MATH 201 - INTRODUCTION TO
PROBABILITY AND STATISTICS -
Combinatorial Analysis
Question Bank - Set 2
Liberty University
Question 1
Question
In a group of 10 people, how many ways can you choose a committee of 4 people
with 2 people chosen as chairperson and vice-chairperson?
Solution
Step 1: Calculate the number of ways to choose a committee of 4 people from
a group of 10 people. To choose a committee of 4 people from 10, we use the
combination formula:
(10
4)=10!
4!(10 −4)! =10 ×9×8×7
4×3×2×1= 210
So, there are 210 ways to choose a committee of 4 people.
Step 2: Calculate the number of ways to choose a chairperson and vice-
chairperson from the 4 committee members. To choose 2 people out of 4 for the
roles of chairperson and vice-chairperson, we use the permutation formula:
P(4,2) = 4!
2! =4×3×2×1
2×1= 12
So, there are 12 ways to choose a chairperson and vice-chairperson from the
committee members.
Step 3: Multiply the results from Step 1 and Step 2 to find the total number
of ways. Total number of ways = 210 (ways to choose a committee) ×12 (ways
to choose chairperson and vice-chairperson) Total number of ways = 2520
Therefore, there are 2520 ways to choose a committee of 4 people with 2
people as chairperson and vice-chairperson from a group of 10 people.
Question 2
Question
A committee of 5 people is to be formed from a group of 10 men and 8 women.
Find the number of ways the committee can be formed if it must consist of 3
men and 2 women.
Solution
To find the number of ways the committee can be formed, we need to use
combinatorial analysis.
Step 1: Find the number of ways to choose 3 men from the group
of 10 men.
10C3=10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
Step 2: Find the number of ways to choose 2 women from the
group of 8 women.
8C2=8!
2!(8 −2)! =8×7
2×1= 28
Step 3: Find the total number of ways the committee can be
formed. Since we want to choose 3 men and 2 women for the committee,
we multiply the results from Step 1 and Step 2. Total number of ways =
120 ×28 = 3360
Therefore, there are 3360 ways to form a committee of 5 people consisting
of 3 men and 2 women from the group of 10 men and 8 women.
Question 3
Question
In how many ways can 6 different books be arranged on a bookshelf if two of
the books must always be placed next to each other?
Solution
To solve this problem, we can treat the two books that must always be together
as one unit. This reduces the problem to arranging 5 units (including the pair
of books) on the bookshelf.
Step 1: Determine the number of ways to arrange the 5 units (including
the pair of books).
There are 5! ways to arrange the 5 units.
Step 2: Consider the pair of books that must always be together as a single
unit.
Within this unit, the two books can be arranged in 2! ways.
2
Step 3: Multiply the results from Step 1 and Step 2 to find the total number
of ways to arrange the books.
Therefore, the total number of ways to arrange the 6 different books on the
bookshelf such that two of the books are always placed next to each other is
5! ×2! = 120 ×2 = 240 ways.
Question 4
Question
In a group of 8 people, how many different committees of 3 people can be formed
if one person, Alice, must always be included in the committee?
Solution
Step 1: Since Alice must always be included in the committee, we already have
1 person chosen. We need to choose 2 more people to form the committee.
Step 2: To choose the remaining 2 people out of the 7 remaining people, we
use the combination formula. The number of ways to choose 2 people from 7 is
given by (7
2).
Step 3: Calculate (7
2).
(7
2)=7!
2!(7 −2)! =7×6
2×1= 21.
Step 4: Therefore, there are 21 different committees of 3 people that can be
formed with Alice always included.
Question 5
Question
A committee of 5 people is to be formed from a group of 10 individuals (5 men
and 5 women). If at least 1 man must be on the committee, how many different
committees can be formed?
Solution
Step 1: First, let’s calculate the total number of ways to form a committee
of 5 people from a group of 10 individuals. This can be calculated using the
combination formula: (n
r)=n!
r!(n−r)!
3
where nis the total number of individuals and ris the number of people in the
committee. In this case, n= 10 and r= 5, so the total number of ways is:
(10
5)=10!
5!5! =10 ×9×8×7×6
5×4×3×2×1= 252
Step 2: Next, let’s calculate the number of ways to form a committee of
5 people with no men on it. Since there are 5 women, we need to choose
all 5 committee members from the women. This can be calculated using the
combination formula: (5
5)= 1
There is only 1 way to choose all 5 women for the committee.
Step 3: Finally, we will calculate the number of ways to form a committee of
5 people with at least 1 man on it. This can be done by subtracting the number
of committees with no men from the total number of committees:
252 −1 = 251
Therefore, there are 251 different committees that can be formed with at least
1 man on it.
Question 6
Question
A group of 8 students are going to be split into two committees, each with 4
students. How many different ways can this be done?
Solution
Step 1: We first need to choose 4 students from the 8 to be on the first com-
mittee. There are (8
4)ways to choose 4 students out of 8.
Step 2: Once the first committee is chosen, the remaining 4 students auto-
matically form the second committee. Therefore, there is only 1 way to assign
the remaining students to the second committee.
Step 3: Compute the total number of ways to form the two committees by
multiplying the number of ways in Step 1 and Step 2. So, the total number of
ways is (8
4)×1.
Step 4: Calculate the resulting value.
(8
4)=8!
4!(8 −4)! =8×7×6×5
4×3×2×1= 70.
Therefore, there are 70 different ways to split the group of 8 students into
two committees, each with 4 students.
4
Question 7
Question
In a group of 10 people, how many ways can we choose a committee of 3 people
to represent the group?
Solution
Step 1: To solve this problem, we can use the combination formula. The number
of ways to choose a committee of rmembers from a group of npeople is given
by the formula:
(n
r)=n!
r!(n−r)!
where n!represents the factorial of n.
Step 2: Substituting n= 10 and r= 3 into the formula, we get:
(10
3)=10!
3!(10 −3)!
Step 3: Calculating the factorials:
(10
3)=10 ×9×8
3×2×1
(10
3)=720
6
(10
3)= 120
Step 4: Therefore, there are 120 ways to choose a committee of 3 people
from a group of 10 people.
Question 8
Question
How many different 7-letter arrangements can be formed using all the letters of
the word ”UNIVERSITY” such that no two vowels are adjacent?
5
Solution
Step 1: Find the total number of ways to arrange the letters of ”UNIVERSITY”
without any restrictions.
The word ”UNIVERSITY” has 10 letters, but there are repetitions of some
letters:-2U’s-2I’s-2E’s-1N-1V-1R-1S-1T
Therefore, the total number of ways to arrange the letters without restric-
tions is the factorial of the total number of letters divided by the factorials of
the number of repetitions:
10!/(2! ×2! ×2! ×1! ×1! ×1! ×1! ×1!)
Step 2: Find the number of ways to arrange the letters of ”UNIVERSITY”
such that the vowels are adjacent.
Since there are 4 vowels in ”UNIVERSITY” (U, I, E, and I), we can consider
them as one unit. Therefore, the letters ”UNVRSITY” together form 7 units.
The number of ways to arrange these 7 units is 7!.
Step 3: Subtract the result from Step 2 from the result of Step 1 to find the
number of arrangements where no two vowels are adjacent.
The number of ways to arrange the letters such that no two vowels are
adjacent is given by:
10!/(2! ×2! ×2! ×1! ×1! ×1! ×1!) −7!
Thus, the above expression represents the total number of different 7-letter
arrangements of the word ”UNIVERSITY” where no two vowels are adjacent.
Question 9
Question
A company needs to assemble a committee of 5 employees from a pool of 10
potential candidates. If 3 of the candidates are engineers and 7 are accountants,
how many ways can the committee be formed if it must consist of at least 1
engineer and 2 accountants?
Solution
Step 1: Calculate the number of ways to choose 1 engineer and 2 accountants:
There are 3 engineers to choose from and 7 accountants. We can choose 1
engineer in 3 ways and 2 accountants in (7
2)ways. So, the number of ways to
choose 1 engineer and 2 accountants is 3×(7
2).
Step 2: Calculate the number of ways to choose 2 engineers and 3 accoun-
tants: There are 3 engineers to choose from and 7 accountants. We can choose
2 engineers in (3
2)ways and 3 accountants in (7
3)ways. So, the number of ways
to choose 2 engineers and 3 accountants is (3
2)×(7
3).
6
Step 3: Add the results from step 1 and step 2 to find the total number of
ways to form the committee: Total number of ways = 3×(7
2)+(3
2)×(7
3).
Step 4: Calculate the total number of ways to form the committee: 3×(7
2)=
3×7!
2!(7−2)! = 3 ×7×6
2×1= 63 ways
(3
2)×(7
3)=3!
2!(3−2)! ×7!
3!(7−3)! = 3 ×35 = 105 ways
Total number of ways = 63 + 105 = 168.
Therefore, there are 168 ways to form the committee consisting of at least 1
engineer and 2 accountants.
Question 10
Question
A committee of 5 people is to be formed from a group of 7 men and 4 women.
If the committee must consist of at least 3 men, how many different committees
can be formed?
Solution
Step 1: Find the total number of committees without any restrictions. The
total number of ways to form a committee of 5 people from 7 men and 4 women
is given by the combination formula:
(11
5)=11!
5!(11 −5)! =11!
5!6! = 462
Step 2: Find the number of committees with less than 3 men. The number
of committees with less than 3 men can be found by: - Choosing 0 men from
7 men and 5 women from 4 women, or - Choosing 1 man from 7 men and 4
women from 4 women, or - Choosing 2 men from 7 men and 3 women from 4
women.
(7
0)×(4
5)+(7
1)×(4
4)+(7
2)×(4
3)= 1×4+7×1 + 21 ×4 = 4+7 + 84 = 95
Step 3: Calculate the number of committees with at least 3 men. The
number of committees with at least 3 men is the total number of committees
minus the number of committees with less than 3 men:
462 −95 = 367
Therefore, there are 367 different committees that can be formed with at
least 3 men.
7
Question 11
Question
A committee of 5 people is to be formed from a group of 8 men and 6 women.
How many ways can the committee be formed if it must consist of at least 3
women?
Solution
Step 1: Calculate the number of ways to select exactly 3 women from the 6
available. There are (6
3)ways to choose 3 women.
Step 2: Calculate the number of ways to choose the remaining 2 committee
members (either men or women). Since the committee must consist of at least
3 women, we can choose the remaining 2 members from the 8 men or the 3
remaining women. We will compute both possibilities: - If we choose 2 men
from the 8 available, there are (8
2)ways to select them. - If we choose 2 women
from the 3 remaining, there are (3
2)ways to select them.
Step 3: Find the total number of ways to form the committee. The total
number of ways to form the committee with at least 3 women is the product
of the number of ways from Step 1 and Step 2. Add the results of the two
possibilities: (6
3)×(8
2)+(6
3)×(3
2)ways.
Step 4: Calculate the final answer. - (6
3)=6!
3!(6−3)! =6×5×4
3×2×1= 20 ways. -
(8
2)=8!
2!(8−2)! =8×7
2×1= 28 ways. - (3
2)=3!
2!(3−2)! =3
2×1= 3 ways.
Therefore, the total number of ways to form the committee is 20 ×28 + 20 ×
3 = 640 + 60 = 700 ways.
Question 12
Question
In a group of 10 students, how many ways can we select a president, vice presi-
dent, and treasurer if no student can hold more than one position?
Solution
To solve this problem, we will use the fundamental principle of counting, also
known as the multiplication principle.
Step 1: Find the number of ways to select a president from 10 students.
Since each student can be selected as the president only once, there are 10
choices for president.
Step 2: Find the number of ways to select a vice president from the re-
maining 9 students. After selecting a president, there are 9 students remaining
to choose from for vice president.
8
Step 3: Find the number of ways to select a treasurer from the remaining
8 students. After selecting a president and vice president, there are 8 students
remaining to choose from for treasurer.
