of ways to choose a committee of 5 students with 2 particular
students included is 12
2×10
3.
Solving this expression, we get: 12
2=12!
2!(12−2)! =12×11
2×1= 66
10
3=10!
3!(10−3)! =10×9×8
3×2×1= 120
Therefore, the total number of ways to choose the committee is: 66 ×120 =
7920
Hence, there are 7920 ways to choose a committee of 5 students with 2
particular students included from a group of 12 students.
Question 4
Question
In a group of 10 people, how many ways can we choose a president, a vice
president, and a treasurer if no person can hold more than one position?
Solution
Step 1: To find the number of ways to choose a president, we simply have 10
choices.
Step 2: After choosing a president, there are 9 remaining people to choose
from for the position of vice president.
Step 3: Once the president and vice president have been chosen, there are 8
remaining people to choose from for the position of treasurer.
Step 4: To find the total number of ways to choose a president, a vice
president, and a treasurer, we multiply the number of choices at each step:
10 ×9×8 = 720
3
Therefore, there are 720 ways to choose a president, a vice president, and
a treasurer in a group of 10 people.
Question 5
Question
A committee of 5 people is to be formed from a group of 10 women and 8 men. If
the committee must consist of at least 3 women, how many different committees
can be formed?
Solution
Step 1: Calculate the number of committees with exactly 3 women: There are
10
3ways to choose 3 women from the 10 available women, and 8
2ways to
choose 2 men from the 8 available men. Therefore, there are 10
3×8
2ways to
form a committee with exactly 3 women.
Step 2: Calculate the number of committees with exactly 4 women: There
are 10
4ways to choose 4 women from the 10 available women, and 8
1ways
to choose 1 man from the 8 available men. Hence, there are 10
4×8
1ways to
create a committee with exactly 4 women.
Step 3: Calculate the number of committees with all 5 women: There is only
10
5way to choose all 5 women from the 10 available women.
Step 4: Add the results from Step 1, Step 2, and Step 3 to find the total
number of valid committees:
10
3×8
2+10
4×8
1+10
5
Calculating the above expression will give us the total number of different
committees that can be formed in this scenario.
Question 6
Question
In a group of 10 people, how many ways can we choose a committee of 4 people
such that 2 of them are women and 2 are men?
Solution
To solve this problem, we can break it down into two steps: selecting the 2
women from a group of 5 women and selecting the 2 men from a group of 5
men.
Step 1: Selecting 2 women from 5 women There are 5
2ways to select 2
women from a group of 5 women.
4
Step 2: Selecting 2 men from 5 men Similarly, there are 5
2ways to select 2
men from a group of 5 men.
To find the total number of ways to choose a committee of 4 people (2 women
and 2 men), we multiply the number of ways from Step 1 and Step 2.
Therefore, the total number of ways to choose the committee is:
5
2×5
2=5!
2!3! ×5!
2!3! =5×4
2×5×4
2= 10 ×10 = 100
So, there are 100 ways to choose a committee of 4 people consisting of 2
women and 2 men from a group of 10 people.
Question 7
Question
A committee of 5 people is to be formed from a group of 10 men and 8 women. If
the committee must consist of at least 2 women, how many different committees
can be formed?
Solution
Step 1: First, we calculate the number of ways to form a committee with exactly
2 women. There are 8 women to choose 2 from, and 10 men to choose 3 from.
Thus, the number of committees with exactly 2 women is 8
2×10
3.
Step 2: Next, we calculate the number of ways to form a committee with
exactly 3 women. There are 8 women to choose 3 from, and 10 men to choose
2 from. Thus, the number of committees with exactly 3 women is 8
3×10
2.
Step 3: Now, we calculate the total number of committees by adding the
number of committees with exactly 2 women and exactly 3 women. Total num-
ber of committees = 8
2×10
3+8
3×10
2.
Step 4: Plug in the values and simplify. Total number of committees =
8
2×10
3+8
3×10
2
Total number of committees = 8!
2!6! ×10!
3!7! +8!
3!5! ×10!
2!8!
Total number of committees = 28 ×120 + 56 ×45
Total number of committees = 3360 + 2520
Total number of committees = 5880.
Therefore, there are 5880 different committees that can be formed with at
least 2 women.
Question 8
Question
In how many ways can 5 distinct books be arranged on a bookshelf such that
two particular books must always be next to each other?
5
Solution
Step 1: Treat the two particular books as a single entity. Step 2: Consider the
arrangement as a combination problem.
There are two particular books that must be next to each other. Let’s treat
these two books as a single entity. So, we now have 4 entities to arrange: AB,
C, D, and E. Step 3: We have 4 entities to arrange in total: AB, C, D, and E.
Step 4: There are 4! ways to arrange these entities, but within the AB entity,
the books can be arranged in 2! ways. Step 5: Therefore, the total number of
ways to arrange the books is 4! ×2! = 24 ×2 = 48 ways.
Hence, there are 48 ways to arrange the 5 distinct books on the bookshelf
such that two particular books must always be next to each other.
Question 9
Question
A committee of 6 people is to be formed from a group of 10 men and 5 women. If
the committee must consist of at least 2 men and 2 women, how many different
committees can be formed?
Solution
Step 1: Calculate the number of ways to choose 2 men out of 10. There are 10
2
ways to choose 2 men from the group of 10 men.
Step 2: Calculate the number of ways to choose 2 women out of 5. There
are 5
2ways to choose 2 women from the group of 5 women.
Step 3: Calculate the number of ways to choose the remaining 2 people (1
man and 1 woman) out of the remaining 8 people (8=10-2 men; 8=5-2 women).
There are 8
2ways to choose the remaining 1 man and 1 woman.
Step 4: Calculate the total number of ways to form the committee. The total
number of ways to form the committee is the product of the ways to choose 2
men, 2 women, and 1 man and 1 woman: Total number of ways = 10
2×5
2×8
2
Calculating this expression gives: Total number of ways = 10!
2!8! ×5!
2!3! ×8!
2!6!
Total number of ways = 45 ×10 ×28 Total number of ways = 12600
Therefore, there are 12,600 different committees that can be formed.
Question 10
Question
A committee of 5 people is to be formed from a group of 10 math students and
6 computer science students. If the committee must consist of at least 2 math
students and 2 computer science students, how many different committees can
be formed?
6
Solution
Step 1: Calculate the number of ways to choose 2 math students out of 10.
There are 10
2ways to choose 2 math students.
Step 2: Calculate the number of ways to choose 2 computer science students
out of 6. There are 6
2ways to choose 2 computer science students.
Step 3: Calculate the number of ways to choose the remaining committee
member. There are 8 students remaining (4 math students and 4 computer
science students). We can choose one person from this group in 8
1ways.
Step 4: Calculate the total number of committees that can be formed. The
total number of committees can be formed by multiplying the results from Step
1, Step 2, and Step 3:
10
2×6
2×8
1= 45 ×15 ×8 = 5400.
Therefore, there are 5400 different committees that can be formed.
Question 11
Question
A committee of 5 people is to be formed from a group of 10 people. If 3 of the
people are friends and refuse to serve on the committee together, how many
different committees can be formed?
Solution
Step 1: To find the number of ways to form a committee, we first calculate the
total number of ways to choose 5 people from a group of 10. This is represented
by 10
5.
Total number of ways to choose 5 people from 10 = 10
5=10!
5!(10 −5)! = 252
Step 2: Now, let’s calculate the number of ways to form a committee where
the 3 friends cannot serve together. - We break this problem into cases: Case 1:
0 friends are on the committee - The number of ways to choose 5 people from
the remaining 7 (10-3 friends) is 7
5. Case 2: 1 friend is on the committee - The
number of ways to choose 4 people from the remaining 7 (10-3 friends and 1
friend already on the committee) is 7
4. Case 3: 2 friends are on the committee
- The number of ways to choose 3 people from the remaining 7 (10-3 friends and
2 friends already on the committee) is 7
3. - We sum the results from all these
cases to get the total number of committees that can be formed.
Number of committees without the friends together = 7
5+7
4+7
3
7
=7!
5!(7 −5)! +7!
4!(7 −4)! +7!
3!(7 −3)!
= 21 + 35 + 35 = 91
Therefore, there are 91 different committees that can be formed from the
group of 10 people with the condition that 3 friends cannot serve together.
Question 12
Question
A committee of 4 people is to be formed from a group of 8 men and 6 women.
In how many ways can the committee be formed if it must consist of 2 men and
2 women?
Solution
Step 1: Calculate the number of ways to choose 2 men from 8. Step 2: Calculate
the number of ways to choose 2 women from 6. Step 3: Multiply the results
from Step 1 and Step 2 to get the total number of ways to form the committee.
Step 1: Choosing 2 men from 8 can be done using a combination. The
number of ways to choose 2 men from 8 is given by
8
2=8!
2!(8 −2)! =8×7
2×1= 28.
Step 2: Choosing 2 women from 6 can be done using a combination. The
number of ways to choose 2 women from 6 is given by
6
2=6!
2!(6 −2)! =6×5
2×1= 15.
Step 3: The total number of ways to form the committee consisting of 2 men
and 2 women is the product of the results from Step 1 and Step 2,
28 ×15 = 420.
Therefore, there are 420 ways to form the committee with 2 men and 2
women from the group of 8 men and 6 women.
Question 13
Question
A group of 6 people are going on a road trip and plan to stop at 4 different
cities along the way. If each person can visit any city, how many different ways
can they split up into groups to visit the cities?
8
Solution
We can solve this problem using combinations.
Step 1: Since the order in which the cities are visited does not matter, we
will use the concept of combinations. The number of ways to split 6 people into
groups to visit 4 cities is equal to the number of ways to distribute the people
among the cities with repetition allowed.
Step 2: We can visualize this problem as placing 3 dividers among the 6
people, representing the 4 different cities. The number of ways to place the
dividers determines the number of ways to split the people into groups.
Step 3: Thus, the total number of ways to split the 6 people into groups to
visit the 4 cities is given by the formula for combinations with repetition:
n+r−1
r=6+4−1
4=9
4=9×8×7×6
4×3×2×1= 126
Step 4: Therefore, there are 126 different ways for the group of 6 people to
split up into groups to visit the 4 cities along the road trip.
Question 14
Question
In how many ways can 5 men and 5 women be seated around a circular table if
each man must be seated next to at least one woman?
Solution
Step 1: First, we will fix one woman in a seat. Then, we can count the number
of ways the other 4 women can be seated around the table.
Step 2: There are 4 women remaining who can be seated in 4! ways.
Step 3: Now, we will arrange the 5 men among the 5 available slots between
the women.
Step 4: There are 5 slots between the women where the men can be seated.
The men can be arranged in 5! ways in these slots.
Step 5: Finally, since the table is circular, we must divide by 5 to account
for the rotations of the same arrangement.
Therefore, the total number of ways the 5 men and 5 women can be seated
around a circular table with each man seated next to at least one woman is
given by: 4! ×5!
5= 2880
9
Question 15
Question
A basketball team has 12 players. In how many ways can a starting lineup of
5 players be chosen if the lineup must include the team captain and the point
guard?
Solution
We can approach this problem by considering the positions of the players in the
starting lineup and filling them one by one. Step 1: Choose the team captain
(1 way) Step 2: Choose the point guard from the remaining 11 players (11
ways) Step 3: Choose the remaining 3 players from the remaining 10 players
(10 choices for the first remaining player, 9 for the second, and 8 for the third)
Therefore, the total number of ways to choose the starting lineup is:
1×11 ×10 ×9×8=7,920
So, there are 7,920 ways to choose the starting lineup.
Question 16
Question
In how many ways can you choose a committee of 4 people from a group of 10
people if 2 of the group members refuse to serve on the committee together?
Solution
Step 1: First, we calculate the total number of ways to choose a committee of 4
people from 10 people without any restrictions. This is a combination problem,
and can be calculated using the formula for combinations:
n
r=n!
r!(n−r)!
where nis the total number of people and ris the number of people we want
to choose.
For this problem, n= 10 and r= 4. So, the total number of ways to choose
a committee without restrictions is:
10
4=10!
4!(10 −4)!
=10 ×9×8×7
4×3×2×1
10
= 210
So, there are 210 ways to choose a committee of 4 people from a group of 10
people without restrictions.
Step 2: Next, we calculate the number of ways that the 2 group members
who refuse to serve on the committee together are chosen. Let’s call these 2
people A and B. There are 8 remaining people to choose from for the committee,
plus we need to choose 2 more people from the remaining 8.
This can be calculated by finding the number of ways to choose a committee
of 4 people from the 8 remaining people:
8
2=8!
2!(8 −2)!
