MATH 201 - INTRODUCTION TO
PROBABILITY AND STATISTICS -
Combinatorial Analysis
Question Bank - Set 10
Liberty University
Question 1
Question
In how many ways can you select a committee of 4 people from a group of 10
students?
Solution
Step 1: To solve this problem, we can use the combination formula, which
calculates the number of ways to choose robjects from a set of nobjects without
regard to the order of selection. The combination formula is given by:
C(n, r) = n!
r!(n−r)!
Step 2: In this case, we have 10 students from which we want to choose
a committee of 4 people. Plugging n= 10 and r= 4 into the combination
formula, we get:
C(10,4) = 10!
4!(10 −4)!
Step 3: Simplifying the expression:
C(10,4) = 10!
4!6! =10 ×9×8×7
4×3×2×1
Step 4: Calculating the value:
C(10,4) = 5040
24 = 210
Step 5: Therefore, there are 210 ways to select a committee of 4 people from
a group of 10 students.
Question 2
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 choose 2 women from 8 women and 3
men from 10 men. Then, calculate the number of ways to choose 3 women from
8 women and 2 men from 10 men.
Step 2: Calculate the number of ways for each combination and sum them
to find the total number of different committees that can be formed.
Let’s calculate the number of ways for each combination.
Step 1: Choosing 2 women from 8 women:
(8
2)=8!
2!(8 −2)! = 28
Choosing 3 men from 10 men:
(10
3)=10!
3!(10 −3)! = 120
Now, we calculate the number of committees with 2 women and 3 men:
28 ×120 = 3360
Step 2: Choosing 3 women from 8 women:
(8
3)=8!
3!(8 −3)! = 56
Choosing 2 men from 10 men:
(10
2)=10!
2!(10 −2)! = 45
Now, we calculate the number of committees with 3 women and 2 men:
56 ×45 = 2520
Therefore, the total number of different committees that can be formed is:
3360 + 2520 = 5880
Thus, there are 5880 different committees that can be formed.
2
Question 3
Question
A committee of 5 people needs to be formed from a group of 10 women and 8
men. If the committee must consist of at least 3 women, how many different
committees can be formed?
Solution
Step 1: Count the number of ways to choose 3 women from 10. Step 2: Count
the number of ways to choose 2 additional members from the remaining 8 people
(men). Step 3: Multiply the results from Step 1 and Step 2 to find the total
number of possible committees.
Step 1: To choose 3 women from 10, we use the combination formula. The
number of ways to choose 3 women from 10 is given by (10
3)=10!
3!(10−3)! =
10×9×8
3×2×1= 120.
Step 2: We need to choose 2 additional members from the remaining 8 people
(men). The number of ways to choose 2 men from 8 is given by (8
2)=8!
2!(8−2)! =
8×7
2×1= 28.
Step 3: To find the total number of different committees that can be formed
consisting of at least 3 women, we multiply the results from Steps 1 and 2:
120 ×28 = 3360.
Therefore, there are 3360 different committees that can be formed.
Question 4
Question
In how many ways can you arrange the letters in the word ”UNIVERSITY”
such that no two vowels are adjacent?
Solution
Step 1: Let’s first count the number of ways to arrange the vowels in the word
”UNIVERSITY”. The word ”UNIVERSITY” has 4 vowels (U, I, E, I) and 6
consonants (N, V, R, S, T, Y). We will treat the vowels as a single entity, so
there are 5 entities to arrange (Vowels, N, V, R, S, T, Y).
Step 2: There are 5! ways to arrange these entities.
Step 3: Within the ”vowels” entity, there are 4! ways to arrange the vowels
themselves.
Step 4: So, the total number of ways to arrange the vowels in the word
”UNIVERSITY” is 5! ×4!.
Step 5: Now, let’s count the number of ways to arrange the 6 consonants in
the word ”UNIVERSITY”. Since there are 6 consonants, we will treat them as
a single entity.
3
Step 6: There are 6! ways to arrange these consonants.
Step 7: So, the total number of ways to arrange the consonants in the word
”UNIVERSITY” is 6!.