Therefore, the total number of ways to select a president, vice president, and
treasurer from a group of 10 students is:
10 ×9×8 = 720
So, there are 720 different ways to select a president, vice president, and
treasurer from the group of 10 students.
Question 13
Question
A committee of 5 members is to be selected from a group of 9 men and 8 women.
In how many ways can this committee be formed if it must consist of 3 men and
2 women?
Solution
Step 1: Find the number of ways to choose 3 men from 9 men. Step 2: Find the
number of ways to choose 2 women from 8 women. Step 3: Multiply the results
from Step 1 and Step 2 to find the total number of ways to form the committee.
Step 1: To choose 3 men from 9 men, we use the combination formula:
(n
k)=n!
k!(n−k)! . Therefore, the number of ways to choose 3 men from 9 men is
(9
3)=9!
3!(9−3)! =9×8×7
3×2×1= 84 ways.
Step 2: To choose 2 women from 8 women, we use the combination formula:
(n
k)=n!
k!(n−k)! . Therefore, the number of ways to choose 2 women from 8 women
is (8
2)=8!
2!(8−2)! =8×7
2×1= 28 ways.
Step 3: Multiply the results from Step 1 and Step 2. Total number of ways
to form the committee = 84 ×28 = 2352 ways.
Therefore, there are 2352 ways to form a committee consisting of 3 men and
2 women from a group of 9 men and 8 women.
Question 14
Question
In a group of 12 people, how many ways can we choose 5 people to form a
committee if one person, John, must be included?
9
Solution
Step 1: Since one person, John, must be included, we need to choose the re-
maining 4 people from the remaining 11 people. Step 2: The number of ways to
choose 4 people out of 11 is given by the combination formula nCk=n!
k!(n−k)! .
Step 3: Substituting n= 11 and k= 4 into the formula, we get:
11C4=11!
4!(11 −4)!
Step 4: Calculating the factorials, we get:
11C4=11!
4!7! =11 ×10 ×9×8
4×3×2×1= 330
Step 5: Therefore, there are 330 ways to choose 5 people to form a committee
if one person, John, must be included.
Question 15
Question
In a graduate statistics class, there are 8 students. The professor needs to choose
a team of 3 students to work on a research project together. How many different
ways can the professor choose the team if one student, Sarah, must be on it?
Solution
Step 1: Since Sarah must be on the team, we only need to choose 2 more
students from the remaining 7 students.
Step 2: We can calculate the number of ways to choose 2 students from 7
using the combination formula C(n, k) = n!
k!(n−k)! .
Step 3: Substitute n= 7 and k= 2 into the formula to find the number of
ways to choose 2 students from 7.
C(7,2) = 7!
2!(7 −2)! =7×6
2×1= 21
Step 4: Therefore, there are 21 different ways the professor can choose a
team of 3 students with Sarah on it.
Question 16
Question
In a group of 10 people, how many ways can you choose a committee of 3 people
and a president from the remaining 7 people?
10
Solution
Step 1: Calculate the number of ways to choose a committee of 3 people from
the group of 10. To choose a committee of 3 people from a group of 10, we use
the combination formula:
C(n, k) = n!
k!(n−k)!
where nis the total number of people and kis the number of people we want
to choose. In this case, n= 10 and k= 3, so the number of ways to choose a
committee of 3 people is:
C(10,3) = 10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
Step 2: Calculate the number of ways to choose a president from the re-
maining 7 people. After choosing the committee of 3 people, we have 7 people
remaining. We need to choose 1 person as the president from these 7 people.
This can be done in 7 ways.
Step 3: Multiply the results from Steps 1 and 2 to find the total number
of ways. The total number of ways to choose a committee of 3 people and a
president is the product of the number of ways to choose the committee and the
number of ways to choose the president:
Total ways = 120 ×7 = 840
Therefore, there are 840 ways to choose a committee of 3 people and a
president from a group of 10 people.
Question 17
Question
In a group of 10 people, how many ways can we choose a committee of 3 people
where one person is designated as the chair of the committee?
Solution
Step 1: First, we choose the person who will be the chair of the committee.
There are 10 ways to choose the chairperson.
Step 2: Next, we choose 2 people from the remaining 9 people to join the
committee. We can choose 2 people out of 9 in (9
2)ways.
Step 3: Multiply the results from Step 1 and Step 2 to find the total number
of ways to choose the committee. So, the total number of ways to choose a
committee of 3 people with a designated chair is: 10 ×(9
2)= 10 ×9!
2!(9−2)! =
10 ×9×8
2= 10 ×36 = 360.
Therefore, there are 360 ways to choose a committee of 3 people with one
person designated as the chair.
11
Question 18
Question
A committee of 5 people is to be formed from a group of 10 men and 12 women.
How many different committees can be formed if at least 2 men must be included
in the committee?
Solution
Step 1: Determine the number of ways to choose 2, 3, 4, or 5 men for the
committee.
There are (10
2)ways to choose 2 men, (10
3)ways to choose 3 men, (10
4)ways
to choose 4 men, and (10
5)ways to choose all 5 men.
Step 2: Determine the number of ways to choose the remaining committee
members from the women.
After choosing 2, 3, 4, or 5 men, the committee must be completed with 3,
2, 1, or 0 women, respectively.
There are (12
3)ways to choose 3 women, (12
2)ways to choose 2 women, (12
1)
ways to choose 1 woman, and (12
0)= 1 way to choose 0 women.
Step 3: Calculate the total number of ways to form the committee.
The total number of ways to form the committee is the sum of the number
of ways from Step 1 and Step 2.
(10
2)×(12
3)+(10
3)×(12
2)+(10
4)×(12
1)+(10
5)×(12
0)
Performing the calculations will give us the final answer.
Question 19
Question
In a group of 10 students, how many ways can we choose a committee of 5
students with a president, a vice president, and a secretary from the group?
Solution
Step 1: To find the number of ways to choose a committee of 5 students from a
group of 10, we will use the combination formula (n
k)=n!
k!(n−k)! . Therefore, the
number of ways to choose 5 students from a group of 10 is (10
5)=10!
5!5! = 252.
Step 2: Among the 5 selected students, there are 5 ways to choose the pres-
ident, then 4 ways to choose the vice president from the remaining 4 students,
and finally 3 ways to choose the secretary from the remaining 3 students.
12
Step 3: Therefore, the total number of ways to choose the committee of 5
students with a president, a vice president, and a secretary is 252 ×5×4×3 =
60480.
So, there are 60,480 ways to choose a committee of 5 students with a presi-
dent, a vice president, and a secretary from a group of 10 students.
Question 20
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 3 men and 2 women, how many different ways
can the committee be formed?
Solution
Step 1: Determine the number of ways to choose 3 men out of 10. There are
(10
3)ways to choose 3 men from a group of 10 men. Calculating (10
3):
(10
3)=10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
So, there are 120 ways to choose 3 men.
Step 2: Determine the number of ways to choose 2 women out of 8. There
are (8
2)ways to choose 2 women from a group of 8 women. Calculating (8
2):
(8
2)=8!
2!(8 −2)! =8×7
2×1= 28
So, there are 28 ways to choose 2 women.
Step 3: Multiply the number of ways to choose men and women. To find the
total number of ways to form a committee consisting of 3 men and 2 women,
multiply the number of ways to choose men and women together: 120 ×28 =
3360.
Therefore, there are 3360 different ways to form the committee.
Question 21
Question
A committee of 5 people must be formed from a group of 8 men and 4 women.
How many different committees can be formed if the committee must contain
at least 2 men and 2 women?
13
Solution
Step 1: Calculate the number of ways to form a committee with 2 men and 2
women. For this case, we choose 2 men out of 8 and 2 women out of 4.
(8
2)×(4
2)= 28 ×6 = 168
Step 2: Calculate the number of ways to form a committee with 3 men and
1 woman. For this case, we choose 3 men out of 8 and 1 woman out of 4.
(8
3)×(4
1)= 56 ×4 = 224
Step 3: Calculate the number of ways to form a committee with 4 men and
1 woman. For this case, we choose 4 men out of 8 and 1 woman out of 4.
(8
4)×(4
1)= 70 ×4 = 280
Step 4: Calculate the total number of committees that can be formed with
at least 2 men and 2 women. Add the results from Step 1, Step 2, and Step 3.
168 + 224 + 280 = 672
Therefore, there are 672 different committees that can be formed if the
committee must contain at least 2 men and 2 women.
Question 22
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 3 men and at least 2 women, how many
different committees can be formed?
Solution
Step 1: Calculate the number of ways to choose exactly 3 men and 2 women for
the committee. There are (10
3)ways to choose 3 men from a group of 10 and
(8
2)ways to choose 2 women from a group of 8. Therefore, the number of ways
to choose 3 men and 2 women is given by:
(10
3)×(8
2)
Step 2: Calculate the number of ways to choose exactly 4 men and 1 woman
for the committee. There are (10
4)ways to choose 4 men from a group of 10 and
14
(8
1)ways to choose 1 woman from a group of 8. Therefore, the number of ways
to choose 4 men and 1 woman is given by:
(10
4)×(8
1)
Step 3: Calculate the number of ways to choose exactly 5 men for the com-
mittee. There are (10
5)ways to choose 5 men from a group of 10. Since the
committee must consist of at least 3 men, we only need to calculate this once.
Step 4: Add up the number of ways from Step 1, Step 2, and Step 3 to find
the total number of different committees that can be formed. The total number
of committees is:
(10
3)×(8
2)+(10
4)×(8
1)+(10
5)
Calculating the values for each combination:
(10
3)= 120,(8
2)= 28,(10
4)= 210,(8
1)= 8,(10
5)= 252
Therefore, the total number of different committees that can be formed is:
120 ×28 + 210 ×8 + 252 = 3360 + 1680 + 252 = 5292
So, there are 5292 different committees that can be formed.
Question 23
Question
In a mathematics class of 24 students, a teacher is selecting a team of 4 students
to participate in a quiz bee. How many ways can this team be formed if the
teacher wants to ensure that both Jane and Mark are included in the team?
Solution
Let’s first find the number of ways to select the other two students to join Jane
and Mark in the team.
Step 1: Since Jane and Mark are already selected, we need to choose 2
students out of the remaining 22 students. The number of ways to select 2
students out of 22 is given by the combination formula (n
r)=n!
r!(n−r)! , where
nis the total number of students and ris the number of students we want to
select. So, in this case, (22
2)=22!
2!(22−2)! .
Step 2: Calculate (22
2):
(22
2)=22!
2!(22 −2)! =22 ×21
2= 231.
15
Step 3: Now, we have the number of ways to select the other two students.
To include Jane and Mark, we need to multiply this result by 1 (since Jane
and Mark are fixed). Therefore, the total number of ways to form the team is
231 ×1 = 231 ways.
So, there are 231 ways to form a team of 4 students where both Jane and
Mark are included.
Question 24
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: Calculate the number of ways to form a committee with exactly 2 women
and 3 men.
• Choose 2 women out of 8: (8
2)= 28 ways
• Choose 3 men out of 10: (10
3)= 120 ways
Therefore, the number of ways to form a committee with exactly 2 women and
3 men is 28 ×120 = 3360 ways.
Step 2: Calculate the number of ways to form a committee with exactly 3
women and 2 men.
• Choose 3 women out of 8: (8
3)= 56 ways
• Choose 2 men out of 10: (10
2)= 45 ways
Therefore, the number of ways to form a committee with exactly 3 women and
2 men is 56 ×45 = 2520 ways.
Step 3: Calculate the number of ways to form a committee with 4 women
and 1 man.
• Choose 4 women out of 8: (8
4)= 70 ways
• Choose 1 man out of 10: (10
1)= 10 ways
Therefore, the number of ways to form a committee with 4 women and 1 man
is 70 ×10 = 700 ways.
Step 4: Calculate the number of ways to form a committee with 5 women
and 0 men.
• There are (8
5)= 56 ways to choose 5 women.