=8×7
2×1
= 28
Step 3: Finally, we subtract the number of committees that include both A
and B from the total number of committees to get the final answer. This can
be calculated as:
Total ways −Ways with A and B together = 210 −28 = 182
Therefore, there are 182 ways to choose a committee of 4 people from a group
of 10 people if 2 group members refuse to serve on the committee together.
Question 17
Question
A committee of 5 people is to be chosen from a group of 10 men and 8 women.
If the committee must have at least 3 men, how many different committees can
be formed?
Solution
Step 1: Find the number of ways to choose a committee with exactly 3 men.
There are 10
3ways to choose 3 men from 10 men, and 8
2ways to choose 2
women from 8 women. Therefore, the number of ways to choose a committee
with exactly 3 men is:
10
3×8
2
Step 2: Find the number of ways to choose a committee with 4 men. There
are 10
4ways to choose 4 men from 10 men, and 8
1ways to choose 1 woman
11
from 8 women. Therefore, the number of ways to choose a committee with 4
men is: 10
4×8
1
Step 3: Find the number of ways to choose a committee with 5 men. There
are 10
5ways to choose 5 men from 10 men. Therefore, the number of ways to
choose a committee with 5 men is:
10
5
Step 4: Add the results from Steps 1, 2, and 3 to find the total number of
different committees that can be formed with at least 3 men. Total number of
committees = 10
3×8
2+10
4×8
1+10
5
Question 18
Question
A committee of 5 students is to be formed from a group of 10 students. If
3 of the students are math majors and 7 are computer science majors, how
many different committees can be formed if: 1. The committee must have 3
math majors and 2 computer science majors? 2. The committee can have any
combination of math and computer science majors?
Solution
1. To find the number of different committees that can be formed with 3 math
majors and 2 computer science majors, we calculate the number of ways to
choose 3 math majors out of 3 and 2 computer science majors out of 7.
Step 1: Choose 3 math majors out of 3 Since there are only 3 math
majors, the number of ways to choose 3 math majors out of 3 is 1.
Step 2: Choose 2 computer science majors out of 7 We need to
choose 2 computer science majors out of 7. This can be done in 7
2= 21 ways.
Step 3: Calculate the total number of committees The total number
of committees that can be formed is the product of the number of ways to
choose math majors and computer science majors. Therefore, the total number
of committees is 1 ∗21 = 21.
Therefore, there are 21 different committees that can be formed with 3 math
majors and 2 computer science majors.
2. To find the total number of different committees that can be formed
with any combination of math and computer science majors, we can consider
all possible cases separately and then add them up.
Case 1: All members are math majors In this case, we need to choose
5 math majors out of 3. However, since there are only 3 math majors, this case
is not possible.
12
Case 2: 4 math majors and 1 computer science major We need to
choose 4 math majors out of 3 and 1 computer science major out of 7. The
number of ways to do this is 3
4×7
1= 0.
Case 3: 3 math majors and 2 computer science majors We have
already calculated this case in part 1. The number of ways is 21.
Case 4: 2 math majors and 3 computer science majors We need to
choose 2 math majors out of 3 and 3 computer science majors out of 7. The
number of ways to do this is 3
2×7
3= 63.
Case 5: 1 math major and 4 computer science majors We need to
choose 1 math major out of 3 and 4 computer science majors out of 7. The
number of ways to do this is 3
1×7
4= 105.
Case 6: All members are computer science majors In this case, we
need to choose 5 computer science majors out of 7. The number of ways to do
this is 7
5= 21.
Step 4: Calculate the total number of committees The total number
of committees is the sum of all possible cases. Therefore, the total number of
different committees is 0 + 21 + 63 + 105 + 21 = 210.
Therefore, there are 210 different committees that can be formed with any
combination of math and computer science majors.
Question 19
Question
In a group of 10 people, how many ways are there to form a committee of 3
people with a president, a secretary, and a treasurer?
Solution
Step 1: First, choose the 3 people to be on the committee. Since the order in
which the people are chosen matters (president, secretary, treasurer), we use
permutations. There are 10 people to choose from, and we want to choose 3:
10P3.
Step 2: Calculate the number of ways to assign the positions of president,
secretary, and treasurer. Once the committee members have been chosen, we
need to assign the specific roles. There are 3 positions to fill and 3 people to
assign to them. Therefore, the number of ways to assign the positions is 3!.
Step 3: Multiply the results from Step 1 and Step 2. The total number of
ways to form the committee of 3 people with specific roles is given by:
10P3×3! = 10!
(10 −3)! ×3! = 10 ×9×8×7!
7! ×3×2×1 = 720 ×6 = 4320
Therefore, there are 4320 ways to form a committee of 3 people with a
president, a secretary, and a treasurer from a group of 10 people.
13
Question 20
Question
In a group of 12 people, how many ways can we select a committee of 4 people
if 2 particular people must always be included in the committee?
Solution
Step 1: First, we select the 2 particular people that must be included in the
committee. This can be done in 2
2= 1 way.
Step 2: Next, we need to select the remaining 2 people from the remaining
10 people. This can be done in 10
2ways.
Step 3: To find the total number of ways to form the committee, we multiply
the results of Step 1 and Step 2.
Therefore, the total number of ways to form a committee of 4 people with 2
particular people always included is 1 ×10
2= 1 ×45 = 45 ways.
Question 21
Question
In a group of 10 friends, how many ways can we choose a committee of 4 people
if one of the friends, Alice, insists on being part of the committee?
Solution
Step 1: Since one person, Alice, is already guaranteed to be part of the com-
mittee, we need to choose the remaining 3 people from the remaining 9 friends.
Step 2: To find the number of ways to choose these 3 people, we use the combi-
nation formula n
r=n!
r!(n−r)! . Step 3: Substituting n= 9 (9 remaining friends)
and r= 3 (to choose 3 people) into the formula:
9
3=9!
3!(9 −3)! =9×8×7
3×2×1= 84.
Therefore, there are 84 ways to choose a committee of 4 people with Alice being
one of them.
Question 22
Question
In how many ways can 5 students be seated in a row of 8 chairs such that two
particular students, Alice and Bob, must sit next to each other?
14
Solution
Step 1: Treat Alice and Bob as one entity. Step 2: Calculate the number of
ways to arrange the remaining entities. Step 3: There are 7 entities to arrange
(including the entity composed of Alice and Bob) in 7 seats.
Step 1: Treat Alice and Bob as one entity. Step 2: There are 7 entities in
total to be arranged. Step 3: The number of ways to arrange these 7 entities in
7 seats is 7!.
Therefore, the total number of ways for 5 students to be seated in a row of
8 chairs such that Alice and Bob must sit next to each other is 2 ·7!.
Question 23
Question
How many five-letter words can be formed using the letters in the word ”AL-
GEBRA” if no letter can be used more times than it appears in ”ALGEBRA”?
Solution
Step 1: Determine the number of ways to choose 5 letters from ”ALGEBRA”.
Since we cannot repeat any letters, we need to consider the unique letters in
”ALGEBRA”. The word ”ALGEBRA” has 6 different letters, so there are 6
5
ways to choose 5 letters from ”ALGEBRA”.
6
5=6!
5!(6−5)! = 6
Step 2: Calculate the number of ways to arrange the chosen 5 letters.
Since we have chosen 5 letters, there are 5 positions to fill in the 5-letter
word. The number of ways to arrange the 5 chosen letters is 5!.
Step 3: Compute the total number of five-letter words.
The total number of five-letter words that can be formed using the letters
in ”ALGEBRA” is the product of the number of ways to choose 5 letters and
the number of ways to arrange those chosen letters.
Total number of five-letter words = 6
5×5!
Total number of five-letter words = 6 ×5!
Total number of five-letter words = 6 ×5×4×3×2
Total number of five-letter words = 720
Therefore, there are 720 five-letter words that can be formed using the letters
in the word ”ALGEBRA” if no letter can be used more times than it appears
in ”ALGEBRA”.
Question 24
Question
A committee of 5 people is to be formed from a group of 10 students and 8
professors. If the committee must consist of 3 students and 2 professors, how
15
many different committees can be formed?
Solution
Step 1: Calculate the number of ways to choose 3 students out of 10. Step 2:
Calculate the number of ways to choose 2 professors out of 8. Step 3: Mul-
tiply the results from Step 1 and Step 2 to find the total number of different
committees that can be formed.
Step 1: The number of ways to choose 3 students out of 10 is given by 10
3.
10
3=10!
3!(10−3)! =10×9×8
3×2×1= 120.
Therefore, there are 120 ways to choose 3 students.
Step 2: The number of ways to choose 2 professors out of 8 is given by 8
2.
8
2=8!
2!(8−2)! =8×7
2×1= 28.
Therefore, there are 28 ways to choose 2 professors.
Step 3: To find the total number of different committees that can be formed,
we multiply the results from Step 1 and Step 2.
Total number of committees = 120 ×28 = 3360.
Therefore, there are 3360 different committees that can be formed consisting
of 3 students and 2 professors from the given group.
Question 25
Question
In how many ways can 5 red, 3 blue, and 4 white identical balls be arranged in
a row?
Solution
Step 1: Find the total number of arrangements without considering the colors.
Since there are 5 red balls, 3 blue balls, and 4 white balls, the total number
of balls is 5 + 3 + 4 = 12.
Therefore, the total number of ways to arrange the balls without considering
colors is 12 factorial, denoted as 12!.
Step 2: Consider the indistinguishable arrangement of each color.
Since the red balls are identical, dividing the total number of arrangements
by the factorial of the number of red balls (5! for red balls) will give us the
number of distinct arrangements of the red balls.
Similarly, dividing the resulting number by the factorial of the number of
blue balls (3! for blue balls) and the factorial of the number of white balls (4!
for white balls) will give us the number of distinct arrangements of the blue
balls and white balls, respectively.
Hence, the total number of ways to arrange the balls, taking their colors into
account, is:
16
12!
5! ×3! ×4!
Question 26
Question
In a group of 10 people, how many ways can 4 people be chosen to form a
committee with a president, a vice president, and two regular members?
Solution
Step 1: Choose the president There are 10 ways to choose the president from
the group of 10 people.
Step 2: Choose the vice president After choosing the president, there are 9
people left for the vice president role. Therefore, there are 9 ways to choose the
vice president.
Step 3: Choose the two regular members After choosing the president and
vice president, there are 8 people left for choosing the first regular member, and
7 people left for choosing the second regular member. So, there are 8
2×1 ways
to choose the two regular members.
Step 4: Calculate the total number of ways The total number of ways to
choose the committee is the product of the number of ways to choose the pres-
ident, vice president, and two regular members. Therefore, the total number of
ways is:
10 ×9×8
2×1 = 10 ×9×8×7
2×1= 10 ×9×28 = 2520
Therefore, there are 2520 ways to form a committee with a president, a vice
president, and two regular members from a group of 10 people.
Question 27
Question
A committee of 4 people is to be formed from a group of 8 men and 5 women.
If the committee must contain at least one man and at least one woman, how
many different committees can be formed?
Solution
Step 1: Find the total number of ways to form a committee of 4 people. To
form a committee of 4 people from a group of 8 men and 5 women, we will use
combinations. The total number of ways to choose 4 people from 13 people is
given by 13
4.
17
Step 2: Find the number of ways to form a committee with only men or only
women. We need to find separately the number of ways to form a committee
with only men and only women. For a committee with only men, we choose 4
men from 8, which can be done in 8
4ways. For a committee with only women,
we choose 4 women from 5, which can be done in 5
4ways. Therefore, the total
number of committees with only men or only women is given by 8
4+5
4.
Step 3: Find the number of ways to form a committee with at least one man
and at least one woman. The total number of ways to form a committee with at
least one man and at least one woman is the total number of committees minus
the number of committees with only men or only women. So, the number of
different committees that can be formed is:
13
4−8
4+5
4
Question 28
Question
In a class of 20 students, how many ways are there to form a committee of 5
students if 2 specific students refuse to work together on the same committee?
Solution
To determine the number of ways to form a committee of 5 students with the
given restriction, we need to consider two cases: 1. The two specific students
are in the committee. 2. The two specific students are not in the committee.
Step 1: Case 1 - The two specific students are in the committee.
Choose the other 3 students from the remaining 18 students (20−2 = 18). This
can be done in 18
3ways.
Step 2: Case 2 - The two specific students are not in the committee.
Choose 5 students from the remaining 18 students, excluding the 2 specific
students. This can be done in 18
5ways.
Step 3: Determine the total number of ways. Therefore, the total
number of ways to form a committee of 5 students where the two specific stu-
dents refuse to work together is the sum of the two cases:
18
3+18
5
Calculating these values:
18
3=18!
3!(18 −3)! =18 ×17 ×16
3×2×1= 816
18
5=18!