Step 8: The total number of ways to arrange the letters in the word ”UNI-
VERSITY” such that no two vowels are adjacent is the product of the number of
ways to arrange the vowels and the number of ways to arrange the consonants.
Step 9: Therefore, the total number of ways to arrange the letters in the word
”UNIVERSITY” such that no two vowels are adjacent is 5! ×4! ×6! = 172,800.
Question 5
Question
In a group of 10 students, how many ways can we choose a committee of 4
students?
Solution
Step 1: To solve this problem, we can use the formula for combinations, which
is given by:
C(n, k) = n!
k!(n−k)!
where nis the total number of items to choose from and kis the number of
items to choose.
Step 2: Plugging in n= 10 and k= 4 into the formula, we get:
C(10,4) = 10!
4!(10 −4)!
Step 3: Calculating the factorials:
C(10,4) = 10 ×9×8×7×6×5×4×3×2×1
(4 ×3×2×1) ×(6 ×5×4×3×2×1)
Step 4: Simplifying the expression:
C(10,4) = 10 ×9×8×7
4×3×2×1= 210
Step 5: Therefore, there are 210 ways to choose a committee of 4 students
from a group of 10.
Question 6
Question
In a group of 10 people, how many ways are there to select a committee of 4
people with exactly 2 men and 2 women if there are 4 men and 6 women in the
group?
4
Solution
Step 1: Calculate the number of ways to choose 2 men from 4 men. There are
(4
2)ways to choose 2 men from the 4 men in the group.
(4
2)=4!
2!(4 −2)! =4×3
2×1= 6
Step 2: Calculate the number of ways to choose 2 women from 6 women.
There are (6
2)ways to choose 2 women from the 6 women in the group.
(6
2)=6!
2!(6 −2)! =6×5
2×1= 15
Step 3: Multiply the number of ways to choose 2 men and 2 women. The
total number of ways to choose 2 men and 2 women from the group of 10 people
is 6×15.
6×15 = 90
Therefore, there are 90 ways to select a committee of 4 people with exactly
2 men and 2 women from a group of 10 people.
Question 7
Question
In a group of 10 students, how many ways can a committee of 4 students be
selected, where one student is the president, one is the vice-president, and the
other two are members?
Solution
To find the number of ways a committee of 4 students can be selected with spe-
cific roles assigned (president, vice-president, members), we can use the concept
of permutations.
Step 1: Choose the president from 10 students. There are 10 choices for
the president.
Step 2: Choose the vice-president from the remaining 9 students. There
are 9 choices for the vice-president.
Step 3: Choose the first member from the remaining 8 students. There are
8 choices for the first member.
Step 4: Choose the second member from the remaining 7 students. There
are 7 choices for the second member.
Therefore, the total number of ways to select a committee of 4 students with
specific roles assigned is given by:
10 ×9×8×7 = 5040
So, there are 5040 ways to select a committee of 4 students with one presi-
dent, one vice-president, and two members from a group of 10 students.
5
Question 8
Question
A committee of 5 people is to be formed from a group of 8 men and 6 women.
How many different committees can be formed if there must be at least 3 men
on the committee?
Solution
Step 1: Find the number of ways to choose 3 men out of 8. There are (8
3)ways
to choose 3 men from 8.
Step 2: Find the number of ways to choose 2 people from the remaining men
and all 6 women. There are (6
2)ways to choose 2 people from the remaining
men and (6
0)ways to choose all 6 women.
Step 3: Calculate the total number of possible committees with at least
3 men. The total number of possible committees with at least 3 men is the
product of the choices in Step 1 and Step 2:
(8
3)×(6
2)×(6
0)= 56 ×15 ×1 = 840
Therefore, there are 840 different committees that can be formed if there
must be at least 3 men on the committee.
Question 9
Question
A committee of 5 people is to be formed from a group of 10 men and 8 women.
How many different committees can be formed if the committee must consist of
at least 2 men and at least 2 women?
Solution
Step 1: Calculate the number of ways to choose 2 men out of 10. There are (10
2)
ways to choose 2 men from a group of 10.
Step 2: Calculate 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.