16
Step 5: Add the results from Steps 1 to 4 to find the total number of different
committees that can be formed.
3360 + 2520 + 700 + 56 = 6636
Therefore, there are 6636 different committees that can be formed with at least
2 women.
Question 25
Question
How many ways are there to choose a committee of 5 people from a group of
10 people, where two specific people, Alice and Bob, must be on the committee
together?
Solution
Step 1: First, we choose Alice and Bob to be on the committee together. There
is only 1 way to choose Alice and Bob since they must be together.
Step 2: Now, we need to choose 3 more people to complete the committee of
5. Since we have already selected 2 people (Alice and Bob), we need to choose 3
more people from the remaining 8 people. This can be done in (8
3)=8!
3!5! = 56
ways.
Step 3: Multiply the results from Step 1 and Step 2 to find the total number
of ways to choose the committee. 1×56 = 56
Therefore, there are 56 ways to choose a committee of 5 people from a group
of 10 people, where Alice and Bob must be on the committee together.
17
Question 2
Question
A committee of 5 people is to be formed from a group of 10 men and 8 women.
Find the number of ways the committee can be formed if it must consist of 3
men and 2 women.
Solution
To find the number of ways the committee can be formed, we need to use
combinatorial analysis.
Step 1: Find the number of ways to choose 3 men from the group
of 10 men.
10C3=10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
Step 2: Find the number of ways to choose 2 women from the
group of 8 women.
8C2=8!
2!(8 −2)! =8×7
2×1= 28
Step 3: Find the total number of ways the committee can be
formed. Since we want to choose 3 men and 2 women for the committee,
we multiply the results from Step 1 and Step 2. Total number of ways =
120 ×28 = 3360
Therefore, there are 3360 ways to form a committee of 5 people consisting
of 3 men and 2 women from the group of 10 men and 8 women.
Question 3
Question
In how many ways can 6 different books be arranged on a bookshelf if two of
the books must always be placed next to each other?
Solution
To solve this problem, we can treat the two books that must always be together
as one unit. This reduces the problem to arranging 5 units (including the pair
of books) on the bookshelf.
Step 1: Determine the number of ways to arrange the 5 units (including
the pair of books).
There are 5! ways to arrange the 5 units.
Step 2: Consider the pair of books that must always be together as a single
unit.
Within this unit, the two books can be arranged in 2! ways.
2
Step 3: Multiply the results from Step 1 and Step 2 to find the total number
of ways to arrange the books.
Therefore, the total number of ways to arrange the 6 different books on the
bookshelf such that two of the books are always placed next to each other is
5! ×2! = 120 ×2 = 240 ways.
Question 4
Question
In a group of 8 people, how many different committees of 3 people can be formed
if one person, Alice, must always be included in the committee?
Solution
Step 1: Since Alice must always be included in the committee, we already have
1 person chosen. We need to choose 2 more people to form the committee.
Step 2: To choose the remaining 2 people out of the 7 remaining people, we
use the combination formula. The number of ways to choose 2 people from 7 is
given by (7
2).
Step 3: Calculate (7
2).
(7
2)=7!
2!(7 −2)! =7×6
2×1= 21.
Step 4: Therefore, there are 21 different committees of 3 people that can be
formed with Alice always included.
Question 5
Question
A committee of 5 people is to be formed from a group of 10 individuals (5 men
and 5 women). If at least 1 man must be on the committee, how many different
committees can be formed?
Solution
Step 1: First, let’s calculate the total number of ways to form a committee
of 5 people from a group of 10 individuals. This can be calculated using the
combination formula: (n
r)=n!
r!(n−r)!
3
where nis the total number of individuals and ris the number of people in the
committee. In this case, n= 10 and r= 5, so the total number of ways is:
(10
5)=10!
5!5! =10 ×9×8×7×6
5×4×3×2×1= 252
Step 2: Next, let’s calculate the number of ways to form a committee of
5 people with no men on it. Since there are 5 women, we need to choose
all 5 committee members from the women. This can be calculated using the
combination formula: (5
5)= 1
There is only 1 way to choose all 5 women for the committee.
Step 3: Finally, we will calculate the number of ways to form a committee of
5 people with at least 1 man on it. This can be done by subtracting the number
of committees with no men from the total number of committees:
252 −1 = 251
Therefore, there are 251 different committees that can be formed with at least
1 man on it.
Question 6
Question
A group of 8 students are going to be split into two committees, each with 4
students. How many different ways can this be done?
Solution
Step 1: We first need to choose 4 students from the 8 to be on the first com-
mittee. There are (8
4)ways to choose 4 students out of 8.
Step 2: Once the first committee is chosen, the remaining 4 students auto-
matically form the second committee. Therefore, there is only 1 way to assign
the remaining students to the second committee.
Step 3: Compute the total number of ways to form the two committees by
multiplying the number of ways in Step 1 and Step 2. So, the total number of
ways is (8
4)×1.
Step 4: Calculate the resulting value.
(8
4)=8!
4!(8 −4)! =8×7×6×5
4×3×2×1= 70.
Therefore, there are 70 different ways to split the group of 8 students into
two committees, each with 4 students.
4
Question 7
Question
In a group of 10 people, how many ways can we choose a committee of 3 people
to represent the group?
Solution
Step 1: To solve this problem, we can use the combination formula. The number
of ways to choose a committee of rmembers from a group of npeople is given
by the formula:
(n
r)=n!
r!(n−r)!
where n!represents the factorial of n.
Step 2: Substituting n= 10 and r= 3 into the formula, we get:
(10
3)=10!
3!(10 −3)!
Step 3: Calculating the factorials:
(10
3)=10 ×9×8
3×2×1
(10
3)=720
6
(10
3)= 120
Step 4: Therefore, there are 120 ways to choose a committee of 3 people
from a group of 10 people.
Question 8
Question
How many different 7-letter arrangements can be formed using all the letters of
the word ”UNIVERSITY” such that no two vowels are adjacent?
5
Solution
Step 1: Find the total number of ways to arrange the letters of ”UNIVERSITY”
without any restrictions.
The word ”UNIVERSITY” has 10 letters, but there are repetitions of some
letters:-2U’s-2I’s-2E’s-1N-1V-1R-1S-1T
Therefore, the total number of ways to arrange the letters without restric-
tions is the factorial of the total number of letters divided by the factorials of
the number of repetitions:
10!/(2! ×2! ×2! ×1! ×1! ×1! ×1! ×1!)
Step 2: Find the number of ways to arrange the letters of ”UNIVERSITY”
such that the vowels are adjacent.
Since there are 4 vowels in ”UNIVERSITY” (U, I, E, and I), we can consider
them as one unit. Therefore, the letters ”UNVRSITY” together form 7 units.
The number of ways to arrange these 7 units is 7!.
Step 3: Subtract the result from Step 2 from the result of Step 1 to find the
number of arrangements where no two vowels are adjacent.
The number of ways to arrange the letters such that no two vowels are
adjacent is given by:
10!/(2! ×2! ×2! ×1! ×1! ×1! ×1!) −7!
Thus, the above expression represents the total number of different 7-letter
arrangements of the word ”UNIVERSITY” where no two vowels are adjacent.
Question 9
Question
A company needs to assemble a committee of 5 employees from a pool of 10
potential candidates. If 3 of the candidates are engineers and 7 are accountants,
how many ways can the committee be formed if it must consist of at least 1
engineer and 2 accountants?
Solution
Step 1: Calculate the number of ways to choose 1 engineer and 2 accountants:
There are 3 engineers to choose from and 7 accountants. We can choose 1
engineer in 3 ways and 2 accountants in (7
2)ways. So, the number of ways to
choose 1 engineer and 2 accountants is 3×(7
2).
Step 2: Calculate the number of ways to choose 2 engineers and 3 accoun-
tants: There are 3 engineers to choose from and 7 accountants. We can choose
2 engineers in (3
2)ways and 3 accountants in (7
3)ways. So, the number of ways
to choose 2 engineers and 3 accountants is (3
2)×(7
3).
6
Step 3: Add the results from step 1 and step 2 to find the total number of
ways to form the committee: Total number of ways = 3×(7
2)+(3
2)×(7
3).
Step 4: Calculate the total number of ways to form the committee: 3×(7
2)=
3×7!
2!(7−2)! = 3 ×7×6
2×1= 63 ways
(3
2)×(7
3)=3!
2!(3−2)! ×7!
3!(7−3)! = 3 ×35 = 105 ways
Total number of ways = 63 + 105 = 168.
Therefore, there are 168 ways to form the committee consisting of at least 1
engineer and 2 accountants.
Question 10
Question
A committee of 5 people is to be formed from a group of 7 men and 4 women.
If the committee must consist of at least 3 men, how many different committees
can be formed?
Solution
Step 1: Find the total number of committees without any restrictions. The
total number of ways to form a committee of 5 people from 7 men and 4 women
is given by the combination formula:
(11
5)=11!
5!(11 −5)! =11!
5!6! = 462
Step 2: Find the number of committees with less than 3 men. The number
of committees with less than 3 men can be found by: - Choosing 0 men from
7 men and 5 women from 4 women, or - Choosing 1 man from 7 men and 4
women from 4 women, or - Choosing 2 men from 7 men and 3 women from 4
women.
(7
0)×(4
5)+(7
1)×(4
4)+(7
2)×(4
3)= 1×4+7×1 + 21 ×4 = 4+7 + 84 = 95
Step 3: Calculate the number of committees with at least 3 men. The
number of committees with at least 3 men is the total number of committees
minus the number of committees with less than 3 men:
462 −95 = 367
Therefore, there are 367 different committees that can be formed with at
least 3 men.
7
Question 11
Question
A committee of 5 people is to be formed from a group of 8 men and 6 women.
How many ways can the committee be formed if it must consist of at least 3
women?
Solution
Step 1: Calculate the number of ways to select exactly 3 women from the 6
available. There are (6
3)ways to choose 3 women.
Step 2: Calculate the number of ways to choose the remaining 2 committee
members (either men or women). Since the committee must consist of at least
3 women, we can choose the remaining 2 members from the 8 men or the 3
remaining women. We will compute both possibilities: - If we choose 2 men
from the 8 available, there are (8
2)ways to select them. - If we choose 2 women
from the 3 remaining, there are (3
2)ways to select them.
Step 3: Find the total number of ways to form the committee. The total
number of ways to form the committee with at least 3 women is the product
of the number of ways from Step 1 and Step 2. Add the results of the two
possibilities: (6
3)×(8
2)+(6
3)×(3
2)ways.
Step 4: Calculate the final answer. - (6
3)=6!
3!(6−3)! =6×5×4
3×2×1= 20 ways. -
(8
2)=8!
2!(8−2)! =8×7
2×1= 28 ways. - (3
2)=3!
2!(3−2)! =3
2×1= 3 ways.
Therefore, the total number of ways to form the committee is 20 ×28 + 20 ×
3 = 640 + 60 = 700 ways.
Question 12
Question
In a group of 10 students, how many ways can we select a president, vice presi-
dent, and treasurer if no student can hold more than one position?
Solution
To solve this problem, we will use the fundamental principle of counting, also
known as the multiplication principle.
Step 1: Find the number of ways to select a president from 10 students.
Since each student can be selected as the president only once, there are 10
choices for president.
Step 2: Find the number of ways to select a vice president from the re-
maining 9 students. After selecting a president, there are 9 students remaining
to choose from for vice president.
8
Step 3: Find the number of ways to select a treasurer from the remaining
8 students. After selecting a president and vice president, there are 8 students
remaining to choose from for treasurer.
Therefore, the total number of ways to select a president, vice president, and
treasurer from a group of 10 students is:
10 ×9×8 = 720
So, there are 720 different ways to select a president, vice president, and
treasurer from the group of 10 students.
Question 13
Question
A committee of 5 members is to be selected from a group of 9 men and 8 women.
In how many ways can this committee be formed if it must consist of 3 men and
2 women?