5!(18 −5)! =18 ×17 ×16 ×15 ×14
5×4×3×2×1= 8568
18
Adding these two values:
816 + 8568 = 9384
Therefore, there are 9,384 ways to form a committee of 5 students with the
given restriction.
Question 29
Question
A committee of 5 people is to be formed from a group of 10 men and 8 women.
1. In how many ways can the committee be formed if there must be at least
2 men and 2 women?
2. In how many ways can the committee be formed if there must be exactly
3 men?
Solution
Let’s solve each part of the question separately.
Part (a)
1. Choose 2 men out of 10: 10
2
2. Choose 2 women out of 8: 8
2
3. Choose the remaining person (can be a man or a woman): 10 +8 −2 = 16
The total number of ways the committee can be formed with at least 2 men and
2 women is: 10
2·8
2·16
Part (b)
1. Choose 3 men out of 10: 10
3
2. Choose 2 women out of 8: 8
2
The total number of ways the committee can be formed with exactly 3 men is:
10
3·8
2
19
Question 30
Question
In a group of 8 friends, how many ways can you choose a committee consisting
of a president, a vice-president, and a treasurer?
Solution
To solve this problem, we can use the concept of permutations. Since the order
in which the friends are chosen matters (president, vice-president, treasurer),
we will use the permutations formula.
Step 1: Find the total number of ways to choose a president from 8 friends.
There are 8 friends to choose from for the president position.
Step 2: Find the total number of ways to choose a vice-president from the
remaining 7 friends. After choosing the president, there are 7 friends remaining
for the vice-president position.
Step 3: Find the total number of ways to choose a treasurer from the
remaining 6 friends. After choosing the president and vice-president, there are
6 friends remaining for the treasurer position.
Step 4: Calculate the total number of ways to choose the committee. To
find the total number of ways to choose the committee consisting of a president,
a vice-president, and a treasurer, we multiply the number of choices at each
step. Total ways = Number of ways to choose president ×Number of ways to
choose vice-president ×Number of ways to choose treasurer
Total ways = 8 ×7×6 = 336
Therefore, there are 336 ways to choose a committee consisting of a president,
a vice-president, and a treasurer from a group of 8 friends.
Question 31
Question
A committee of 7 people is to be formed from a group of 12 individuals. If 3 of
the individuals refuse to serve on the committee together, how many different
committees can be formed?
Solution
Step 1: Find the number of ways to form a committee without any restrictions.
Since there are 12 individuals and we are forming a committee of 7 people, we
can calculate the number of ways as:
12
7=12!
7!(12 −7)! = 792
20
Step 2: Find the number of ways when 3 individuals refuse to serve on the
committee together.
Let’s consider the 3 individuals who refuse to serve as a single entity. Then we
have 10 individuals remaining to choose from. Since we still need to choose 7
people for the committee, the number of ways can be calculated as:
10
7=10!
7!(10 −7)! = 120
Step 3: Subtract the number of committees that include all 3 refusing indi-
viduals from the total number of committees.
Therefore, the number of different committees that can be formed when 3 of
the individuals refuse to serve on the committee together is:
792 −120 = 672
Thus, there are 672 different committees that can be formed under the given
conditions.
Question 32
Question
In a committee of 8 people, there are 4 mathematicians and 4 computer sci-
entists. If the committee needs to select a president, a vice president, and a
treasurer, how many ways can this be done if the president cannot be a mathe-
matician and the treasurer cannot be a computer scientist?
Solution
Step 1: Calculate the number of ways to choose the president from the computer
scientists. Since the president cannot be a mathematician, we must choose the
president from the 4 computer scientists. This can be done in 4
1= 4 ways.
Step 2: Calculate the number of ways to choose the treasurer from the
mathematicians. Similarly, since the treasurer cannot be a computer scientist,
we must choose the treasurer from the 4 mathematicians. This can be done in
4
1= 4 ways.
Step 3: Calculate the number of ways to choose the vice president. After
choosing the president and treasurer, the vice president will be chosen from the
remaining pool of 6 people (2 mathematicians and 4 computer scientists). This
can be done in 6
1= 6 ways.
Step 4: Multiply the number of choices for each position to find the total
number of ways to select the committee positions. The total number of ways to
select a president, vice president, and treasurer following the given restrictions
is 4 ×4×6 = 96 ways.
21
Question 33
Question
In a survey of 10 students, each student was asked to rank their top 3 favorite
books. If the order in which the books are ranked matters, how many possible
rankings are there in total?
Solution
Step 1: Determine the number of ways to choose the first-ranked book. Since
each student has 3 choices and the order matters, there are 3 ways to choose
the first-ranked book.
Step 2: Determine the number of ways to choose the second-ranked book.
After selecting the first-ranked book, there are 2 remaining books to choose
from for the second rank. Thus, there are 2 ways to choose the second-ranked
book.
Step 3: Determine the number of ways to choose the third-ranked book.
After selecting the first and second-ranked books, there is 1 book remaining
to choose from for the third rank. Thus, there is only 1 way to choose the
third-ranked book.
Step 4: Calculate the total number of possible rankings. To find the total
number of possible rankings, we multiply the number of ways for each step.
Total number of possible rankings = 3 (ways for first rank) ×2 (ways for
second rank) ×1 (ways for third rank) = 6
Therefore, there are 6possible rankings in total.
Question 34
Question
A committee of 5 people is to be formed from a group of 9 students and 6
professors. How many ways can the committee be formed if it must consist of
3 students and 2 professors?
Solution
Step 1: Calculate the number of ways to choose 3 students out of 9. Step 2:
Calculate the number of ways to choose 2 professors out of 6. Step 3: Multiply
the results from Step 1 and Step 2 to find the total number of ways to form the
committee.
Step 1: The number of ways to choose 3 students out of 9 can be calculated
using combinations. The number of ways is given by 9
3=9!
3!6! = 84.
Step 2: The number of ways to choose 2 professors out of 6 can be calculated
using combinations. The number of ways is given by 6
2=6!
2!4! = 15.
22
Step 3: To find the total number of ways to form the committee, we multiply
the results from Step 1 and Step 2: Total number of ways = 84 ×15 = 1260.
Therefore, there are 1260 ways to form a committee of 3 students and 2
professors from a group of 9 students and 6 professors.
Question 35
Question
A committee of 5 people is to be formed from a group of 9 men and 7 women.
If the committee must contain at least 2 men and at least 2 women, how many
different committees could be formed?
Solution
Step 1: Calculate the number of ways to form a committee with 2 men and 3
women. The number of ways to choose 2 men from 9 is 9
2and the number of
ways to choose 3 women from 7 is 7
3. Therefore, the number of ways to form
a committee with 2 men and 3 women is 9
2×7
3.
Step 2: Calculate the number of ways to form a committee with 3 men and
2 women. The number of ways to choose 3 men from 9 is 9
3and the number of
ways to choose 2 women from 7 is 7
2. Therefore, the number of ways to form
a committee with 3 men and 2 women is 9
3×7
2.
Step 3: Calculate the total number of committees that can be formed. Add
the number of committees with 2 men and 3 women to the number of committees
with 3 men and 2 women: 9
2×7
3+9
3×7
2.
Step 4: Compute the final answer. Calculate the values of the combinations
and sum them to get the total number of different committees that could be
formed.
Therefore, the total number of different committees that could be formed is
9
2×7
3+9
3×7
2.
23
Question 2
Question
A committee of 4 people is to be formed from a group of 7 women and 5 men.
If at least one man must be included in the committee, how many different
committees can be formed?
Solution
Step 1: Calculate the total number of committees without any restrictions.
First, we calculate the total number of committees that can be formed with-
out any restrictions. This can be done using combinations formula:
Total number of committees = 12
4=12!
4!(12 −4)! =12!
4!8! = 495
So, there are 495 different committees that can be formed without any re-
strictions.
Step 2: Calculate the number of committees with only women.
Next, we calculate the number of committees that consist only of women.
This can be done using combinations formula:
Number of committees with only women = 7
4=7!
4!(7 −4)! =7!
4!3! = 35
Step 3: Calculate the number of committees with no men.
Now, we calculate the number of committees with no men. This can be
done by subtracting the number of committees with only women from the total
number of committees:
Number of committees with no men = Total number of committees−Number of committees with only women = 495−35 = 460
Step 4: Calculate the number of committees with at least one man.
Since we want to find the number of committees with at least one man, we
can subtract the number of committees with no men from the total number of
committees:
Number of committees with at least one man = Total number of committees−Number of committees with no men = 495−460 = 35
Therefore, there are 35 different committees that can be formed with at least
one man included.
2
Question 3
Question
In a group of 12 students, how many ways can we choose a committee of 5
students if 2 particular students must be included?
Solution
Step 1: First, we select the 2 particular students who must be included in the
committee. This can be done in 12
2ways.
Step 2: After selecting the 2 particular students, we need to select 3 more
students to complete the committee. Since we have already selected 2 students,
we are left with 10 students to choose from. We can select 3 more students in
10
3ways.
Step 3: To find the total number of ways to choose the committee of 5
students, we multiply the number of ways from Step 1 and Step 2. Therefore,
the total number of ways to choose a committee of 5 students with 2 particular
students included is 12
2×10
3.
Solving this expression, we get: 12
2=12!
2!(12−2)! =12×11
2×1= 66
10
3=10!
3!(10−3)! =10×9×8
3×2×1= 120
Therefore, the total number of ways to choose the committee is: 66 ×120 =
7920
Hence, there are 7920 ways to choose a committee of 5 students with 2
particular students included from a group of 12 students.
Question 4
Question
In a group of 10 people, how many ways can we choose a president, a vice
president, and a treasurer if no person can hold more than one position?
Solution
Step 1: To find the number of ways to choose a president, we simply have 10
choices.
Step 2: After choosing a president, there are 9 remaining people to choose
from for the position of vice president.
Step 3: Once the president and vice president have been chosen, there are 8
remaining people to choose from for the position of treasurer.
Step 4: To find the total number of ways to choose a president, a vice
president, and a treasurer, we multiply the number of choices at each step:
10 ×9×8 = 720
3
Therefore, there are 720 ways to choose a president, a vice president, and
a treasurer in a group of 10 people.
Question 5
Question
A committee of 5 people is to be formed from a group of 10 women and 8 men. If
the committee must consist of at least 3 women, how many different committees
can be formed?
Solution
Step 1: Calculate the number of committees with exactly 3 women: There are
10
3ways to choose 3 women from the 10 available women, and 8
2ways to
choose 2 men from the 8 available men. Therefore, there are 10
3×8
2ways to
form a committee with exactly 3 women.
Step 2: Calculate the number of committees with exactly 4 women: There
are 10
4ways to choose 4 women from the 10 available women, and 8
1ways
to choose 1 man from the 8 available men. Hence, there are 10
4×8
1ways to
create a committee with exactly 4 women.
Step 3: Calculate the number of committees with all 5 women: There is only
10
5way to choose all 5 women from the 10 available women.
Step 4: Add the results from Step 1, Step 2, and Step 3 to find the total
number of valid committees:
10
3×8
2+10
4×8
1+10
5
Calculating the above expression will give us the total number of different
committees that can be formed in this scenario.
Question 6
Question
In a group of 10 people, how many ways can we choose a committee of 4 people
such that 2 of them are women and 2 are men?
Solution
To solve this problem, we can break it down into two steps: selecting the 2
women from a group of 5 women and selecting the 2 men from a group of 5
men.
Step 1: Selecting 2 women from 5 women There are 5
2ways to select 2
women from a group of 5 women.
4
Step 2: Selecting 2 men from 5 men Similarly, there are 5
2ways to select 2
men from a group of 5 men.
To find the total number of ways to choose a committee of 4 people (2 women
and 2 men), we multiply the number of ways from Step 1 and Step 2.
Therefore, the total number of ways to choose the committee is:
5
2×5
2=5!
2!3! ×5!
2!3! =5×4
2×5×4
2= 10 ×10 = 100
So, there are 100 ways to choose a committee of 4 people consisting of 2
women and 2 men from a group of 10 people.
Question 7
Question
A committee of 5 people is to be formed from a group of 10 men and 8 women. If
the committee must consist of at least 2 women, how many different committees
can be formed?
Solution
Step 1: First, we calculate the number of ways to form a committee with exactly
2 women. There are 8 women to choose 2 from, and 10 men to choose 3 from.
Thus, the number of committees with exactly 2 women is 8
2×10
3.
Step 2: Next, we calculate the number of ways to form a committee with
exactly 3 women. There are 8 women to choose 3 from, and 10 men to choose
2 from. Thus, the number of committees with exactly 3 women is 8
3×10
2.
Step 3: Now, we calculate the total number of committees by adding the
number of committees with exactly 2 women and exactly 3 women. Total num-
ber of committees = 8
2×10
3+8
3×10
2.
Step 4: Plug in the values and simplify. Total number of committees =
8
2×10
3+8
3×10
2
Total number of committees = 8!
2!6! ×10!
3!7! +8!
3!5! ×10!
2!8!
Total number of committees = 28 ×120 + 56 ×45
Total number of committees = 3360 + 2520
Total number of committees = 5880.
Therefore, there are 5880 different committees that can be formed with at
least 2 women.
Question 8
Question
In how many ways can 5 distinct books be arranged on a bookshelf such that
two particular books must always be next to each other?
5
Solution
Step 1: Treat the two particular books as a single entity. Step 2: Consider the
arrangement as a combination problem.
There are two particular books that must be next to each other. Let’s treat
these two books as a single entity. So, we now have 4 entities to arrange: AB,
C, D, and E. Step 3: We have 4 entities to arrange in total: AB, C, D, and E.
Step 4: There are 4! ways to arrange these entities, but within the AB entity,
the books can be arranged in 2! ways. Step 5: Therefore, the total number of
ways to arrange the books is 4! ×2! = 24 ×2 = 48 ways.
Hence, there are 48 ways to arrange the 5 distinct books on the bookshelf
such that two particular books must always be next to each other.
Question 9
Question
A committee of 6 people is to be formed from a group of 10 men and 5 women. If
the committee must consist of at least 2 men and 2 women, how many different
committees can be formed?
Solution
Step 1: Calculate the number of ways to choose 2 men out of 10. There are 10
2
ways to choose 2 men from the group of 10 men.
Step 2: Calculate the number of ways to choose 2 women out of 5. There
are 5
2ways to choose 2 women from the group of 5 women.
Step 3: Calculate the number of ways to choose the remaining 2 people (1
man and 1 woman) out of the remaining 8 people (8=10-2 men; 8=5-2 women).
There are 8
2ways to choose the remaining 1 man and 1 woman.
Step 4: Calculate the total number of ways to form the committee. The total
number of ways to form the committee is the product of the ways to choose 2
men, 2 women, and 1 man and 1 woman: Total number of ways = 10
2×5
2×8
2
Calculating this expression gives: Total number of ways = 10!
2!8! ×5!
2!3! ×8!
2!6!
Total number of ways = 45 ×10 ×28 Total number of ways = 12600
Therefore, there are 12,600 different committees that can be formed.
Question 10
Question
A committee of 5 people is to be formed from a group of 10 math students and
6 computer science students. If the committee must consist of at least 2 math
students and 2 computer science students, how many different committees can
be formed?
6
Solution
Step 1: Calculate the number of ways to choose 2 math students out of 10.
There are 10
2ways to choose 2 math students.
Step 2: Calculate the number of ways to choose 2 computer science students
out of 6. There are 6
2ways to choose 2 computer science students.
Step 3: Calculate the number of ways to choose the remaining committee
member. There are 8 students remaining (4 math students and 4 computer
science students). We can choose one person from this group in 8
1ways.
Step 4: Calculate the total number of committees that can be formed. The
total number of committees can be formed by multiplying the results from Step
1, Step 2, and Step 3:
10
2×6
2×8
1= 45 ×15 ×8 = 5400.
Therefore, there are 5400 different committees that can be formed.
Question 11
Question
A committee of 5 people is to be formed from a group of 10 people. If 3 of the
people are friends and refuse to serve on the committee together, how many
different committees can be formed?
Solution
Step 1: To find the number of ways to form a committee, we first calculate the
total number of ways to choose 5 people from a group of 10. This is represented
by 10
5.
Total number of ways to choose 5 people from 10 = 10
5=10!
5!(10 −5)! = 252
Step 2: Now, let’s calculate the number of ways to form a committee where
the 3 friends cannot serve together. - We break this problem into cases: Case 1:
0 friends are on the committee - The number of ways to choose 5 people from
the remaining 7 (10-3 friends) is 7
5. Case 2: 1 friend is on the committee - The
number of ways to choose 4 people from the remaining 7 (10-3 friends and 1
friend already on the committee) is 7
4. Case 3: 2 friends are on the committee
- The number of ways to choose 3 people from the remaining 7 (10-3 friends and
2 friends already on the committee) is 7
3. - We sum the results from all these
cases to get the total number of committees that can be formed.
Number of committees without the friends together = 7
5+7
4+7
3
7
=7!
5!(7 −5)! +7!
4!(7 −4)! +7!
3!(7 −3)!
= 21 + 35 + 35 = 91
Therefore, there are 91 different committees that can be formed from the
group of 10 people with the condition that 3 friends cannot serve together.
Question 12
Question
A committee of 4 people is to be formed from a group of 8 men and 6 women.
In how many ways can the committee be formed if it must consist of 2 men and
2 women?
Solution
Step 1: Calculate the number of ways to choose 2 men from 8. Step 2: Calculate
the number of ways to choose 2 women from 6. Step 3: Multiply the results
from Step 1 and Step 2 to get the total number of ways to form the committee.
Step 1: Choosing 2 men from 8 can be done using a combination. The
number of ways to choose 2 men from 8 is given by
8
2=8!
2!(8 −2)! =8×7
2×1= 28.
Step 2: Choosing 2 women from 6 can be done using a combination. The
number of ways to choose 2 women from 6 is given by
6
2=6!
2!(6 −2)! =6×5
2×1= 15.
Step 3: The total number of ways to form the committee consisting of 2 men
and 2 women is the product of the results from Step 1 and Step 2,
28 ×15 = 420.
Therefore, there are 420 ways to form the committee with 2 men and 2
women from the group of 8 men and 6 women.
Question 13
Question
A group of 6 people are going on a road trip and plan to stop at 4 different
cities along the way. If each person can visit any city, how many different ways
can they split up into groups to visit the cities?
8
Solution
We can solve this problem using combinations.
Step 1: Since the order in which the cities are visited does not matter, we
will use the concept of combinations. The number of ways to split 6 people into
groups to visit 4 cities is equal to the number of ways to distribute the people
among the cities with repetition allowed.
Step 2: We can visualize this problem as placing 3 dividers among the 6
people, representing the 4 different cities. The number of ways to place the
dividers determines the number of ways to split the people into groups.
Step 3: Thus, the total number of ways to split the 6 people into groups to
visit the 4 cities is given by the formula for combinations with repetition:
n+r−1
r=6+4−1
4=9
4=9×8×7×6
4×3×2×1= 126
Step 4: Therefore, there are 126 different ways for the group of 6 people to
split up into groups to visit the 4 cities along the road trip.
Question 14
Question
In how many ways can 5 men and 5 women be seated around a circular table if
each man must be seated next to at least one woman?
Solution
Step 1: First, we will fix one woman in a seat. Then, we can count the number
of ways the other 4 women can be seated around the table.
Step 2: There are 4 women remaining who can be seated in 4! ways.
Step 3: Now, we will arrange the 5 men among the 5 available slots between
the women.
Step 4: There are 5 slots between the women where the men can be seated.
The men can be arranged in 5! ways in these slots.
Step 5: Finally, since the table is circular, we must divide by 5 to account
for the rotations of the same arrangement.
Therefore, the total number of ways the 5 men and 5 women can be seated
around a circular table with each man seated next to at least one woman is
given by: 4! ×5!
5= 2880
9
Question 15
Question
A basketball team has 12 players. In how many ways can a starting lineup of
5 players be chosen if the lineup must include the team captain and the point
guard?
Solution
We can approach this problem by considering the positions of the players in the
starting lineup and filling them one by one. Step 1: Choose the team captain
(1 way) Step 2: Choose the point guard from the remaining 11 players (11
ways) Step 3: Choose the remaining 3 players from the remaining 10 players
(10 choices for the first remaining player, 9 for the second, and 8 for the third)
Therefore, the total number of ways to choose the starting lineup is:
1×11 ×10 ×9×8=7,920
So, there are 7,920 ways to choose the starting lineup.
Question 16
Question
In how many ways can you choose a committee of 4 people from a group of 10
people if 2 of the group members refuse to serve on the committee together?
Solution
Step 1: First, we calculate the total number of ways to choose a committee of 4
people from 10 people without any restrictions. This is a combination problem,
and can be calculated using the formula for combinations:
n
r=n!
r!(n−r)!
where nis the total number of people and ris the number of people we want
to choose.
For this problem, n= 10 and r= 4. So, the total number of ways to choose
a committee without restrictions is:
10
4=10!
4!(10 −4)!
=10 ×9×8×7
4×3×2×1
10
= 210
So, there are 210 ways to choose a committee of 4 people from a group of 10
people without restrictions.
Step 2: Next, we calculate the number of ways that the 2 group members
who refuse to serve on the committee together are chosen. Let’s call these 2
people A and B. There are 8 remaining people to choose from for the committee,
plus we need to choose 2 more people from the remaining 8.
This can be calculated by finding the number of ways to choose a committee
of 4 people from the 8 remaining people:
8
2=8!
2!(8 −2)!
=8×7
2×1
= 28
Step 3: Finally, we subtract the number of committees that include both A
and B from the total number of committees to get the final answer. This can
be calculated as:
Total ways −Ways with A and B together = 210 −28 = 182
Therefore, there are 182 ways to choose a committee of 4 people from a group
of 10 people if 2 group members refuse to serve on the committee together.
Question 17
Question
A committee of 5 people is to be chosen from a group of 10 men and 8 women.
If the committee must have at least 3 men, how many different committees can
be formed?
Solution
Step 1: Find the number of ways to choose a committee with exactly 3 men.
There are 10
3ways to choose 3 men from 10 men, and 8
2ways to choose 2
women from 8 women. Therefore, the number of ways to choose a committee
with exactly 3 men is:
10
3×8
2
Step 2: Find the number of ways to choose a committee with 4 men. There
are 10
4ways to choose 4 men from 10 men, and 8
1ways to choose 1 woman
11
from 8 women. Therefore, the number of ways to choose a committee with 4
men is: 10
4×8
1
Step 3: Find the number of ways to choose a committee with 5 men. There
are 10
5ways to choose 5 men from 10 men. Therefore, the number of ways to
choose a committee with 5 men is:
10
5
Step 4: Add the results from Steps 1, 2, and 3 to find the total number of
different committees that can be formed with at least 3 men. Total number of
committees = 10
3×8
2+10
4×8
1+10
5
Question 18
Question
A committee of 5 students is to be formed from a group of 10 students. If
3 of the students are math majors and 7 are computer science majors, how
many different committees can be formed if: 1. The committee must have 3
math majors and 2 computer science majors? 2. The committee can have any
combination of math and computer science majors?
Solution
1. To find the number of different committees that can be formed with 3 math
majors and 2 computer science majors, we calculate the number of ways to
choose 3 math majors out of 3 and 2 computer science majors out of 7.
Step 1: Choose 3 math majors out of 3 Since there are only 3 math
majors, the number of ways to choose 3 math majors out of 3 is 1.
Step 2: Choose 2 computer science majors out of 7 We need to
choose 2 computer science majors out of 7. This can be done in 7
2= 21 ways.
Step 3: Calculate the total number of committees The total number
of committees that can be formed is the product of the number of ways to
choose math majors and computer science majors. Therefore, the total number
of committees is 1 ∗21 = 21.
Therefore, there are 21 different committees that can be formed with 3 math
majors and 2 computer science majors.
2. To find the total number of different committees that can be formed
with any combination of math and computer science majors, we can consider
all possible cases separately and then add them up.
Case 1: All members are math majors In this case, we need to choose
5 math majors out of 3. However, since there are only 3 math majors, this case
is not possible.
12
Case 2: 4 math majors and 1 computer science major We need to
choose 4 math majors out of 3 and 1 computer science major out of 7. The
number of ways to do this is 3
4×7
1= 0.
Case 3: 3 math majors and 2 computer science majors We have
already calculated this case in part 1. The number of ways is 21.
Case 4: 2 math majors and 3 computer science majors We need to
choose 2 math majors out of 3 and 3 computer science majors out of 7. The
number of ways to do this is 3
2×7
3= 63.
Case 5: 1 math major and 4 computer science majors We need to
choose 1 math major out of 3 and 4 computer science majors out of 7. The
number of ways to do this is 3
1×7
4= 105.
Case 6: All members are computer science majors In this case, we
need to choose 5 computer science majors out of 7. The number of ways to do
this is 7
5= 21.
Step 4: Calculate the total number of committees The total number
of committees is the sum of all possible cases. Therefore, the total number of
different committees is 0 + 21 + 63 + 105 + 21 = 210.
Therefore, there are 210 different committees that can be formed with any
combination of math and computer science majors.