Solution
Step 1: Find the number of ways to choose 3 men from 9 men. Step 2: Find the
number of ways to choose 2 women from 8 women. Step 3: Multiply the results
from Step 1 and Step 2 to find the total number of ways to form the committee.
Step 1: To choose 3 men from 9 men, we use the combination formula:
(n
k)=n!
k!(n−k)! . Therefore, the number of ways to choose 3 men from 9 men is
(9
3)=9!
3!(9−3)! =9×8×7
3×2×1= 84 ways.
Step 2: To choose 2 women from 8 women, we use the combination formula:
(n
k)=n!
k!(n−k)! . Therefore, the number of ways to choose 2 women from 8 women
is (8
2)=8!
2!(8−2)! =8×7
2×1= 28 ways.
Step 3: Multiply the results from Step 1 and Step 2. Total number of ways
to form the committee = 84 ×28 = 2352 ways.
Therefore, there are 2352 ways to form a committee consisting of 3 men and
2 women from a group of 9 men and 8 women.
Question 14
Question
In a group of 12 people, how many ways can we choose 5 people to form a
committee if one person, John, must be included?
9
Solution
Step 1: Since one person, John, must be included, we need to choose the re-
maining 4 people from the remaining 11 people. Step 2: The number of ways to
choose 4 people out of 11 is given by the combination formula nCk=n!
k!(n−k)! .
Step 3: Substituting n= 11 and k= 4 into the formula, we get:
11C4=11!
4!(11 −4)!
Step 4: Calculating the factorials, we get:
11C4=11!
4!7! =11 ×10 ×9×8
4×3×2×1= 330
Step 5: Therefore, there are 330 ways to choose 5 people to form a committee
if one person, John, must be included.
Question 15
Question
In a graduate statistics class, there are 8 students. The professor needs to choose
a team of 3 students to work on a research project together. How many different
ways can the professor choose the team if one student, Sarah, must be on it?
Solution
Step 1: Since Sarah must be on the team, we only need to choose 2 more
students from the remaining 7 students.
Step 2: We can calculate the number of ways to choose 2 students from 7
using the combination formula C(n, k) = n!
k!(n−k)! .
Step 3: Substitute n= 7 and k= 2 into the formula to find the number of
ways to choose 2 students from 7.
C(7,2) = 7!
2!(7 −2)! =7×6
2×1= 21
Step 4: Therefore, there are 21 different ways the professor can choose a
team of 3 students with Sarah on it.
Question 16
Question
In a group of 10 people, how many ways can you choose a committee of 3 people
and a president from the remaining 7 people?
10
Solution
Step 1: Calculate the number of ways to choose a committee of 3 people from
the group of 10. To choose a committee of 3 people from a group of 10, we use
the combination formula:
C(n, k) = n!
k!(n−k)!
where nis the total number of people and kis the number of people we want
to choose. In this case, n= 10 and k= 3, so the number of ways to choose a
committee of 3 people is:
C(10,3) = 10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
Step 2: Calculate the number of ways to choose a president from the re-
maining 7 people. After choosing the committee of 3 people, we have 7 people
remaining. We need to choose 1 person as the president from these 7 people.
This can be done in 7 ways.
Step 3: Multiply the results from Steps 1 and 2 to find the total number
of ways. The total number of ways to choose a committee of 3 people and a
president is the product of the number of ways to choose the committee and the
number of ways to choose the president:
Total ways = 120 ×7 = 840
Therefore, there are 840 ways to choose a committee of 3 people and a
president from a group of 10 people.
Question 17
Question
In a group of 10 people, how many ways can we choose a committee of 3 people
where one person is designated as the chair of the committee?
Solution
Step 1: First, we choose the person who will be the chair of the committee.
There are 10 ways to choose the chairperson.
Step 2: Next, we choose 2 people from the remaining 9 people to join the
committee. We can choose 2 people out of 9 in (9
2)ways.
Step 3: Multiply the results from Step 1 and Step 2 to find the total number
of ways to choose the committee. So, the total number of ways to choose a
committee of 3 people with a designated chair is: 10 ×(9
2)= 10 ×9!
2!(9−2)! =
10 ×9×8
2= 10 ×36 = 360.
Therefore, there are 360 ways to choose a committee of 3 people with one
person designated as the chair.
11
Question 18
Question
A committee of 5 people is to be formed from a group of 10 men and 12 women.
How many different committees can be formed if at least 2 men must be included
in the committee?
Solution
Step 1: Determine the number of ways to choose 2, 3, 4, or 5 men for the
committee.
There are (10
2)ways to choose 2 men, (10
3)ways to choose 3 men, (10
4)ways
to choose 4 men, and (10
5)ways to choose all 5 men.
Step 2: Determine the number of ways to choose the remaining committee
members from the women.
After choosing 2, 3, 4, or 5 men, the committee must be completed with 3,
2, 1, or 0 women, respectively.
There are (12
3)ways to choose 3 women, (12
2)ways to choose 2 women, (12
1)
ways to choose 1 woman, and (12
0)= 1 way to choose 0 women.
Step 3: Calculate the total number of ways to form the committee.
The total number of ways to form the committee is the sum of the number
of ways from Step 1 and Step 2.
(10
2)×(12
3)+(10
3)×(12
2)+(10
4)×(12
1)+(10
5)×(12
0)
Performing the calculations will give us the final answer.
Question 19
Question
In a group of 10 students, how many ways can we choose a committee of 5
students with a president, a vice president, and a secretary from the group?
Solution
Step 1: To find the number of ways to choose a committee of 5 students from a
group of 10, we will use the combination formula (n
k)=n!
k!(n−k)! . Therefore, the
number of ways to choose 5 students from a group of 10 is (10
5)=10!
5!5! = 252.
Step 2: Among the 5 selected students, there are 5 ways to choose the pres-
ident, then 4 ways to choose the vice president from the remaining 4 students,
and finally 3 ways to choose the secretary from the remaining 3 students.
12
Step 3: Therefore, the total number of ways to choose the committee of 5
students with a president, a vice president, and a secretary is 252 ×5×4×3 =
60480.
So, there are 60,480 ways to choose a committee of 5 students with a presi-
dent, a vice president, and a secretary from a group of 10 students.
Question 20
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 3 men and 2 women, how many different ways
can the committee be formed?
Solution
Step 1: Determine the number of ways to choose 3 men out of 10. There are
(10
3)ways to choose 3 men from a group of 10 men. Calculating (10
3):
(10
3)=10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
So, there are 120 ways to choose 3 men.
Step 2: Determine the number of ways to choose 2 women out of 8. There
are (8
2)ways to choose 2 women from a group of 8 women. Calculating (8
2):
(8
2)=8!
2!(8 −2)! =8×7
2×1= 28
So, there are 28 ways to choose 2 women.
Step 3: Multiply the number of ways to choose men and women. To find the
total number of ways to form a committee consisting of 3 men and 2 women,
multiply the number of ways to choose men and women together: 120 ×28 =
3360.
Therefore, there are 3360 different ways to form the committee.
Question 21
Question
A committee of 5 people must be formed from a group of 8 men and 4 women.
How many different committees can be formed if the committee must contain
at least 2 men and 2 women?
13
Solution
Step 1: Calculate the number of ways to form a committee with 2 men and 2
women. For this case, we choose 2 men out of 8 and 2 women out of 4.
(8
2)×(4
2)= 28 ×6 = 168
Step 2: Calculate the number of ways to form a committee with 3 men and
1 woman. For this case, we choose 3 men out of 8 and 1 woman out of 4.
(8
3)×(4
1)= 56 ×4 = 224
Step 3: Calculate the number of ways to form a committee with 4 men and
1 woman. For this case, we choose 4 men out of 8 and 1 woman out of 4.
(8
4)×(4
1)= 70 ×4 = 280
Step 4: Calculate the total number of committees that can be formed with
at least 2 men and 2 women. Add the results from Step 1, Step 2, and Step 3.
168 + 224 + 280 = 672
Therefore, there are 672 different committees that can be formed if the
committee must contain at least 2 men and 2 women.
Question 22
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 3 men and at least 2 women, how many
different committees can be formed?
Solution
Step 1: Calculate the number of ways to choose exactly 3 men and 2 women for
the committee. There are (10
3)ways to choose 3 men from a group of 10 and
(8
2)ways to choose 2 women from a group of 8. Therefore, the number of ways
to choose 3 men and 2 women is given by:
(10
3)×(8
2)
Step 2: Calculate the number of ways to choose exactly 4 men and 1 woman
for the committee. There are (10
4)ways to choose 4 men from a group of 10 and
14
(8
1)ways to choose 1 woman from a group of 8. Therefore, the number of ways
to choose 4 men and 1 woman is given by:
(10
4)×(8
1)
Step 3: Calculate the number of ways to choose exactly 5 men for the com-
mittee. There are (10
5)ways to choose 5 men from a group of 10. Since the
committee must consist of at least 3 men, we only need to calculate this once.
Step 4: Add up the number of ways from Step 1, Step 2, and Step 3 to find
the total number of different committees that can be formed. The total number
of committees is:
(10
3)×(8
2)+(10
4)×(8
1)+(10
5)
Calculating the values for each combination:
(10
3)= 120,(8
2)= 28,(10
4)= 210,(8
1)= 8,(10
5)= 252
Therefore, the total number of different committees that can be formed is:
120 ×28 + 210 ×8 + 252 = 3360 + 1680 + 252 = 5292
So, there are 5292 different committees that can be formed.
Question 23
Question
In a mathematics class of 24 students, a teacher is selecting a team of 4 students
to participate in a quiz bee. How many ways can this team be formed if the
teacher wants to ensure that both Jane and Mark are included in the team?
Solution
Let’s first find the number of ways to select the other two students to join Jane
and Mark in the team.
Step 1: Since Jane and Mark are already selected, we need to choose 2
students out of the remaining 22 students. The number of ways to select 2
students out of 22 is given by the combination formula (n
r)=n!
r!(n−r)! , where
nis the total number of students and ris the number of students we want to
select. So, in this case, (22
2)=22!
2!(22−2)! .
Step 2: Calculate (22
2):
(22
2)=22!
2!(22 −2)! =22 ×21
2= 231.
15
Step 3: Now, we have the number of ways to select the other two students.
To include Jane and Mark, we need to multiply this result by 1 (since Jane
and Mark are fixed). Therefore, the total number of ways to form the team is
231 ×1 = 231 ways.
So, there are 231 ways to form a team of 4 students where both Jane and
Mark are included.
Question 24
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: Calculate the number of ways to form a committee with exactly 2 women
and 3 men.
• Choose 2 women out of 8: (8
2)= 28 ways
• Choose 3 men out of 10: (10
3)= 120 ways
Therefore, the number of ways to form a committee with exactly 2 women and
3 men is 28 ×120 = 3360 ways.
Step 2: Calculate the number of ways to form a committee with exactly 3
women and 2 men.
• Choose 3 women out of 8: (8
3)= 56 ways
• Choose 2 men out of 10: (10
2)= 45 ways
Therefore, the number of ways to form a committee with exactly 3 women and
2 men is 56 ×45 = 2520 ways.
Step 3: Calculate the number of ways to form a committee with 4 women
and 1 man.
• Choose 4 women out of 8: (8
4)= 70 ways
• Choose 1 man out of 10: (10
1)= 10 ways
Therefore, the number of ways to form a committee with 4 women and 1 man
is 70 ×10 = 700 ways.
Step 4: Calculate the number of ways to form a committee with 5 women
and 0 men.
• There are (8
5)= 56 ways to choose 5 women.
16
Step 5: Add the results from Steps 1 to 4 to find the total number of different
committees that can be formed.
3360 + 2520 + 700 + 56 = 6636
Therefore, there are 6636 different committees that can be formed with at least
2 women.
Question 25
Question
How many ways are there to choose a committee of 5 people from a group of
10 people, where two specific people, Alice and Bob, must be on the committee
together?