Question 19
Question
In a group of 10 people, how many ways are there to form a committee of 3
people with a president, a secretary, and a treasurer?
Solution
Step 1: First, choose the 3 people to be on the committee. Since the order in
which the people are chosen matters (president, secretary, treasurer), we use
permutations. There are 10 people to choose from, and we want to choose 3:
10P3.
Step 2: Calculate the number of ways to assign the positions of president,
secretary, and treasurer. Once the committee members have been chosen, we
need to assign the specific roles. There are 3 positions to fill and 3 people to
assign to them. Therefore, the number of ways to assign the positions is 3!.
Step 3: Multiply the results from Step 1 and Step 2. The total number of
ways to form the committee of 3 people with specific roles is given by:
10P3×3! = 10!
(10 −3)! ×3! = 10 ×9×8×7!
7! ×3×2×1 = 720 ×6 = 4320
Therefore, there are 4320 ways to form a committee of 3 people with a
president, a secretary, and a treasurer from a group of 10 people.
13
Question 20
Question
In a group of 12 people, how many ways can we select a committee of 4 people
if 2 particular people must always be included in the committee?
Solution
Step 1: First, we select the 2 particular people that must be included in the
committee. This can be done in 2
2= 1 way.
Step 2: Next, we need to select the remaining 2 people from the remaining
10 people. This can be done in 10
2ways.
Step 3: To find the total number of ways to form the committee, we multiply
the results of Step 1 and Step 2.
Therefore, the total number of ways to form a committee of 4 people with 2
particular people always included is 1 ×10
2= 1 ×45 = 45 ways.
Question 21
Question
In a group of 10 friends, how many ways can we choose a committee of 4 people
if one of the friends, Alice, insists on being part of the committee?
Solution
Step 1: Since one person, Alice, is already guaranteed to be part of the com-
mittee, we need to choose the remaining 3 people from the remaining 9 friends.
Step 2: To find the number of ways to choose these 3 people, we use the combi-
nation formula n
r=n!
r!(n−r)! . Step 3: Substituting n= 9 (9 remaining friends)
and r= 3 (to choose 3 people) into the formula:
9
3=9!
3!(9 −3)! =9×8×7
3×2×1= 84.
Therefore, there are 84 ways to choose a committee of 4 people with Alice being
one of them.
Question 22
Question
In how many ways can 5 students be seated in a row of 8 chairs such that two
particular students, Alice and Bob, must sit next to each other?
14
Solution
Step 1: Treat Alice and Bob as one entity. Step 2: Calculate the number of
ways to arrange the remaining entities. Step 3: There are 7 entities to arrange
(including the entity composed of Alice and Bob) in 7 seats.
Step 1: Treat Alice and Bob as one entity. Step 2: There are 7 entities in
total to be arranged. Step 3: The number of ways to arrange these 7 entities in
7 seats is 7!.
Therefore, the total number of ways for 5 students to be seated in a row of
8 chairs such that Alice and Bob must sit next to each other is 2 ·7!.
Question 23
Question
How many five-letter words can be formed using the letters in the word ”AL-
GEBRA” if no letter can be used more times than it appears in ”ALGEBRA”?
Solution
Step 1: Determine the number of ways to choose 5 letters from ”ALGEBRA”.
Since we cannot repeat any letters, we need to consider the unique letters in
”ALGEBRA”. The word ”ALGEBRA” has 6 different letters, so there are 6
5
ways to choose 5 letters from ”ALGEBRA”.
6
5=6!
5!(6−5)! = 6
Step 2: Calculate the number of ways to arrange the chosen 5 letters.
Since we have chosen 5 letters, there are 5 positions to fill in the 5-letter
word. The number of ways to arrange the 5 chosen letters is 5!.
Step 3: Compute the total number of five-letter words.
The total number of five-letter words that can be formed using the letters
in ”ALGEBRA” is the product of the number of ways to choose 5 letters and
the number of ways to arrange those chosen letters.
Total number of five-letter words = 6
5×5!
Total number of five-letter words = 6 ×5!
Total number of five-letter words = 6 ×5×4×3×2
Total number of five-letter words = 720
Therefore, there are 720 five-letter words that can be formed using the letters
in the word ”ALGEBRA” if no letter can be used more times than it appears
in ”ALGEBRA”.
Question 24
Question
A committee of 5 people is to be formed from a group of 10 students and 8
professors. If the committee must consist of 3 students and 2 professors, how
15
many different committees can be formed?
Solution
Step 1: Calculate the number of ways to choose 3 students out of 10. Step 2:
Calculate the number of ways to choose 2 professors out of 8. Step 3: Mul-
tiply the results from Step 1 and Step 2 to find the total number of different
committees that can be formed.
Step 1: The number of ways to choose 3 students out of 10 is given by 10
3.
10
3=10!
3!(10−3)! =10×9×8
3×2×1= 120.
Therefore, there are 120 ways to choose 3 students.
Step 2: The number of ways to choose 2 professors out of 8 is given by 8
2.
8
2=8!
2!(8−2)! =8×7
2×1= 28.
Therefore, there are 28 ways to choose 2 professors.
Step 3: To find the total number of different committees that can be formed,
we multiply the results from Step 1 and Step 2.
Total number of committees = 120 ×28 = 3360.
Therefore, there are 3360 different committees that can be formed consisting
of 3 students and 2 professors from the given group.
Question 25
Question
In how many ways can 5 red, 3 blue, and 4 white identical balls be arranged in
a row?
Solution
Step 1: Find the total number of arrangements without considering the colors.
Since there are 5 red balls, 3 blue balls, and 4 white balls, the total number
of balls is 5 + 3 + 4 = 12.
Therefore, the total number of ways to arrange the balls without considering
colors is 12 factorial, denoted as 12!.
Step 2: Consider the indistinguishable arrangement of each color.
Since the red balls are identical, dividing the total number of arrangements
by the factorial of the number of red balls (5! for red balls) will give us the
number of distinct arrangements of the red balls.
Similarly, dividing the resulting number by the factorial of the number of
blue balls (3! for blue balls) and the factorial of the number of white balls (4!
for white balls) will give us the number of distinct arrangements of the blue
balls and white balls, respectively.
Hence, the total number of ways to arrange the balls, taking their colors into
account, is:
16
12!
5! ×3! ×4!
Question 26
Question
In a group of 10 people, how many ways can 4 people be chosen to form a
committee with a president, a vice president, and two regular members?
Solution
Step 1: Choose the president There are 10 ways to choose the president from
the group of 10 people.
Step 2: Choose the vice president After choosing the president, there are 9
people left for the vice president role. Therefore, there are 9 ways to choose the
vice president.
Step 3: Choose the two regular members After choosing the president and
vice president, there are 8 people left for choosing the first regular member, and
7 people left for choosing the second regular member. So, there are 8
2×1 ways
to choose the two regular members.
Step 4: Calculate the total number of ways The total number of ways to
choose the committee is the product of the number of ways to choose the pres-
ident, vice president, and two regular members. Therefore, the total number of
ways is:
10 ×9×8
2×1 = 10 ×9×8×7
2×1= 10 ×9×28 = 2520
Therefore, there are 2520 ways to form a committee with a president, a vice
president, and two regular members from a group of 10 people.
Question 27
Question
A committee of 4 people is to be formed from a group of 8 men and 5 women.
If the committee must contain at least one man and at least one woman, how
many different committees can be formed?
Solution
Step 1: Find the total number of ways to form a committee of 4 people. To
form a committee of 4 people from a group of 8 men and 5 women, we will use
combinations. The total number of ways to choose 4 people from 13 people is
given by 13
4.
17
Step 2: Find the number of ways to form a committee with only men or only
women. We need to find separately the number of ways to form a committee
with only men and only women. For a committee with only men, we choose 4
men from 8, which can be done in 8
4ways. For a committee with only women,
we choose 4 women from 5, which can be done in 5
4ways. Therefore, the total
number of committees with only men or only women is given by 8
4+5
4.
Step 3: Find the number of ways to form a committee with at least one man
and at least one woman. The total number of ways to form a committee with at
least one man and at least one woman is the total number of committees minus
the number of committees with only men or only women. So, the number of
different committees that can be formed is:
13
4−8
4+5
4
Question 28
Question
In a class of 20 students, how many ways are there to form a committee of 5
students if 2 specific students refuse to work together on the same committee?
Solution
To determine the number of ways to form a committee of 5 students with the
given restriction, we need to consider two cases: 1. The two specific students
are in the committee. 2. The two specific students are not in the committee.
Step 1: Case 1 - The two specific students are in the committee.
Choose the other 3 students from the remaining 18 students (20−2 = 18). This
can be done in 18
3ways.
Step 2: Case 2 - The two specific students are not in the committee.
Choose 5 students from the remaining 18 students, excluding the 2 specific
students. This can be done in 18
5ways.
Step 3: Determine the total number of ways. Therefore, the total
number of ways to form a committee of 5 students where the two specific stu-
dents refuse to work together is the sum of the two cases:
18
3+18
5
Calculating these values:
18
3=18!
3!(18 −3)! =18 ×17 ×16
3×2×1= 816
18
5=18!
5!(18 −5)! =18 ×17 ×16 ×15 ×14
5×4×3×2×1= 8568
18
Adding these two values:
816 + 8568 = 9384
Therefore, there are 9,384 ways to form a committee of 5 students with the
given restriction.
Question 29
Question
A committee of 5 people is to be formed from a group of 10 men and 8 women.
1. In how many ways can the committee be formed if there must be at least
2 men and 2 women?
2. In how many ways can the committee be formed if there must be exactly
3 men?
Solution
Let’s solve each part of the question separately.
Part (a)
1. Choose 2 men out of 10: 10
2
2. Choose 2 women out of 8: 8
2
3. Choose the remaining person (can be a man or a woman): 10 + 8 −2 = 16
The total number of ways the committee can be formed with at least 2 men and
2 women is: 10
2·8
2·16
Part (b)
1. Choose 3 men out of 10: 10
3
2. Choose 2 women out of 8: 8
2
The total number of ways the committee can be formed with exactly 3 men is:
10
3·8
2
19
Question 30
Question
In a group of 8 friends, how many ways can you choose a committee consisting
of a president, a vice-president, and a treasurer?
Solution
To solve this problem, we can use the concept of permutations. Since the order
in which the friends are chosen matters (president, vice-president, treasurer),
we will use the permutations formula.
Step 1: Find the total number of ways to choose a president from 8 friends.
There are 8 friends to choose from for the president position.
Step 2: Find the total number of ways to choose a vice-president from the
remaining 7 friends. After choosing the president, there are 7 friends remaining
for the vice-president position.
Step 3: Find the total number of ways to choose a treasurer from the
remaining 6 friends. After choosing the president and vice-president, there are
6 friends remaining for the treasurer position.
Step 4: Calculate the total number of ways to choose the committee. To
find the total number of ways to choose the committee consisting of a president,
a vice-president, and a treasurer, we multiply the number of choices at each
step. Total ways = Number of ways to choose president ×Number of ways to
choose vice-president ×Number of ways to choose treasurer
Total ways = 8 ×7×6 = 336
Therefore, there are 336 ways to choose a committee consisting of a president,
a vice-president, and a treasurer from a group of 8 friends.
Question 31
Question
A committee of 7 people is to be formed from a group of 12 individuals. If 3 of
the individuals refuse to serve on the committee together, how many different
committees can be formed?
Solution
Step 1: Find the number of ways to form a committee without any restrictions.
Since there are 12 individuals and we are forming a committee of 7 people, we
can calculate the number of ways as:
12
7=12!
7!(12 −7)! = 792
20
Step 2: Find the number of ways when 3 individuals refuse to serve on the
committee together.
Let’s consider the 3 individuals who refuse to serve as a single entity. Then we
have 10 individuals remaining to choose from. Since we still need to choose 7
people for the committee, the number of ways can be calculated as:
10
7=10!
7!(10 −7)! = 120
Step 3: Subtract the number of committees that include all 3 refusing indi-
viduals from the total number of committees.
Therefore, the number of different committees that can be formed when 3 of
the individuals refuse to serve on the committee together is:
792 −120 = 672
Thus, there are 672 different committees that can be formed under the given
conditions.
Question 32
Question
In a committee of 8 people, there are 4 mathematicians and 4 computer sci-
entists. If the committee needs to select a president, a vice president, and a
treasurer, how many ways can this be done if the president cannot be a mathe-
matician and the treasurer cannot be a computer scientist?
Solution
Step 1: Calculate the number of ways to choose the president from the computer
scientists. Since the president cannot be a mathematician, we must choose the
president from the 4 computer scientists. This can be done in 4
1= 4 ways.