Solution
Step 1: First, we choose Alice and Bob to be on the committee together. There
is only 1 way to choose Alice and Bob since they must be together.
Step 2: Now, we need to choose 3 more people to complete the committee of
5. Since we have already selected 2 people (Alice and Bob), we need to choose 3
more people from the remaining 8 people. This can be done in (8
3)=8!
3!5! = 56
ways.
Step 3: Multiply the results from Step 1 and Step 2 to find the total number
of ways to choose the committee. 1×56 = 56
Therefore, there are 56 ways to choose a committee of 5 people from a group
of 10 people, where Alice and Bob must be on the committee together.
17
Question 2
Question
A committee of 5 people is to be formed from a group of 10 men and 8 women.
Find the number of ways the committee can be formed if it must consist of 3
men and 2 women.
Solution
To find the number of ways the committee can be formed, we need to use
combinatorial analysis.
Step 1: Find the number of ways to choose 3 men from the group
of 10 men.
10C3=10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
Step 2: Find the number of ways to choose 2 women from the
group of 8 women.
8C2=8!
2!(8 −2)! =8×7
2×1= 28
Step 3: Find the total number of ways the committee can be
formed. Since we want to choose 3 men and 2 women for the committee,
we multiply the results from Step 1 and Step 2. Total number of ways =
120 ×28 = 3360
Therefore, there are 3360 ways to form a committee of 5 people consisting
of 3 men and 2 women from the group of 10 men and 8 women.
Question 3
Question
In how many ways can 6 different books be arranged on a bookshelf if two of
the books must always be placed next to each other?
Solution
To solve this problem, we can treat the two books that must always be together
as one unit. This reduces the problem to arranging 5 units (including the pair
of books) on the bookshelf.
Step 1: Determine the number of ways to arrange the 5 units (including
the pair of books).
There are 5! ways to arrange the 5 units.
Step 2: Consider the pair of books that must always be together as a single
unit.
Within this unit, the two books can be arranged in 2! ways.
2
Step 3: Multiply the results from Step 1 and Step 2 to find the total number
of ways to arrange the books.
Therefore, the total number of ways to arrange the 6 different books on the
bookshelf such that two of the books are always placed next to each other is
5! ×2! = 120 ×2 = 240 ways.
Question 4
Question
In a group of 8 people, how many different committees of 3 people can be formed
if one person, Alice, must always be included in the committee?
Solution
Step 1: Since Alice must always be included in the committee, we already have
1 person chosen. We need to choose 2 more people to form the committee.
Step 2: To choose the remaining 2 people out of the 7 remaining people, we
use the combination formula. The number of ways to choose 2 people from 7 is
given by (7
2).
Step 3: Calculate (7
2).
(7
2)=7!
2!(7 −2)! =7×6
2×1= 21.
Step 4: Therefore, there are 21 different committees of 3 people that can be
formed with Alice always included.
Question 5
Question
A committee of 5 people is to be formed from a group of 10 individuals (5 men
and 5 women). If at least 1 man must be on the committee, how many different
committees can be formed?
Solution
Step 1: First, let’s calculate the total number of ways to form a committee
of 5 people from a group of 10 individuals. This can be calculated using the
combination formula: (n
r)=n!
r!(n−r)!
3
where nis the total number of individuals and ris the number of people in the
committee. In this case, n= 10 and r= 5, so the total number of ways is:
(10
5)=10!
5!5! =10 ×9×8×7×6
5×4×3×2×1= 252
Step 2: Next, let’s calculate the number of ways to form a committee of
5 people with no men on it. Since there are 5 women, we need to choose
all 5 committee members from the women. This can be calculated using the
combination formula: (5
5)= 1
There is only 1 way to choose all 5 women for the committee.
Step 3: Finally, we will calculate the number of ways to form a committee of
5 people with at least 1 man on it. This can be done by subtracting the number
of committees with no men from the total number of committees:
252 −1 = 251
Therefore, there are 251 different committees that can be formed with at least
1 man on it.
Question 6
Question
A group of 8 students are going to be split into two committees, each with 4
students. How many different ways can this be done?
Solution
Step 1: We first need to choose 4 students from the 8 to be on the first com-
mittee. There are (8
4)ways to choose 4 students out of 8.
Step 2: Once the first committee is chosen, the remaining 4 students auto-
matically form the second committee. Therefore, there is only 1 way to assign
the remaining students to the second committee.
Step 3: Compute the total number of ways to form the two committees by
multiplying the number of ways in Step 1 and Step 2. So, the total number of
ways is (8
4)×1.
Step 4: Calculate the resulting value.
(8
4)=8!
4!(8 −4)! =8×7×6×5
4×3×2×1= 70.
Therefore, there are 70 different ways to split the group of 8 students into
two committees, each with 4 students.
4
Question 7
Question
In a group of 10 people, how many ways can we choose a committee of 3 people
to represent the group?
Solution
Step 1: To solve this problem, we can use the combination formula. The number
of ways to choose a committee of rmembers from a group of npeople is given
by the formula:
(n
r)=n!
r!(n−r)!
where n!represents the factorial of n.
Step 2: Substituting n= 10 and r= 3 into the formula, we get:
(10
3)=10!
3!(10 −3)!
Step 3: Calculating the factorials:
(10
3)=10 ×9×8
3×2×1
(10
3)=720
6
(10
3)= 120
Step 4: Therefore, there are 120 ways to choose a committee of 3 people
from a group of 10 people.
Question 8
Question
How many different 7-letter arrangements can be formed using all the letters of
the word ”UNIVERSITY” such that no two vowels are adjacent?
5
Solution
Step 1: Find the total number of ways to arrange the letters of ”UNIVERSITY”
without any restrictions.
The word ”UNIVERSITY” has 10 letters, but there are repetitions of some
letters:-2U’s-2I’s-2E’s-1N-1V-1R-1S-1T
Therefore, the total number of ways to arrange the letters without restric-
tions is the factorial of the total number of letters divided by the factorials of
the number of repetitions:
10!/(2! ×2! ×2! ×1! ×1! ×1! ×1! ×1!)
Step 2: Find the number of ways to arrange the letters of ”UNIVERSITY”
such that the vowels are adjacent.
Since there are 4 vowels in ”UNIVERSITY” (U, I, E, and I), we can consider
them as one unit. Therefore, the letters ”UNVRSITY” together form 7 units.
The number of ways to arrange these 7 units is 7!.
Step 3: Subtract the result from Step 2 from the result of Step 1 to find the
number of arrangements where no two vowels are adjacent.
The number of ways to arrange the letters such that no two vowels are
adjacent is given by:
10!/(2! ×2! ×2! ×1! ×1! ×1! ×1!) −7!
Thus, the above expression represents the total number of different 7-letter
arrangements of the word ”UNIVERSITY” where no two vowels are adjacent.
Question 9
Question
A company needs to assemble a committee of 5 employees from a pool of 10
potential candidates. If 3 of the candidates are engineers and 7 are accountants,
how many ways can the committee be formed if it must consist of at least 1
engineer and 2 accountants?
Solution
Step 1: Calculate the number of ways to choose 1 engineer and 2 accountants:
There are 3 engineers to choose from and 7 accountants. We can choose 1
engineer in 3 ways and 2 accountants in (7
2)ways. So, the number of ways to
choose 1 engineer and 2 accountants is 3×(7
2).
Step 2: Calculate the number of ways to choose 2 engineers and 3 accoun-
tants: There are 3 engineers to choose from and 7 accountants. We can choose
2 engineers in (3
2)ways and 3 accountants in (7
3)ways. So, the number of ways
to choose 2 engineers and 3 accountants is (3
2)×(7
3).
6
Step 3: Add the results from step 1 and step 2 to find the total number of
ways to form the committee: Total number of ways = 3×(7
2)+(3
2)×(7
3).
Step 4: Calculate the total number of ways to form the committee: 3×(7
2)=
3×7!
2!(7−2)! = 3 ×7×6
2×1= 63 ways
(3
2)×(7
3)=3!
2!(3−2)! ×7!
3!(7−3)! = 3 ×35 = 105 ways
Total number of ways = 63 + 105 = 168.
Therefore, there are 168 ways to form the committee consisting of at least 1
engineer and 2 accountants.
Question 10
Question
A committee of 5 people is to be formed from a group of 7 men and 4 women.
If the committee must consist of at least 3 men, how many different committees
can be formed?
Solution
Step 1: Find the total number of committees without any restrictions. The
total number of ways to form a committee of 5 people from 7 men and 4 women
is given by the combination formula:
(11
5)=11!
5!(11 −5)! =11!
5!6! = 462
Step 2: Find the number of committees with less than 3 men. The number
of committees with less than 3 men can be found by: - Choosing 0 men from
7 men and 5 women from 4 women, or - Choosing 1 man from 7 men and 4
women from 4 women, or - Choosing 2 men from 7 men and 3 women from 4
women.
(7
0)×(4
5)+(7
1)×(4
4)+(7
2)×(4
3)= 1×4+7×1 + 21 ×4 = 4+7 + 84 = 95
Step 3: Calculate the number of committees with at least 3 men. The
number of committees with at least 3 men is the total number of committees
minus the number of committees with less than 3 men:
462 −95 = 367
Therefore, there are 367 different committees that can be formed with at
least 3 men.
7
Question 11
Question
A committee of 5 people is to be formed from a group of 8 men and 6 women.
How many ways can the committee be formed if it must consist of at least 3
women?
Solution
Step 1: Calculate the number of ways to select exactly 3 women from the 6
available. There are (6
3)ways to choose 3 women.
Step 2: Calculate the number of ways to choose the remaining 2 committee
members (either men or women). Since the committee must consist of at least
3 women, we can choose the remaining 2 members from the 8 men or the 3
remaining women. We will compute both possibilities: - If we choose 2 men
from the 8 available, there are (8
2)ways to select them. - If we choose 2 women
from the 3 remaining, there are (3
2)ways to select them.
Step 3: Find the total number of ways to form the committee. The total
number of ways to form the committee with at least 3 women is the product
of the number of ways from Step 1 and Step 2. Add the results of the two
possibilities: (6
3)×(8
2)+(6
3)×(3
2)ways.
Step 4: Calculate the final answer. - (6
3)=6!
3!(6−3)! =6×5×4
3×2×1= 20 ways. -
(8
2)=8!
2!(8−2)! =8×7
2×1= 28 ways. - (3
2)=3!
2!(3−2)! =3
2×1= 3 ways.
Therefore, the total number of ways to form the committee is 20 ×28 + 20 ×
3 = 640 + 60 = 700 ways.
Question 12
Question
In a group of 10 students, how many ways can we select a president, vice presi-
dent, and treasurer if no student can hold more than one position?
Solution
To solve this problem, we will use the fundamental principle of counting, also
known as the multiplication principle.
Step 1: Find the number of ways to select a president from 10 students.
Since each student can be selected as the president only once, there are 10
choices for president.
Step 2: Find the number of ways to select a vice president from the re-
maining 9 students. After selecting a president, there are 9 students remaining
to choose from for vice president.
8
Step 3: Find the number of ways to select a treasurer from the remaining
8 students. After selecting a president and vice president, there are 8 students
remaining to choose from for treasurer.
Therefore, the total number of ways to select a president, vice president, and
treasurer from a group of 10 students is:
10 ×9×8 = 720
So, there are 720 different ways to select a president, vice president, and
treasurer from the group of 10 students.
Question 13
Question
A committee of 5 members is to be selected from a group of 9 men and 8 women.
In how many ways can this committee be formed if it must consist of 3 men and
2 women?
Solution
Step 1: Find the number of ways to choose 3 men from 9 men. Step 2: Find the
number of ways to choose 2 women from 8 women. Step 3: Multiply the results
from Step 1 and Step 2 to find the total number of ways to form the committee.
Step 1: To choose 3 men from 9 men, we use the combination formula:
(n
k)=n!
k!(n−k)! . Therefore, the number of ways to choose 3 men from 9 men is
(9
3)=9!
3!(9−3)! =9×8×7
3×2×1= 84 ways.
Step 2: To choose 2 women from 8 women, we use the combination formula:
(n
k)=n!
k!(n−k)! . Therefore, the number of ways to choose 2 women from 8 women
is (8
2)=8!
2!(8−2)! =8×7
2×1= 28 ways.
Step 3: Multiply the results from Step 1 and Step 2. Total number of ways
to form the committee = 84 ×28 = 2352 ways.
Therefore, there are 2352 ways to form a committee consisting of 3 men and
2 women from a group of 9 men and 8 women.
Question 14
Question
In a group of 12 people, how many ways can we choose 5 people to form a
committee if one person, John, must be included?
9
Solution
Step 1: Since one person, John, must be included, we need to choose the re-
maining 4 people from the remaining 11 people. Step 2: The number of ways to
choose 4 people out of 11 is given by the combination formula nCk=n!
k!(n−k)! .
Step 3: Substituting n= 11 and k= 4 into the formula, we get:
11C4=11!
4!(11 −4)!
Step 4: Calculating the factorials, we get:
11C4=11!
4!7! =11 ×10 ×9×8
4×3×2×1= 330
Step 5: Therefore, there are 330 ways to choose 5 people to form a committee
if one person, John, must be included.
Question 15
Question
In a graduate statistics class, there are 8 students. The professor needs to choose
a team of 3 students to work on a research project together. How many different
ways can the professor choose the team if one student, Sarah, must be on it?
Solution
Step 1: Since Sarah must be on the team, we only need to choose 2 more
students from the remaining 7 students.
Step 2: We can calculate the number of ways to choose 2 students from 7
using the combination formula C(n, k) = n!
k!(n−k)! .
Step 3: Substitute n= 7 and k= 2 into the formula to find the number of
ways to choose 2 students from 7.
C(7,2) = 7!
2!(7 −2)! =7×6
2×1= 21
Step 4: Therefore, there are 21 different ways the professor can choose a
team of 3 students with Sarah on it.
Question 16
Question
In a group of 10 people, how many ways can you choose a committee of 3 people
and a president from the remaining 7 people?
10
Solution
Step 1: Calculate the number of ways to choose a committee of 3 people from
the group of 10. To choose a committee of 3 people from a group of 10, we use
the combination formula:
C(n, k) = n!
k!(n−k)!
where nis the total number of people and kis the number of people we want
to choose. In this case, n= 10 and k= 3, so the number of ways to choose a
committee of 3 people is:
C(10,3) = 10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
Step 2: Calculate the number of ways to choose a president from the re-
maining 7 people. After choosing the committee of 3 people, we have 7 people
remaining. We need to choose 1 person as the president from these 7 people.
This can be done in 7 ways.
Step 3: Multiply the results from Steps 1 and 2 to find the total number
of ways. The total number of ways to choose a committee of 3 people and a
president is the product of the number of ways to choose the committee and the
number of ways to choose the president:
Total ways = 120 ×7 = 840
Therefore, there are 840 ways to choose a committee of 3 people and a
president from a group of 10 people.
Question 17
Question
In a group of 10 people, how many ways can we choose a committee of 3 people
where one person is designated as the chair of the committee?
Solution
Step 1: First, we choose the person who will be the chair of the committee.
There are 10 ways to choose the chairperson.
Step 2: Next, we choose 2 people from the remaining 9 people to join the
committee. We can choose 2 people out of 9 in (9
2)ways.
Step 3: Multiply the results from Step 1 and Step 2 to find the total number
of ways to choose the committee. So, the total number of ways to choose a
committee of 3 people with a designated chair is: 10 ×(9
2)= 10 ×9!
2!(9−2)! =
10 ×9×8
2= 10 ×36 = 360.
Therefore, there are 360 ways to choose a committee of 3 people with one
person designated as the chair.
11
Question 18
Question
A committee of 5 people is to be formed from a group of 10 men and 12 women.
How many different committees can be formed if at least 2 men must be included
in the committee?
Solution
Step 1: Determine the number of ways to choose 2, 3, 4, or 5 men for the
committee.
There are (10
2)ways to choose 2 men, (10
3)ways to choose 3 men, (10
4)ways
to choose 4 men, and (10
5)ways to choose all 5 men.
Step 2: Determine the number of ways to choose the remaining committee
members from the women.
After choosing 2, 3, 4, or 5 men, the committee must be completed with 3,
2, 1, or 0 women, respectively.
There are (12
3)ways to choose 3 women, (12
2)ways to choose 2 women, (12
1)
ways to choose 1 woman, and (12
0)= 1 way to choose 0 women.
Step 3: Calculate the total number of ways to form the committee.
The total number of ways to form the committee is the sum of the number
of ways from Step 1 and Step 2.
(10
2)×(12
3)+(10
3)×(12
2)+(10
4)×(12
1)+(10
5)×(12
0)
Performing the calculations will give us the final answer.
Question 19
Question
In a group of 10 students, how many ways can we choose a committee of 5
students with a president, a vice president, and a secretary from the group?
Solution
Step 1: To find the number of ways to choose a committee of 5 students from a
group of 10, we will use the combination formula (n
k)=n!
k!(n−k)! . Therefore, the
number of ways to choose 5 students from a group of 10 is (10
5)=10!
5!5! = 252.
Step 2: Among the 5 selected students, there are 5 ways to choose the pres-
ident, then 4 ways to choose the vice president from the remaining 4 students,
and finally 3 ways to choose the secretary from the remaining 3 students.
12
Step 3: Therefore, the total number of ways to choose the committee of 5
students with a president, a vice president, and a secretary is 252 ×5×4×3 =
60480.
So, there are 60,480 ways to choose a committee of 5 students with a presi-
dent, a vice president, and a secretary from a group of 10 students.
Question 20
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 3 men and 2 women, how many different ways
can the committee be formed?
Solution
Step 1: Determine the number of ways to choose 3 men out of 10. There are
(10
3)ways to choose 3 men from a group of 10 men. Calculating (10
3):
(10
3)=10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
So, there are 120 ways to choose 3 men.
Step 2: Determine the number of ways to choose 2 women out of 8. There
are (8
2)ways to choose 2 women from a group of 8 women. Calculating (8
2):
(8
2)=8!
2!(8 −2)! =8×7
2×1= 28
So, there are 28 ways to choose 2 women.
Step 3: Multiply the number of ways to choose men and women. To find the
total number of ways to form a committee consisting of 3 men and 2 women,
multiply the number of ways to choose men and women together: 120 ×28 =
3360.
Therefore, there are 3360 different ways to form the committee.
Question 21
Question
A committee of 5 people must be formed from a group of 8 men and 4 women.
How many different committees can be formed if the committee must contain
at least 2 men and 2 women?
13
Solution
Step 1: Calculate the number of ways to form a committee with 2 men and 2
women. For this case, we choose 2 men out of 8 and 2 women out of 4.
(8
2)×(4
2)= 28 ×6 = 168
Step 2: Calculate the number of ways to form a committee with 3 men and
1 woman. For this case, we choose 3 men out of 8 and 1 woman out of 4.
(8
3)×(4
1)= 56 ×4 = 224
Step 3: Calculate the number of ways to form a committee with 4 men and
1 woman. For this case, we choose 4 men out of 8 and 1 woman out of 4.
(8
4)×(4
1)= 70 ×4 = 280
Step 4: Calculate the total number of committees that can be formed with
at least 2 men and 2 women. Add the results from Step 1, Step 2, and Step 3.
168 + 224 + 280 = 672
Therefore, there are 672 different committees that can be formed if the
committee must contain at least 2 men and 2 women.
Question 22
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 3 men and at least 2 women, how many
different committees can be formed?
Solution
Step 1: Calculate the number of ways to choose exactly 3 men and 2 women for
the committee. There are (10
3)ways to choose 3 men from a group of 10 and
(8
2)ways to choose 2 women from a group of 8. Therefore, the number of ways
to choose 3 men and 2 women is given by:
(10
3)×(8
2)
Step 2: Calculate the number of ways to choose exactly 4 men and 1 woman
for the committee. There are (10
4)ways to choose 4 men from a group of 10 and
14
(8
1)ways to choose 1 woman from a group of 8. Therefore, the number of ways
to choose 4 men and 1 woman is given by:
(10
4)×(8
1)
Step 3: Calculate the number of ways to choose exactly 5 men for the com-
mittee. There are (10
5)ways to choose 5 men from a group of 10. Since the
committee must consist of at least 3 men, we only need to calculate this once.
Step 4: Add up the number of ways from Step 1, Step 2, and Step 3 to find
the total number of different committees that can be formed. The total number
of committees is:
(10
3)×(8
2)+(10
4)×(8
1)+(10
5)
Calculating the values for each combination:
(10
3)= 120,(8
2)= 28,(10
4)= 210,(8
1)= 8,(10
5)= 252
Therefore, the total number of different committees that can be formed is:
120 ×28 + 210 ×8 + 252 = 3360 + 1680 + 252 = 5292
So, there are 5292 different committees that can be formed.
Question 23
Question
In a mathematics class of 24 students, a teacher is selecting a team of 4 students
to participate in a quiz bee. How many ways can this team be formed if the
teacher wants to ensure that both Jane and Mark are included in the team?
Solution
Let’s first find the number of ways to select the other two students to join Jane
and Mark in the team.
Step 1: Since Jane and Mark are already selected, we need to choose 2
students out of the remaining 22 students. The number of ways to select 2
students out of 22 is given by the combination formula (n
r)=n!
r!(n−r)! , where
nis the total number of students and ris the number of students we want to
select. So, in this case, (22
2)=22!
2!(22−2)! .
Step 2: Calculate (22
2):
(22
2)=22!
2!(22 −2)! =22 ×21
2= 231.
15
Step 3: Now, we have the number of ways to select the other two students.
To include Jane and Mark, we need to multiply this result by 1 (since Jane
and Mark are fixed). Therefore, the total number of ways to form the team is
231 ×1 = 231 ways.
So, there are 231 ways to form a team of 4 students where both Jane and
Mark are included.
Question 24
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: Calculate the number of ways to form a committee with exactly 2 women
and 3 men.
• Choose 2 women out of 8: (8
2)= 28 ways
• Choose 3 men out of 10: (10
3)= 120 ways
Therefore, the number of ways to form a committee with exactly 2 women and
3 men is 28 ×120 = 3360 ways.
Step 2: Calculate the number of ways to form a committee with exactly 3
women and 2 men.
• Choose 3 women out of 8: (8
3)= 56 ways
• Choose 2 men out of 10: (10
2)= 45 ways
Therefore, the number of ways to form a committee with exactly 3 women and
2 men is 56 ×45 = 2520 ways.
Step 3: Calculate the number of ways to form a committee with 4 women
and 1 man.
• Choose 4 women out of 8: (8
4)= 70 ways
• Choose 1 man out of 10: (10
1)= 10 ways
Therefore, the number of ways to form a committee with 4 women and 1 man
is 70 ×10 = 700 ways.
Step 4: Calculate the number of ways to form a committee with 5 women
and 0 men.
• There are (8
5)= 56 ways to choose 5 women.
16
Step 5: Add the results from Steps 1 to 4 to find the total number of different
committees that can be formed.
3360 + 2520 + 700 + 56 = 6636
Therefore, there are 6636 different committees that can be formed with at least
2 women.
Question 25
Question
How many ways are there to choose a committee of 5 people from a group of
10 people, where two specific people, Alice and Bob, must be on the committee
together?
Solution
Step 1: First, we choose Alice and Bob to be on the committee together. There
is only 1 way to choose Alice and Bob since they must be together.
Step 2: Now, we need to choose 3 more people to complete the committee of
5. Since we have already selected 2 people (Alice and Bob), we need to choose 3
more people from the remaining 8 people. This can be done in (8
3)=8!