Step 2: Calculate the number of ways to choose the treasurer from the
mathematicians. Similarly, since the treasurer cannot be a computer scientist,
we must choose the treasurer from the 4 mathematicians. This can be done in
4
1= 4 ways.
Step 3: Calculate the number of ways to choose the vice president. After
choosing the president and treasurer, the vice president will be chosen from the
remaining pool of 6 people (2 mathematicians and 4 computer scientists). This
can be done in 6
1= 6 ways.
Step 4: Multiply the number of choices for each position to find the total
number of ways to select the committee positions. The total number of ways to
select a president, vice president, and treasurer following the given restrictions
is 4 ×4×6 = 96 ways.
21
Question 33
Question
In a survey of 10 students, each student was asked to rank their top 3 favorite
books. If the order in which the books are ranked matters, how many possible
rankings are there in total?
Solution
Step 1: Determine the number of ways to choose the first-ranked book. Since
each student has 3 choices and the order matters, there are 3 ways to choose
the first-ranked book.
Step 2: Determine the number of ways to choose the second-ranked book.
After selecting the first-ranked book, there are 2 remaining books to choose
from for the second rank. Thus, there are 2 ways to choose the second-ranked
book.
Step 3: Determine the number of ways to choose the third-ranked book.
After selecting the first and second-ranked books, there is 1 book remaining
to choose from for the third rank. Thus, there is only 1 way to choose the
third-ranked book.
Step 4: Calculate the total number of possible rankings. To find the total
number of possible rankings, we multiply the number of ways for each step.
Total number of possible rankings = 3 (ways for first rank) ×2 (ways for
second rank) ×1 (ways for third rank) = 6
Therefore, there are 6possible rankings in total.
Question 34
Question
A committee of 5 people is to be formed from a group of 9 students and 6
professors. How many ways can the committee be formed if it must consist of
3 students and 2 professors?
Solution
Step 1: Calculate the number of ways to choose 3 students out of 9. Step 2:
Calculate the number of ways to choose 2 professors out of 6. Step 3: Multiply
the results from Step 1 and Step 2 to find the total number of ways to form the
committee.
Step 1: The number of ways to choose 3 students out of 9 can be calculated
using combinations. The number of ways is given by 9
3=9!
3!6! = 84.
Step 2: The number of ways to choose 2 professors out of 6 can be calculated
using combinations. The number of ways is given by 6
2=6!
2!4! = 15.
22
Step 3: To find the total number of ways to form the committee, we multiply
the results from Step 1 and Step 2: Total number of ways = 84 ×15 = 1260.
Therefore, there are 1260 ways to form a committee of 3 students and 2
professors from a group of 9 students and 6 professors.
Question 35
Question
A committee of 5 people is to be formed from a group of 9 men and 7 women.
If the committee must contain at least 2 men and at least 2 women, how many
different committees could be formed?
Solution
Step 1: Calculate the number of ways to form a committee with 2 men and 3
women. The number of ways to choose 2 men from 9 is 9
2and the number of
ways to choose 3 women from 7 is 7
3. Therefore, the number of ways to form
a committee with 2 men and 3 women is 9
2×7
3.
Step 2: Calculate the number of ways to form a committee with 3 men and
2 women. The number of ways to choose 3 men from 9 is 9
3and the number of
ways to choose 2 women from 7 is 7
2. Therefore, the number of ways to form
a committee with 3 men and 2 women is 9
3×7
2.
Step 3: Calculate the total number of committees that can be formed. Add
the number of committees with 2 men and 3 women to the number of committees
with 3 men and 2 women: 9
2×7
3+9
3×7
2.
Step 4: Compute the final answer. Calculate the values of the combinations
and sum them to get the total number of different committees that could be
formed.
Therefore, the total number of different committees that could be formed is
9
2×7
3+9
3×7
2.
23
Question 2
Question
A committee of 4 people is to be formed from a group of 7 women and 5 men.
If at least one man must be included in the committee, how many different
committees can be formed?
Solution
Step 1: Calculate the total number of committees without any restrictions.
First, we calculate the total number of committees that can be formed with-
out any restrictions. This can be done using combinations formula:
Total number of committees = 12
4=12!
4!(12 −4)! =12!
4!8! = 495
So, there are 495 different committees that can be formed without any re-
strictions.
Step 2: Calculate the number of committees with only women.
Next, we calculate the number of committees that consist only of women.
This can be done using combinations formula:
Number of committees with only women = 7
4=7!
4!(7 −4)! =7!
4!3! = 35
Step 3: Calculate the number of committees with no men.
Now, we calculate the number of committees with no men. This can be
done by subtracting the number of committees with only women from the total
number of committees:
Number of committees with no men = Total number of committees−Number of committees with only women = 495−35 = 460
Step 4: Calculate the number of committees with at least one man.
Since we want to find the number of committees with at least one man, we
can subtract the number of committees with no men from the total number of
committees:
Number of committees with at least one man = Total number of committees−Number of committees with no men = 495−460 = 35
Therefore, there are 35 different committees that can be formed with at least
one man included.
2
Question 3
Question
In a group of 12 students, how many ways can we choose a committee of 5
students if 2 particular students must be included?
Solution
Step 1: First, we select the 2 particular students who must be included in the
committee. This can be done in 12
2ways.
Step 2: After selecting the 2 particular students, we need to select 3 more
students to complete the committee. Since we have already selected 2 students,
we are left with 10 students to choose from. We can select 3 more students in
10
3ways.
Step 3: To find the total number of ways to choose the committee of 5
students, we multiply the number of ways from Step 1 and Step 2. Therefore,
the total number of ways to choose a committee of 5 students with 2 particular
students included is 12
2×10
3.
Solving this expression, we get: 12
2=12!
2!(12−2)! =12×11
2×1= 66
10
3=10!
3!(10−3)! =10×9×8
3×2×1= 120
Therefore, the total number of ways to choose the committee is: 66 ×120 =
7920
Hence, there are 7920 ways to choose a committee of 5 students with 2
particular students included from a group of 12 students.
Question 4
Question
In a group of 10 people, how many ways can we choose a president, a vice
president, and a treasurer if no person can hold more than one position?
Solution
Step 1: To find the number of ways to choose a president, we simply have 10
choices.
Step 2: After choosing a president, there are 9 remaining people to choose
from for the position of vice president.
Step 3: Once the president and vice president have been chosen, there are 8
remaining people to choose from for the position of treasurer.
Step 4: To find the total number of ways to choose a president, a vice
president, and a treasurer, we multiply the number of choices at each step:
10 ×9×8 = 720
3
Therefore, there are 720 ways to choose a president, a vice president, and
a treasurer in a group of 10 people.
Question 5
Question
A committee of 5 people is to be formed from a group of 10 women and 8 men. If
the committee must consist of at least 3 women, how many different committees
can be formed?
Solution
Step 1: Calculate the number of committees with exactly 3 women: There are
10
3ways to choose 3 women from the 10 available women, and 8
2ways to
choose 2 men from the 8 available men. Therefore, there are 10
3×8
2ways to
form a committee with exactly 3 women.
Step 2: Calculate the number of committees with exactly 4 women: There
are 10
4ways to choose 4 women from the 10 available women, and 8
1ways
to choose 1 man from the 8 available men. Hence, there are 10
4×8
1ways to
create a committee with exactly 4 women.
Step 3: Calculate the number of committees with all 5 women: There is only
10
5way to choose all 5 women from the 10 available women.
Step 4: Add the results from Step 1, Step 2, and Step 3 to find the total
number of valid committees:
10
3×8
2+10
4×8
1+10
5
Calculating the above expression will give us the total number of different
committees that can be formed in this scenario.
Question 6
Question
In a group of 10 people, how many ways can we choose a committee of 4 people
such that 2 of them are women and 2 are men?
Solution
To solve this problem, we can break it down into two steps: selecting the 2
women from a group of 5 women and selecting the 2 men from a group of 5
men.
Step 1: Selecting 2 women from 5 women There are 5
2ways to select 2
women from a group of 5 women.
4
Step 2: Selecting 2 men from 5 men Similarly, there are 5
2ways to select 2
men from a group of 5 men.
To find the total number of ways to choose a committee of 4 people (2 women
and 2 men), we multiply the number of ways from Step 1 and Step 2.
Therefore, the total number of ways to choose the committee is:
5
2×5
2=5!
2!3! ×5!
2!3! =5×4
2×5×4
2= 10 ×10 = 100
So, there are 100 ways to choose a committee of 4 people consisting of 2
women and 2 men from a group of 10 people.
Question 7
Question
A committee of 5 people is to be formed from a group of 10 men and 8 women. If
the committee must consist of at least 2 women, how many different committees
can be formed?
Solution
Step 1: First, we calculate the number of ways to form a committee with exactly
2 women. There are 8 women to choose 2 from, and 10 men to choose 3 from.
Thus, the number of committees with exactly 2 women is 8
2×10
3.
Step 2: Next, we calculate the number of ways to form a committee with
exactly 3 women. There are 8 women to choose 3 from, and 10 men to choose
2 from. Thus, the number of committees with exactly 3 women is 8
3×10
2.
Step 3: Now, we calculate the total number of committees by adding the
number of committees with exactly 2 women and exactly 3 women. Total num-
ber of committees = 8
2×10
3+8
3×10
2.
Step 4: Plug in the values and simplify. Total number of committees =
8
2×10
3+8
3×10
2
Total number of committees = 8!
2!6! ×10!
3!7! +8!
3!5! ×10!
2!8!
Total number of committees = 28 ×120 + 56 ×45
Total number of committees = 3360 + 2520
Total number of committees = 5880.
Therefore, there are 5880 different committees that can be formed with at
least 2 women.
Question 8
Question
In how many ways can 5 distinct books be arranged on a bookshelf such that
two particular books must always be next to each other?
5
Solution
Step 1: Treat the two particular books as a single entity. Step 2: Consider the
arrangement as a combination problem.
There are two particular books that must be next to each other. Let’s treat
these two books as a single entity. So, we now have 4 entities to arrange: AB,
C, D, and E. Step 3: We have 4 entities to arrange in total: AB, C, D, and E.
Step 4: There are 4! ways to arrange these entities, but within the AB entity,
the books can be arranged in 2! ways. Step 5: Therefore, the total number of
ways to arrange the books is 4! ×2! = 24 ×2 = 48 ways.
Hence, there are 48 ways to arrange the 5 distinct books on the bookshelf
such that two particular books must always be next to each other.
Question 9
Question
A committee of 6 people is to be formed from a group of 10 men and 5 women. If
the committee must consist of at least 2 men and 2 women, how many different
committees can be formed?
Solution
Step 1: Calculate the number of ways to choose 2 men out of 10. There are 10
2
ways to choose 2 men from the group of 10 men.
Step 2: Calculate the number of ways to choose 2 women out of 5. There
are 5
2ways to choose 2 women from the group of 5 women.
Step 3: Calculate the number of ways to choose the remaining 2 people (1
man and 1 woman) out of the remaining 8 people (8=10-2 men; 8=5-2 women).
There are 8
2ways to choose the remaining 1 man and 1 woman.
Step 4: Calculate the total number of ways to form the committee. The total
number of ways to form the committee is the product of the ways to choose 2
men, 2 women, and 1 man and 1 woman: Total number of ways = 10
2×5
2×8
2
Calculating this expression gives: Total number of ways = 10!
2!8! ×5!
2!3! ×8!
2!6!
Total number of ways = 45 ×10 ×28 Total number of ways = 12600
Therefore, there are 12,600 different committees that can be formed.
Question 10
Question
A committee of 5 people is to be formed from a group of 10 math students and
6 computer science students. If the committee must consist of at least 2 math
students and 2 computer science students, how many different committees can
be formed?
6
Solution
Step 1: Calculate the number of ways to choose 2 math students out of 10.
There are 10
2ways to choose 2 math students.
Step 2: Calculate the number of ways to choose 2 computer science students
out of 6. There are 6
2ways to choose 2 computer science students.
Step 3: Calculate the number of ways to choose the remaining committee
member. There are 8 students remaining (4 math students and 4 computer
science students). We can choose one person from this group in 8
1ways.
Step 4: Calculate the total number of committees that can be formed. The
total number of committees can be formed by multiplying the results from Step
1, Step 2, and Step 3:
10
2×6
2×8
1= 45 ×15 ×8 = 5400.
Therefore, there are 5400 different committees that can be formed.
Question 11
Question
A committee of 5 people is to be formed from a group of 10 people. If 3 of the
people are friends and refuse to serve on the committee together, how many
different committees can be formed?
Solution
Step 1: To find the number of ways to form a committee, we first calculate the
total number of ways to choose 5 people from a group of 10. This is represented
by 10
5.
Total number of ways to choose 5 people from 10 = 10
5=10!