3!5! = 56
ways.
Step 3: Multiply the results from Step 1 and Step 2 to find the total number
of ways to choose the committee. 1×56 = 56
Therefore, there are 56 ways to choose a committee of 5 people from a group
of 10 people, where Alice and Bob must be on the committee together.
17
Question 2
Question
A committee of 5 people is to be formed from a group of 10 men and 8 women.
Find the number of ways the committee can be formed if it must consist of 3
men and 2 women.
Solution
To find the number of ways the committee can be formed, we need to use
combinatorial analysis.
Step 1: Find the number of ways to choose 3 men from the group
of 10 men.
10C3=10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
Step 2: Find the number of ways to choose 2 women from the
group of 8 women.
8C2=8!
2!(8 −2)! =8×7
2×1= 28
Step 3: Find the total number of ways the committee can be
formed. Since we want to choose 3 men and 2 women for the committee,
we multiply the results from Step 1 and Step 2. Total number of ways =
120 ×28 = 3360
Therefore, there are 3360 ways to form a committee of 5 people consisting
of 3 men and 2 women from the group of 10 men and 8 women.
Question 3
Question
In how many ways can 6 different books be arranged on a bookshelf if two of
the books must always be placed next to each other?
Solution
To solve this problem, we can treat the two books that must always be together
as one unit. This reduces the problem to arranging 5 units (including the pair
of books) on the bookshelf.
Step 1: Determine the number of ways to arrange the 5 units (including
the pair of books).
There are 5! ways to arrange the 5 units.
Step 2: Consider the pair of books that must always be together as a single
unit.
Within this unit, the two books can be arranged in 2! ways.
2
Step 3: Multiply the results from Step 1 and Step 2 to find the total number
of ways to arrange the books.
Therefore, the total number of ways to arrange the 6 different books on the
bookshelf such that two of the books are always placed next to each other is
5! ×2! = 120 ×2 = 240 ways.
Question 4
Question
In a group of 8 people, how many different committees of 3 people can be formed
if one person, Alice, must always be included in the committee?
Solution
Step 1: Since Alice must always be included in the committee, we already have
1 person chosen. We need to choose 2 more people to form the committee.
Step 2: To choose the remaining 2 people out of the 7 remaining people, we
use the combination formula. The number of ways to choose 2 people from 7 is
given by (7
2).
Step 3: Calculate (7
2).
(7
2)=7!
2!(7 −2)! =7×6
2×1= 21.
Step 4: Therefore, there are 21 different committees of 3 people that can be
formed with Alice always included.
Question 5
Question
A committee of 5 people is to be formed from a group of 10 individuals (5 men
and 5 women). If at least 1 man must be on the committee, how many different
committees can be formed?
Solution
Step 1: First, let’s calculate the total number of ways to form a committee
of 5 people from a group of 10 individuals. This can be calculated using the
combination formula: (n
r)=n!
r!(n−r)!
3
where nis the total number of individuals and ris the number of people in the
committee. In this case, n= 10 and r= 5, so the total number of ways is:
(10
5)=10!
5!5! =10 ×9×8×7×6
5×4×3×2×1= 252
Step 2: Next, let’s calculate the number of ways to form a committee of
5 people with no men on it. Since there are 5 women, we need to choose
all 5 committee members from the women. This can be calculated using the
combination formula: (5
5)= 1
There is only 1 way to choose all 5 women for the committee.
Step 3: Finally, we will calculate the number of ways to form a committee of
5 people with at least 1 man on it. This can be done by subtracting the number
of committees with no men from the total number of committees:
252 −1 = 251
Therefore, there are 251 different committees that can be formed with at least
1 man on it.
Question 6
Question
A group of 8 students are going to be split into two committees, each with 4
students. How many different ways can this be done?
Solution
Step 1: We first need to choose 4 students from the 8 to be on the first com-
mittee. There are (8
4)ways to choose 4 students out of 8.
Step 2: Once the first committee is chosen, the remaining 4 students auto-
matically form the second committee. Therefore, there is only 1 way to assign
the remaining students to the second committee.
Step 3: Compute the total number of ways to form the two committees by
multiplying the number of ways in Step 1 and Step 2. So, the total number of
ways is (8
4)×1.
Step 4: Calculate the resulting value.
(8
4)=8!
4!(8 −4)! =8×7×6×5
4×3×2×1= 70.
Therefore, there are 70 different ways to split the group of 8 students into
two committees, each with 4 students.
4
Question 7
Question
In a group of 10 people, how many ways can we choose a committee of 3 people
to represent the group?
Solution
Step 1: To solve this problem, we can use the combination formula. The number
of ways to choose a committee of rmembers from a group of npeople is given
by the formula:
(n
r)=n!
r!(n−r)!
where n!represents the factorial of n.
Step 2: Substituting n= 10 and r= 3 into the formula, we get:
(10
3)=10!
3!(10 −3)!
Step 3: Calculating the factorials:
(10
3)=10 ×9×8
3×2×1
(10
3)=720
6
(10
3)= 120
Step 4: Therefore, there are 120 ways to choose a committee of 3 people
from a group of 10 people.
Question 8
Question
How many different 7-letter arrangements can be formed using all the letters of
the word ”UNIVERSITY” such that no two vowels are adjacent?
5
Solution
Step 1: Find the total number of ways to arrange the letters of ”UNIVERSITY”
without any restrictions.
The word ”UNIVERSITY” has 10 letters, but there are repetitions of some
letters:-2U’s-2I’s-2E’s-1N-1V-1R-1S-1T
Therefore, the total number of ways to arrange the letters without restric-
tions is the factorial of the total number of letters divided by the factorials of
the number of repetitions:
10!/(2! ×2! ×2! ×1! ×1! ×1! ×1! ×1!)
Step 2: Find the number of ways to arrange the letters of ”UNIVERSITY”
such that the vowels are adjacent.
Since there are 4 vowels in ”UNIVERSITY” (U, I, E, and I), we can consider
them as one unit. Therefore, the letters ”UNVRSITY” together form 7 units.
The number of ways to arrange these 7 units is 7!.
Step 3: Subtract the result from Step 2 from the result of Step 1 to find the
number of arrangements where no two vowels are adjacent.
The number of ways to arrange the letters such that no two vowels are
adjacent is given by:
10!/(2! ×2! ×2! ×1! ×1! ×1! ×1!) −7!
Thus, the above expression represents the total number of different 7-letter
arrangements of the word ”UNIVERSITY” where no two vowels are adjacent.
Question 9
Question
A company needs to assemble a committee of 5 employees from a pool of 10
potential candidates. If 3 of the candidates are engineers and 7 are accountants,
how many ways can the committee be formed if it must consist of at least 1
engineer and 2 accountants?
Solution
Step 1: Calculate the number of ways to choose 1 engineer and 2 accountants:
There are 3 engineers to choose from and 7 accountants. We can choose 1
engineer in 3 ways and 2 accountants in (7
2)ways. So, the number of ways to
choose 1 engineer and 2 accountants is 3×(7
2).
Step 2: Calculate the number of ways to choose 2 engineers and 3 accoun-
tants: There are 3 engineers to choose from and 7 accountants. We can choose
2 engineers in (3
2)ways and 3 accountants in (7
3)ways. So, the number of ways
to choose 2 engineers and 3 accountants is (3
2)×(7
3).
6
Step 3: Add the results from step 1 and step 2 to find the total number of
ways to form the committee: Total number of ways = 3×(7
2)+(3
2)×(7
3).
Step 4: Calculate the total number of ways to form the committee: 3×(7
2)=
3×7!
2!(7−2)! = 3 ×7×6
2×1= 63 ways
(3
2)×(7
3)=3!
2!(3−2)! ×7!
3!(7−3)! = 3 ×35 = 105 ways
Total number of ways = 63 + 105 = 168.
Therefore, there are 168 ways to form the committee consisting of at least 1
engineer and 2 accountants.
Question 10
Question
A committee of 5 people is to be formed from a group of 7 men and 4 women.
If the committee must consist of at least 3 men, how many different committees
can be formed?
Solution
Step 1: Find the total number of committees without any restrictions. The
total number of ways to form a committee of 5 people from 7 men and 4 women
is given by the combination formula:
(11
5)=11!
5!(11 −5)! =11!
5!6! = 462
Step 2: Find the number of committees with less than 3 men. The number
of committees with less than 3 men can be found by: - Choosing 0 men from
7 men and 5 women from 4 women, or - Choosing 1 man from 7 men and 4
women from 4 women, or - Choosing 2 men from 7 men and 3 women from 4
women.
(7
0)×(4
5)+(7
1)×(4
4)+(7
2)×(4
3)= 1×4+7×1 + 21 ×4 = 4+7 + 84 = 95
Step 3: Calculate the number of committees with at least 3 men. The
number of committees with at least 3 men is the total number of committees
minus the number of committees with less than 3 men:
462 −95 = 367
Therefore, there are 367 different committees that can be formed with at
least 3 men.
7
Question 11
Question
A committee of 5 people is to be formed from a group of 8 men and 6 women.
How many ways can the committee be formed if it must consist of at least 3
women?
Solution
Step 1: Calculate the number of ways to select exactly 3 women from the 6
available. There are (6
3)ways to choose 3 women.
Step 2: Calculate the number of ways to choose the remaining 2 committee
members (either men or women). Since the committee must consist of at least
3 women, we can choose the remaining 2 members from the 8 men or the 3
remaining women. We will compute both possibilities: - If we choose 2 men
from the 8 available, there are (8
2)ways to select them. - If we choose 2 women
from the 3 remaining, there are (3
2)ways to select them.
Step 3: Find the total number of ways to form the committee. The total
number of ways to form the committee with at least 3 women is the product
of the number of ways from Step 1 and Step 2. Add the results of the two
possibilities: (6
3)×(8
2)+(6
3)×(3
2)ways.
Step 4: Calculate the final answer. - (6
3)=6!
3!(6−3)! =6×5×4
3×2×1= 20 ways. -
(8
2)=8!
2!(8−2)! =8×7
2×1= 28 ways. - (3
2)=3!
2!(3−2)! =3
2×1= 3 ways.
Therefore, the total number of ways to form the committee is 20 ×28 + 20 ×
3 = 640 + 60 = 700 ways.
Question 12
Question
In a group of 10 students, how many ways can we select a president, vice presi-
dent, and treasurer if no student can hold more than one position?
Solution
To solve this problem, we will use the fundamental principle of counting, also
known as the multiplication principle.
Step 1: Find the number of ways to select a president from 10 students.
Since each student can be selected as the president only once, there are 10
choices for president.
Step 2: Find the number of ways to select a vice president from the re-
maining 9 students. After selecting a president, there are 9 students remaining
to choose from for vice president.
8
Step 3: Find the number of ways to select a treasurer from the remaining
8 students. After selecting a president and vice president, there are 8 students
remaining to choose from for treasurer.
Therefore, the total number of ways to select a president, vice president, and
treasurer from a group of 10 students is:
10 ×9×8 = 720
So, there are 720 different ways to select a president, vice president, and
treasurer from the group of 10 students.
Question 13
Question
A committee of 5 members is to be selected from a group of 9 men and 8 women.
In how many ways can this committee be formed if it must consist of 3 men and
2 women?
Solution
Step 1: Find the number of ways to choose 3 men from 9 men. Step 2: Find the
number of ways to choose 2 women from 8 women. Step 3: Multiply the results
from Step 1 and Step 2 to find the total number of ways to form the committee.
Step 1: To choose 3 men from 9 men, we use the combination formula:
(n
k)=n!
k!(n−k)! . Therefore, the number of ways to choose 3 men from 9 men is
(9
3)=9!
3!(9−3)! =9×8×7
3×2×1= 84 ways.
Step 2: To choose 2 women from 8 women, we use the combination formula:
(n
k)=n!
k!(n−k)! . Therefore, the number of ways to choose 2 women from 8 women
is (8
2)=8!