5!(10 −5)! = 252
Step 2: Now, let’s calculate the number of ways to form a committee where
the 3 friends cannot serve together. - We break this problem into cases: Case 1:
0 friends are on the committee - The number of ways to choose 5 people from
the remaining 7 (10-3 friends) is 7
5. Case 2: 1 friend is on the committee - The
number of ways to choose 4 people from the remaining 7 (10-3 friends and 1
friend already on the committee) is 7
4. Case 3: 2 friends are on the committee
- The number of ways to choose 3 people from the remaining 7 (10-3 friends and
2 friends already on the committee) is 7
3. - We sum the results from all these
cases to get the total number of committees that can be formed.
Number of committees without the friends together = 7
5+7
4+7
3
7
=7!
5!(7 −5)! +7!
4!(7 −4)! +7!
3!(7 −3)!
= 21 + 35 + 35 = 91
Therefore, there are 91 different committees that can be formed from the
group of 10 people with the condition that 3 friends cannot serve together.
Question 12
Question
A committee of 4 people is to be formed from a group of 8 men and 6 women.
In how many ways can the committee be formed if it must consist of 2 men and
2 women?
Solution
Step 1: Calculate the number of ways to choose 2 men from 8. Step 2: Calculate
the number of ways to choose 2 women from 6. Step 3: Multiply the results
from Step 1 and Step 2 to get the total number of ways to form the committee.
Step 1: Choosing 2 men from 8 can be done using a combination. The
number of ways to choose 2 men from 8 is given by
8
2=8!
2!(8 −2)! =8×7
2×1= 28.
Step 2: Choosing 2 women from 6 can be done using a combination. The
number of ways to choose 2 women from 6 is given by
6
2=6!
2!(6 −2)! =6×5
2×1= 15.
Step 3: The total number of ways to form the committee consisting of 2 men
and 2 women is the product of the results from Step 1 and Step 2,
28 ×15 = 420.
Therefore, there are 420 ways to form the committee with 2 men and 2
women from the group of 8 men and 6 women.
Question 13
Question
A group of 6 people are going on a road trip and plan to stop at 4 different
cities along the way. If each person can visit any city, how many different ways
can they split up into groups to visit the cities?
8
Solution
We can solve this problem using combinations.
Step 1: Since the order in which the cities are visited does not matter, we
will use the concept of combinations. The number of ways to split 6 people into
groups to visit 4 cities is equal to the number of ways to distribute the people
among the cities with repetition allowed.
Step 2: We can visualize this problem as placing 3 dividers among the 6
people, representing the 4 different cities. The number of ways to place the
dividers determines the number of ways to split the people into groups.
Step 3: Thus, the total number of ways to split the 6 people into groups to
visit the 4 cities is given by the formula for combinations with repetition:
n+r−1
r=6+4−1
4=9
4=9×8×7×6
4×3×2×1= 126
Step 4: Therefore, there are 126 different ways for the group of 6 people to
split up into groups to visit the 4 cities along the road trip.
Question 14
Question
In how many ways can 5 men and 5 women be seated around a circular table if
each man must be seated next to at least one woman?
Solution
Step 1: First, we will fix one woman in a seat. Then, we can count the number
of ways the other 4 women can be seated around the table.
Step 2: There are 4 women remaining who can be seated in 4! ways.
Step 3: Now, we will arrange the 5 men among the 5 available slots between
the women.
Step 4: There are 5 slots between the women where the men can be seated.
The men can be arranged in 5! ways in these slots.
Step 5: Finally, since the table is circular, we must divide by 5 to account
for the rotations of the same arrangement.
Therefore, the total number of ways the 5 men and 5 women can be seated
around a circular table with each man seated next to at least one woman is
given by: 4! ×5!
5= 2880
9
Question 15
Question
A basketball team has 12 players. In how many ways can a starting lineup of
5 players be chosen if the lineup must include the team captain and the point
guard?
Solution
We can approach this problem by considering the positions of the players in the
starting lineup and filling them one by one. Step 1: Choose the team captain
(1 way) Step 2: Choose the point guard from the remaining 11 players (11
ways) Step 3: Choose the remaining 3 players from the remaining 10 players
(10 choices for the first remaining player, 9 for the second, and 8 for the third)
Therefore, the total number of ways to choose the starting lineup is:
1×11 ×10 ×9×8=7,920
So, there are 7,920 ways to choose the starting lineup.
Question 16
Question
In how many ways can you choose a committee of 4 people from a group of 10
people if 2 of the group members refuse to serve on the committee together?
Solution
Step 1: First, we calculate the total number of ways to choose a committee of 4
people from 10 people without any restrictions. This is a combination problem,
and can be calculated using the formula for combinations:
n
r=n!
r!(n−r)!
where nis the total number of people and ris the number of people we want
to choose.
For this problem, n= 10 and r= 4. So, the total number of ways to choose
a committee without restrictions is:
10
4=10!
4!(10 −4)!
=10 ×9×8×7
4×3×2×1
10
= 210
So, there are 210 ways to choose a committee of 4 people from a group of 10
people without restrictions.
Step 2: Next, we calculate the number of ways that the 2 group members
who refuse to serve on the committee together are chosen. Let’s call these 2
people A and B. There are 8 remaining people to choose from for the committee,
plus we need to choose 2 more people from the remaining 8.
This can be calculated by finding the number of ways to choose a committee
of 4 people from the 8 remaining people:
8
2=8!
2!(8 −2)!
=8×7
2×1
= 28
Step 3: Finally, we subtract the number of committees that include both A
and B from the total number of committees to get the final answer. This can
be calculated as:
Total ways −Ways with A and B together = 210 −28 = 182
Therefore, there are 182 ways to choose a committee of 4 people from a group
of 10 people if 2 group members refuse to serve on the committee together.
Question 17
Question
A committee of 5 people is to be chosen from a group of 10 men and 8 women.
If the committee must have at least 3 men, how many different committees can
be formed?
Solution
Step 1: Find the number of ways to choose a committee with exactly 3 men.
There are 10
3ways to choose 3 men from 10 men, and 8
2ways to choose 2
women from 8 women. Therefore, the number of ways to choose a committee
with exactly 3 men is:
10
3×8
2
Step 2: Find the number of ways to choose a committee with 4 men. There
are 10
4ways to choose 4 men from 10 men, and 8
1ways to choose 1 woman
11
from 8 women. Therefore, the number of ways to choose a committee with 4
men is: 10
4×8
1
Step 3: Find the number of ways to choose a committee with 5 men. There
are 10
5ways to choose 5 men from 10 men. Therefore, the number of ways to
choose a committee with 5 men is:
10
5
Step 4: Add the results from Steps 1, 2, and 3 to find the total number of
different committees that can be formed with at least 3 men. Total number of
committees = 10
3×8
2+10
4×8
1+10
5
Question 18
Question
A committee of 5 students is to be formed from a group of 10 students. If
3 of the students are math majors and 7 are computer science majors, how
many different committees can be formed if: 1. The committee must have 3
math majors and 2 computer science majors? 2. The committee can have any
combination of math and computer science majors?
Solution
1. To find the number of different committees that can be formed with 3 math
majors and 2 computer science majors, we calculate the number of ways to
choose 3 math majors out of 3 and 2 computer science majors out of 7.
Step 1: Choose 3 math majors out of 3 Since there are only 3 math
majors, the number of ways to choose 3 math majors out of 3 is 1.
Step 2: Choose 2 computer science majors out of 7 We need to
choose 2 computer science majors out of 7. This can be done in 7
2= 21 ways.
Step 3: Calculate the total number of committees The total number
of committees that can be formed is the product of the number of ways to
choose math majors and computer science majors. Therefore, the total number
of committees is 1 ∗21 = 21.
Therefore, there are 21 different committees that can be formed with 3 math
majors and 2 computer science majors.
2. To find the total number of different committees that can be formed
with any combination of math and computer science majors, we can consider
all possible cases separately and then add them up.
Case 1: All members are math majors In this case, we need to choose
5 math majors out of 3. However, since there are only 3 math majors, this case
is not possible.
12
Case 2: 4 math majors and 1 computer science major We need to
choose 4 math majors out of 3 and 1 computer science major out of 7. The
number of ways to do this is 3
4×7
1= 0.
Case 3: 3 math majors and 2 computer science majors We have
already calculated this case in part 1. The number of ways is 21.
Case 4: 2 math majors and 3 computer science majors We need to
choose 2 math majors out of 3 and 3 computer science majors out of 7. The
number of ways to do this is 3
2×7
3= 63.
Case 5: 1 math major and 4 computer science majors We need to
choose 1 math major out of 3 and 4 computer science majors out of 7. The
number of ways to do this is 3
1×7
4= 105.
Case 6: All members are computer science majors In this case, we
need to choose 5 computer science majors out of 7. The number of ways to do
this is 7
5= 21.
Step 4: Calculate the total number of committees The total number
of committees is the sum of all possible cases. Therefore, the total number of
different committees is 0 + 21 + 63 + 105 + 21 = 210.
Therefore, there are 210 different committees that can be formed with any
combination of math and computer science majors.
Question 19
Question
In a group of 10 people, how many ways are there to form a committee of 3
people with a president, a secretary, and a treasurer?
Solution
Step 1: First, choose the 3 people to be on the committee. Since the order in
which the people are chosen matters (president, secretary, treasurer), we use
permutations. There are 10 people to choose from, and we want to choose 3:
10P3.
Step 2: Calculate the number of ways to assign the positions of president,
secretary, and treasurer. Once the committee members have been chosen, we
need to assign the specific roles. There are 3 positions to fill and 3 people to
assign to them. Therefore, the number of ways to assign the positions is 3!.
Step 3: Multiply the results from Step 1 and Step 2. The total number of
ways to form the committee of 3 people with specific roles is given by:
10P3×3! = 10!
(10 −3)! ×3! = 10 ×9×8×7!
7! ×3×2×1 = 720 ×6 = 4320
Therefore, there are 4320 ways to form a committee of 3 people with a
president, a secretary, and a treasurer from a group of 10 people.
13
Question 20
Question
In a group of 12 people, how many ways can we select a committee of 4 people
if 2 particular people must always be included in the committee?
Solution
Step 1: First, we select the 2 particular people that must be included in the
committee. This can be done in 2
2= 1 way.
Step 2: Next, we need to select the remaining 2 people from the remaining
10 people. This can be done in 10
2ways.
Step 3: To find the total number of ways to form the committee, we multiply
the results of Step 1 and Step 2.
Therefore, the total number of ways to form a committee of 4 people with 2
particular people always included is 1 ×10
2= 1 ×45 = 45 ways.
Question 21
Question
In a group of 10 friends, how many ways can we choose a committee of 4 people
if one of the friends, Alice, insists on being part of the committee?
Solution
Step 1: Since one person, Alice, is already guaranteed to be part of the com-
mittee, we need to choose the remaining 3 people from the remaining 9 friends.
Step 2: To find the number of ways to choose these 3 people, we use the combi-
nation formula n
r=n!
r!(n−r)! . Step 3: Substituting n= 9 (9 remaining friends)
and r= 3 (to choose 3 people) into the formula:
9
3=9!
3!(9 −3)! =9×8×7
3×2×1= 84.
Therefore, there are 84 ways to choose a committee of 4 people with Alice being
one of them.
Question 22
Question
In how many ways can 5 students be seated in a row of 8 chairs such that two
particular students, Alice and Bob, must sit next to each other?
14
Solution
Step 1: Treat Alice and Bob as one entity. Step 2: Calculate the number of
ways to arrange the remaining entities. Step 3: There are 7 entities to arrange
(including the entity composed of Alice and Bob) in 7 seats.
Step 1: Treat Alice and Bob as one entity. Step 2: There are 7 entities in
total to be arranged. Step 3: The number of ways to arrange these 7 entities in
7 seats is 7!.
Therefore, the total number of ways for 5 students to be seated in a row of
8 chairs such that Alice and Bob must sit next to each other is 2 ·7!.
Question 23
Question
How many five-letter words can be formed using the letters in the word ”AL-
GEBRA” if no letter can be used more times than it appears in ”ALGEBRA”?
Solution
Step 1: Determine the number of ways to choose 5 letters from ”ALGEBRA”.
Since we cannot repeat any letters, we need to consider the unique letters in
”ALGEBRA”. The word ”ALGEBRA” has 6 different letters, so there are 6
5
ways to choose 5 letters from ”ALGEBRA”.
6
5=6!
5!(6−5)! = 6
Step 2: Calculate the number of ways to arrange the chosen 5 letters.
Since we have chosen 5 letters, there are 5 positions to fill in the 5-letter
word. The number of ways to arrange the 5 chosen letters is 5!.
Step 3: Compute the total number of five-letter words.
The total number of five-letter words that can be formed using the letters
in ”ALGEBRA” is the product of the number of ways to choose 5 letters and
the number of ways to arrange those chosen letters.
Total number of five-letter words = 6
5×5!
Total number of five-letter words = 6 ×5!
Total number of five-letter words = 6 ×5×4×3×2
Total number of five-letter words = 720
Therefore, there are 720 five-letter words that can be formed using the letters
in the word ”ALGEBRA” if no letter can be used more times than it appears
in ”ALGEBRA”.
Question 24
Question
A committee of 5 people is to be formed from a group of 10 students and 8
professors. If the committee must consist of 3 students and 2 professors, how
15
many different committees can be formed?
Solution
Step 1: Calculate the number of ways to choose 3 students out of 10. Step 2:
Calculate the number of ways to choose 2 professors out of 8. Step 3: Mul-
tiply the results from Step 1 and Step 2 to find the total number of different
committees that can be formed.
Step 1: The number of ways to choose 3 students out of 10 is given by 10
3.
10
3=10!
3!(10−3)! =10×9×8
3×2×1= 120.
Therefore, there are 120 ways to choose 3 students.
Step 2: The number of ways to choose 2 professors out of 8 is given by 8
2.
8
2=8!
2!(8−2)! =8×7
2×1= 28.
Therefore, there are 28 ways to choose 2 professors.
Step 3: To find the total number of different committees that can be formed,
we multiply the results from Step 1 and Step 2.
Total number of committees = 120 ×28 = 3360.
Therefore, there are 3360 different committees that can be formed consisting
of 3 students and 2 professors from the given group.
Question 25
Question
In how many ways can 5 red, 3 blue, and 4 white identical balls be arranged in
a row?
Solution
Step 1: Find the total number of arrangements without considering the colors.
Since there are 5 red balls, 3 blue balls, and 4 white balls, the total number
of balls is 5 + 3 + 4 = 12.
Therefore, the total number of ways to arrange the balls without considering
colors is 12 factorial, denoted as 12!.
Step 2: Consider the indistinguishable arrangement of each color.
Since the red balls are identical, dividing the total number of arrangements
by the factorial of the number of red balls (5! for red balls) will give us the
number of distinct arrangements of the red balls.
Similarly, dividing the resulting number by the factorial of the number of
blue balls (3! for blue balls) and the factorial of the number of white balls (4!
for white balls) will give us the number of distinct arrangements of the blue
balls and white balls, respectively.
Hence, the total number of ways to arrange the balls, taking their colors into
account, is:
16
12!
5! ×3! ×4!
Question 26
Question
In a group of 10 people, how many ways can 4 people be chosen to form a
committee with a president, a vice president, and two regular members?
Solution
Step 1: Choose the president There are 10 ways to choose the president from
the group of 10 people.
Step 2: Choose the vice president After choosing the president, there are 9
people left for the vice president role. Therefore, there are 9 ways to choose the
vice president.
Step 3: Choose the two regular members After choosing the president and
vice president, there are 8 people left for choosing the first regular member, and
7 people left for choosing the second regular member. So, there are 8
2×1 ways
to choose the two regular members.
Step 4: Calculate the total number of ways The total number of ways to
choose the committee is the product of the number of ways to choose the pres-
ident, vice president, and two regular members. Therefore, the total number of
ways is:
10 ×9×8
2×1 = 10 ×9×8×7
2×1= 10 ×9×28 = 2520
Therefore, there are 2520 ways to form a committee with a president, a vice
president, and two regular members from a group of 10 people.
Question 27
Question
A committee of 4 people is to be formed from a group of 8 men and 5 women.
If the committee must contain at least one man and at least one woman, how
many different committees can be formed?
Solution
Step 1: Find the total number of ways to form a committee of 4 people. To
form a committee of 4 people from a group of 8 men and 5 women, we will use
combinations. The total number of ways to choose 4 people from 13 people is
given by 13
4.
17
Step 2: Find the number of ways to form a committee with only men or only
women. We need to find separately the number of ways to form a committee
with only men and only women. For a committee with only men, we choose 4
men from 8, which can be done in 8
4ways. For a committee with only women,
we choose 4 women from 5, which can be done in 5
4ways. Therefore, the total
number of committees with only men or only women is given by 8
4+5
4.
Step 3: Find the number of ways to form a committee with at least one man
and at least one woman. The total number of ways to form a committee with at
least one man and at least one woman is the total number of committees minus
the number of committees with only men or only women. So, the number of
different committees that can be formed is:
13
4−8
4+5
4
Question 28
Question
In a class of 20 students, how many ways are there to form a committee of 5
students if 2 specific students refuse to work together on the same committee?
Solution
To determine the number of ways to form a committee of 5 students with the
given restriction, we need to consider two cases: 1. The two specific students
are in the committee. 2. The two specific students are not in the committee.
Step 1: Case 1 - The two specific students are in the committee.
Choose the other 3 students from the remaining 18 students (20−2 = 18). This
can be done in 18
3ways.
Step 2: Case 2 - The two specific students are not in the committee.
Choose 5 students from the remaining 18 students, excluding the 2 specific
students. This can be done in 18
5ways.
Step 3: Determine the total number of ways. Therefore, the total
number of ways to form a committee of 5 students where the two specific stu-
dents refuse to work together is the sum of the two cases:
18
3+18
5
Calculating these values:
18
3=18!
3!(18 −3)! =18 ×17 ×16
3×2×1= 816
18
5=18!
5!(18 −5)! =18 ×17 ×16 ×15 ×14
5×4×3×2×1= 8568
18
Adding these two values:
816 + 8568 = 9384
Therefore, there are 9,384 ways to form a committee of 5 students with the
given restriction.
Question 29
Question
A committee of 5 people is to be formed from a group of 10 men and 8 women.
1. In how many ways can the committee be formed if there must be at least
2 men and 2 women?
2. In how many ways can the committee be formed if there must be exactly
3 men?
Solution
Let’s solve each part of the question separately.
Part (a)
1. Choose 2 men out of 10: 10
2
2. Choose 2 women out of 8: 8
2
3. Choose the remaining person (can be a man or a woman): 10 + 8 −2 = 16
The total number of ways the committee can be formed with at least 2 men and
2 women is: 10
2·8
2·16
Part (b)
1. Choose 3 men out of 10: 10
3
2. Choose 2 women out of 8: 8
2
The total number of ways the committee can be formed with exactly 3 men is:
10
3·8
2
19
Question 30
Question
In a group of 8 friends, how many ways can you choose a committee consisting
of a president, a vice-president, and a treasurer?
Solution
To solve this problem, we can use the concept of permutations. Since the order
in which the friends are chosen matters (president, vice-president, treasurer),
we will use the permutations formula.
Step 1: Find the total number of ways to choose a president from 8 friends.
There are 8 friends to choose from for the president position.
Step 2: Find the total number of ways to choose a vice-president from the
remaining 7 friends. After choosing the president, there are 7 friends remaining
for the vice-president position.
Step 3: Find the total number of ways to choose a treasurer from the
remaining 6 friends. After choosing the president and vice-president, there are
6 friends remaining for the treasurer position.
Step 4: Calculate the total number of ways to choose the committee. To
find the total number of ways to choose the committee consisting of a president,
a vice-president, and a treasurer, we multiply the number of choices at each
step. Total ways = Number of ways to choose president ×Number of ways to
choose vice-president ×Number of ways to choose treasurer
Total ways = 8 ×7×6 = 336
Therefore, there are 336 ways to choose a committee consisting of a president,
a vice-president, and a treasurer from a group of 8 friends.
Question 31
Question
A committee of 7 people is to be formed from a group of 12 individuals. If 3 of
the individuals refuse to serve on the committee together, how many different
committees can be formed?
Solution
Step 1: Find the number of ways to form a committee without any restrictions.
Since there are 12 individuals and we are forming a committee of 7 people, we
can calculate the number of ways as:
12
7=12!
7!(12 −7)! = 792
20
Step 2: Find the number of ways when 3 individuals refuse to serve on the
committee together.
Let’s consider the 3 individuals who refuse to serve as a single entity. Then we
have 10 individuals remaining to choose from. Since we still need to choose 7
people for the committee, the number of ways can be calculated as:
10
7=10!
7!(10 −7)! = 120
Step 3: Subtract the number of committees that include all 3 refusing indi-
viduals from the total number of committees.
Therefore, the number of different committees that can be formed when 3 of
the individuals refuse to serve on the committee together is:
792 −120 = 672
Thus, there are 672 different committees that can be formed under the given
conditions.
Question 32
Question
In a committee of 8 people, there are 4 mathematicians and 4 computer sci-
entists. If the committee needs to select a president, a vice president, and a
treasurer, how many ways can this be done if the president cannot be a mathe-
matician and the treasurer cannot be a computer scientist?
Solution
Step 1: Calculate the number of ways to choose the president from the computer
scientists. Since the president cannot be a mathematician, we must choose the
president from the 4 computer scientists. This can be done in 4
1= 4 ways.
Step 2: Calculate the number of ways to choose the treasurer from the
mathematicians. Similarly, since the treasurer cannot be a computer scientist,
we must choose the treasurer from the 4 mathematicians. This can be done in
4
1= 4 ways.
Step 3: Calculate the number of ways to choose the vice president. After
choosing the president and treasurer, the vice president will be chosen from the
remaining pool of 6 people (2 mathematicians and 4 computer scientists). This
can be done in 6
1= 6 ways.
Step 4: Multiply the number of choices for each position to find the total
number of ways to select the committee positions. The total number of ways to
select a president, vice president, and treasurer following the given restrictions
is 4 ×4×6 = 96 ways.
21
Question 33
Question
In a survey of 10 students, each student was asked to rank their top 3 favorite
books. If the order in which the books are ranked matters, how many possible
rankings are there in total?
Solution
Step 1: Determine the number of ways to choose the first-ranked book. Since
each student has 3 choices and the order matters, there are 3 ways to choose
the first-ranked book.
Step 2: Determine the number of ways to choose the second-ranked book.
After selecting the first-ranked book, there are 2 remaining books to choose
from for the second rank. Thus, there are 2 ways to choose the second-ranked
book.
Step 3: Determine the number of ways to choose the third-ranked book.
After selecting the first and second-ranked books, there is 1 book remaining
to choose from for the third rank. Thus, there is only 1 way to choose the
third-ranked book.
Step 4: Calculate the total number of possible rankings. To find the total
number of possible rankings, we multiply the number of ways for each step.
Total number of possible rankings = 3 (ways for first rank) ×2 (ways for
second rank) ×1 (ways for third rank) = 6
Therefore, there are 6possible rankings in total.
Question 34
Question
A committee of 5 people is to be formed from a group of 9 students and 6
professors. How many ways can the committee be formed if it must consist of
3 students and 2 professors?
Solution
Step 1: Calculate the number of ways to choose 3 students out of 9. Step 2:
Calculate the number of ways to choose 2 professors out of 6. Step 3: Multiply
the results from Step 1 and Step 2 to find the total number of ways to form the
committee.
Step 1: The number of ways to choose 3 students out of 9 can be calculated
using combinations. The number of ways is given by 9
3=9!
3!6! = 84.
Step 2: The number of ways to choose 2 professors out of 6 can be calculated
using combinations. The number of ways is given by 6
2=6!
2!4! = 15.
22
Step 3: To find the total number of ways to form the committee, we multiply
the results from Step 1 and Step 2: Total number of ways = 84 ×15 = 1260.
Therefore, there are 1260 ways to form a committee of 3 students and 2
professors from a group of 9 students and 6 professors.
Question 35
Question
A committee of 5 people is to be formed from a group of 9 men and 7 women.
If the committee must contain at least 2 men and at least 2 women, how many
different committees could be formed?
Solution
Step 1: Calculate the number of ways to form a committee with 2 men and 3
women. The number of ways to choose 2 men from 9 is 9
2and the number of
ways to choose 3 women from 7 is 7
3. Therefore, the number of ways to form
a committee with 2 men and 3 women is 9
2×7
3.
Step 2: Calculate the number of ways to form a committee with 3 men and
2 women. The number of ways to choose 3 men from 9 is 9
3and the number of
ways to choose 2 women from 7 is 7
2. Therefore, the number of ways to form
a committee with 3 men and 2 women is 9
3×7
2.
Step 3: Calculate the total number of committees that can be formed. Add
the number of committees with 2 men and 3 women to the number of committees
with 3 men and 2 women: 9
2×7
3+9
3×7
2.
Step 4: Compute the final answer. Calculate the values of the combinations
and sum them to get the total number of different committees that could be
formed.
Therefore, the total number of different committees that could be formed is
9
2×7
3+9
3×7
2.
23