2!(8−2)! =8×7
2×1= 28 ways.
Step 3: Multiply the results from Step 1 and Step 2. Total number of ways
to form the committee = 84 ×28 = 2352 ways.
Therefore, there are 2352 ways to form a committee consisting of 3 men and
2 women from a group of 9 men and 8 women.
Question 14
Question
In a group of 12 people, how many ways can we choose 5 people to form a
committee if one person, John, must be included?
9
Solution
Step 1: Since one person, John, must be included, we need to choose the re-
maining 4 people from the remaining 11 people. Step 2: The number of ways to
choose 4 people out of 11 is given by the combination formula nCk=n!
k!(n−k)! .
Step 3: Substituting n= 11 and k= 4 into the formula, we get:
11C4=11!
4!(11 −4)!
Step 4: Calculating the factorials, we get:
11C4=11!
4!7! =11 ×10 ×9×8
4×3×2×1= 330
Step 5: Therefore, there are 330 ways to choose 5 people to form a committee
if one person, John, must be included.
Question 15
Question
In a graduate statistics class, there are 8 students. The professor needs to choose
a team of 3 students to work on a research project together. How many different
ways can the professor choose the team if one student, Sarah, must be on it?
Solution
Step 1: Since Sarah must be on the team, we only need to choose 2 more
students from the remaining 7 students.
Step 2: We can calculate the number of ways to choose 2 students from 7
using the combination formula C(n, k) = n!
k!(n−k)! .
Step 3: Substitute n= 7 and k= 2 into the formula to find the number of
ways to choose 2 students from 7.
C(7,2) = 7!
2!(7 −2)! =7×6
2×1= 21
Step 4: Therefore, there are 21 different ways the professor can choose a
team of 3 students with Sarah on it.
Question 16
Question
In a group of 10 people, how many ways can you choose a committee of 3 people
and a president from the remaining 7 people?
10
Solution
Step 1: Calculate the number of ways to choose a committee of 3 people from
the group of 10. To choose a committee of 3 people from a group of 10, we use
the combination formula:
C(n, k) = n!
k!(n−k)!
where nis the total number of people and kis the number of people we want
to choose. In this case, n= 10 and k= 3, so the number of ways to choose a
committee of 3 people is:
C(10,3) = 10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
Step 2: Calculate the number of ways to choose a president from the re-
maining 7 people. After choosing the committee of 3 people, we have 7 people
remaining. We need to choose 1 person as the president from these 7 people.
This can be done in 7 ways.
Step 3: Multiply the results from Steps 1 and 2 to find the total number
of ways. The total number of ways to choose a committee of 3 people and a
president is the product of the number of ways to choose the committee and the
number of ways to choose the president:
Total ways = 120 ×7 = 840
Therefore, there are 840 ways to choose a committee of 3 people and a
president from a group of 10 people.
Question 17
Question
In a group of 10 people, how many ways can we choose a committee of 3 people
where one person is designated as the chair of the committee?
Solution
Step 1: First, we choose the person who will be the chair of the committee.
There are 10 ways to choose the chairperson.
Step 2: Next, we choose 2 people from the remaining 9 people to join the
committee. We can choose 2 people out of 9 in (9
2)ways.
Step 3: Multiply the results from Step 1 and Step 2 to find the total number
of ways to choose the committee. So, the total number of ways to choose a
committee of 3 people with a designated chair is: 10 ×(9
2)= 10 ×9!
2!(9−2)! =
10 ×9×8
2= 10 ×36 = 360.
Therefore, there are 360 ways to choose a committee of 3 people with one
person designated as the chair.
11
Question 18
Question
A committee of 5 people is to be formed from a group of 10 men and 12 women.
How many different committees can be formed if at least 2 men must be included
in the committee?
Solution
Step 1: Determine the number of ways to choose 2, 3, 4, or 5 men for the
committee.
There are (10
2)ways to choose 2 men, (10
3)ways to choose 3 men, (10
4)ways
to choose 4 men, and (10
5)ways to choose all 5 men.
Step 2: Determine the number of ways to choose the remaining committee
members from the women.
After choosing 2, 3, 4, or 5 men, the committee must be completed with 3,
2, 1, or 0 women, respectively.
There are (12
3)ways to choose 3 women, (12
2)ways to choose 2 women, (12
1)
ways to choose 1 woman, and (12
0)= 1 way to choose 0 women.
Step 3: Calculate the total number of ways to form the committee.
The total number of ways to form the committee is the sum of the number
of ways from Step 1 and Step 2.
(10
2)×(12
3)+(10
3)×(12
2)+(10
4)×(12
1)+(10
5)×(12
0)
Performing the calculations will give us the final answer.
Question 19
Question
In a group of 10 students, how many ways can we choose a committee of 5
students with a president, a vice president, and a secretary from the group?
Solution
Step 1: To find the number of ways to choose a committee of 5 students from a
group of 10, we will use the combination formula (n
k)=n!
k!(n−k)! . Therefore, the
number of ways to choose 5 students from a group of 10 is (10
5)=10!
5!5! = 252.
Step 2: Among the 5 selected students, there are 5 ways to choose the pres-
ident, then 4 ways to choose the vice president from the remaining 4 students,
and finally 3 ways to choose the secretary from the remaining 3 students.
12
Step 3: Therefore, the total number of ways to choose the committee of 5
students with a president, a vice president, and a secretary is 252 ×5×4×3 =
60480.
So, there are 60,480 ways to choose a committee of 5 students with a presi-
dent, a vice president, and a secretary from a group of 10 students.
Question 20
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 3 men and 2 women, how many different ways
can the committee be formed?
Solution
Step 1: Determine the number of ways to choose 3 men out of 10. There are
(10
3)ways to choose 3 men from a group of 10 men. Calculating (10
3):
(10
3)=10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
So, there are 120 ways to choose 3 men.
Step 2: Determine the number of ways to choose 2 women out of 8. There
are (8
2)ways to choose 2 women from a group of 8 women. Calculating (8
2):
(8
2)=8!
2!(8 −2)! =8×7
2×1= 28
So, there are 28 ways to choose 2 women.
Step 3: Multiply the number of ways to choose men and women. To find the
total number of ways to form a committee consisting of 3 men and 2 women,
multiply the number of ways to choose men and women together: 120 ×28 =
3360.
Therefore, there are 3360 different ways to form the committee.
Question 21
Question
A committee of 5 people must be formed from a group of 8 men and 4 women.
How many different committees can be formed if the committee must contain
at least 2 men and 2 women?
13
Solution
Step 1: Calculate the number of ways to form a committee with 2 men and 2
women. For this case, we choose 2 men out of 8 and 2 women out of 4.
(8
2)×(4
2)= 28 ×6 = 168
Step 2: Calculate the number of ways to form a committee with 3 men and
1 woman. For this case, we choose 3 men out of 8 and 1 woman out of 4.
(8
3)×(4
1)= 56 ×4 = 224
Step 3: Calculate the number of ways to form a committee with 4 men and
1 woman. For this case, we choose 4 men out of 8 and 1 woman out of 4.
(8
4)×(4
1)= 70 ×4 = 280
Step 4: Calculate the total number of committees that can be formed with
at least 2 men and 2 women. Add the results from Step 1, Step 2, and Step 3.
168 + 224 + 280 = 672
Therefore, there are 672 different committees that can be formed if the
committee must contain at least 2 men and 2 women.
Question 22
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 3 men and at least 2 women, how many
different committees can be formed?
Solution
Step 1: Calculate the number of ways to choose exactly 3 men and 2 women for
the committee. There are (10
3)ways to choose 3 men from a group of 10 and
(8
2)ways to choose 2 women from a group of 8. Therefore, the number of ways
to choose 3 men and 2 women is given by:
(10
3)×(8
2)
Step 2: Calculate the number of ways to choose exactly 4 men and 1 woman
for the committee. There are (10
4)ways to choose 4 men from a group of 10 and
14
(8
1)ways to choose 1 woman from a group of 8. Therefore, the number of ways
to choose 4 men and 1 woman is given by:
(10
4)×(8
1)
Step 3: Calculate the number of ways to choose exactly 5 men for the com-
mittee. There are (10
5)ways to choose 5 men from a group of 10. Since the
committee must consist of at least 3 men, we only need to calculate this once.
Step 4: Add up the number of ways from Step 1, Step 2, and Step 3 to find
the total number of different committees that can be formed. The total number
of committees is:
(10
3)×(8
2)+(10
4)×(8
1)+(10
5)
Calculating the values for each combination:
(10
3)= 120,(8
2)= 28,(10
4)= 210,(8
1)= 8,(10
5)= 252
Therefore, the total number of different committees that can be formed is:
120 ×28 + 210 ×8 + 252 = 3360 + 1680 + 252 = 5292
So, there are 5292 different committees that can be formed.
Question 23
Question
In a mathematics class of 24 students, a teacher is selecting a team of 4 students
to participate in a quiz bee. How many ways can this team be formed if the
teacher wants to ensure that both Jane and Mark are included in the team?
Solution
Let’s first find the number of ways to select the other two students to join Jane
and Mark in the team.
Step 1: Since Jane and Mark are already selected, we need to choose 2
students out of the remaining 22 students. The number of ways to select 2
students out of 22 is given by the combination formula (n
r)=n!
r!(n−r)! , where
nis the total number of students and ris the number of students we want to
select. So, in this case, (22
2)=22!
2!(22−2)! .
Step 2: Calculate (22
2):
(22
2)=22!
2!(22 −2)! =22 ×21
2= 231.
15
Step 3: Now, we have the number of ways to select the other two students.
To include Jane and Mark, we need to multiply this result by 1 (since Jane
and Mark are fixed). Therefore, the total number of ways to form the team is
231 ×1 = 231 ways.
So, there are 231 ways to form a team of 4 students where both Jane and
Mark are included.
Question 24
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: Calculate the number of ways to form a committee with exactly 2 women
and 3 men.
• Choose 2 women out of 8: (8
2)= 28 ways
• Choose 3 men out of 10: (10
3)= 120 ways
Therefore, the number of ways to form a committee with exactly 2 women and
3 men is 28 ×120 = 3360 ways.
Step 2: Calculate the number of ways to form a committee with exactly 3
women and 2 men.
• Choose 3 women out of 8: (8
3)= 56 ways
• Choose 2 men out of 10: (10
2)= 45 ways
Therefore, the number of ways to form a committee with exactly 3 women and
2 men is 56 ×45 = 2520 ways.
Step 3: Calculate the number of ways to form a committee with 4 women
and 1 man.
• Choose 4 women out of 8: (8
4)= 70 ways
• Choose 1 man out of 10: (10
1)= 10 ways
Therefore, the number of ways to form a committee with 4 women and 1 man
is 70 ×10 = 700 ways.
Step 4: Calculate the number of ways to form a committee with 5 women
and 0 men.
• There are (8
5)= 56 ways to choose 5 women.
16
Step 5: Add the results from Steps 1 to 4 to find the total number of different
committees that can be formed.
3360 + 2520 + 700 + 56 = 6636
Therefore, there are 6636 different committees that can be formed with at least
2 women.
Question 25
Question
How many ways are there to choose a committee of 5 people from a group of
10 people, where two specific people, Alice and Bob, must be on the committee
together?
Solution
Step 1: First, we choose Alice and Bob to be on the committee together. There
is only 1 way to choose Alice and Bob since they must be together.
Step 2: Now, we need to choose 3 more people to complete the committee of
5. Since we have already selected 2 people (Alice and Bob), we need to choose 3
more people from the remaining 8 people. This can be done in (8
3)=8!
3!5! = 56
ways.
Step 3: Multiply the results from Step 1 and Step 2 to find the total number
of ways to choose the committee. 1×56 = 56
Therefore, there are 56 ways to choose a committee of 5 people from a group
of 10 people, where Alice and Bob must be on the committee together.
17