Step 3: Calculate the number of ways to choose the remaining person (either
a man or a woman). Since the committee must consist of at least 2 men and
at least 2 women, we need to choose 1 more person after selecting 2 men and 2
women. This person can be either a man or a woman, so there are 10+8−4 = 14
people left to choose from.
Step 4: Calculate the total number of committees that can be formed. The
total number of committees that can be formed is the product of the choices in
6
steps 1, 2, and 3:
(10
2)×(8
2)×14
Step 5: Simplify the expression.
(10
2)=10!
2!(10 −2)! =10 ×9
2×1= 45
(8
2)=8!
2!(8 −2)! =8×7
2×1= 28
Therefore, the total number of committees that can be formed is:
45 ×28 ×14 = 17640
So, there are 17,640 different committees that can be formed consisting of
at least 2 men and at least 2 women.
Question 10
Question
Suppose we have a group of 10 people, including Alice and Bob. In how many
ways can we arrange these 10 people in a line such that Alice and Bob are not
next to each other?
Solution
Let’s first find the total number of ways to arrange the 10 people in a line
without any restrictions, and then subtract the number of ways where Alice
and Bob are next to each other.
Step 1: Find the total number of ways to arrange the 10 people in a line.
Since there are 10 people, there are 10! ways to arrange them in a line.
Step 2: Find the number of ways to arrange the 10 people in a line such
that Alice and Bob are next to each other. Consider Alice and Bob as a single
entity. Then there are 9 entities to arrange (the group of Alice and Bob counted
as one, and the remaining 8 people). Within the group of Alice and Bob, there
are 2 ways to arrange them. Hence, there are 9! ×2! ways to arrange the 10
people in a line with Alice and Bob next to each other.
Step 3: Subtract the number of ways to arrange the 10 people with Alice
and Bob next to each other from the total number of ways to arrange the 10
people. The number of ways to arrange the 10 people in a line such that Alice
and Bob are not next to each other is 10! −9! ×2!.
Therefore, there are 10! −9! ×2! = 3,026,880 −725,760 = 2,301,120 ways
to arrange the 10 people in a line such that Alice and Bob are not next to each
other.
7
Question 11
Question
In a group of 12 people, how many ways can we choose a committee of 5 people
if there are 3 specific people that must be on the committee?
Solution
Step 1: First, we choose the 3 specific people that must be on the committee.
Since these 3 people are already chosen, we only need to choose 2 more people
from the remaining 9 people.
(9
2)=9!
2!(9 −2)! =9×8
2×1= 36
Step 2: So, there are 36 ways to choose 2 more people from the remaining 9
people. Therefore, there are a total of 3×36 = 108 ways to choose a committee
of 5 people with 3 specific people included.
Question 12
Question
In how many ways can 5 people be seated in a row with restrictions that two of
them, A and B, must not sit together?
Solution
Step 1: First we calculate the total number of ways the 5 people can be seated
without any restrictions.
The number of ways to seat 5 people in a row is 5! = 5 ×4×3×2×1 = 120.
Step 2: Next, we calculate the number of ways when A and B sit together.
Since A and B must sit together, consider them as one group. Then, there
are 4 entities (AB, C, D, E) that can be seated in 4! ways. However, A and B can
switch positions among themselves in 2 ways, so the final count is 4! ×2 = 48.
Step 3: Finally, we subtract the number of ways A and B can sit together
from the total number of ways to get the number of ways they must not sit
together.
Therefore, the number of ways the 5 people can be seated with A and B not
sitting together is 120 −48 = 72.
8
Question 13
Question
Suppose you have 5 red balls, 3 blue balls, and 4 yellow balls. If you randomly
select 4 balls without replacement, what is the probability of selecting 2 red
balls, 1 blue ball, and 1 yellow ball in any order?
Solution
Step 1: Find the total number of ways to choose 4 balls out of 12. Since order
does not matter, we use combinations. The total number of ways to choose 4
balls out of 12 is given by (12
4).
Step 2: Find the number of ways to choose 2 red balls, 1 blue ball, and 1
yellow ball. The number of ways to choose 2 red balls out of 5 is (5
2). The
number of ways to choose 1 blue ball out of 3 is (3
1). The number of ways to
choose 1 yellow ball out of 4 is (4
1). Therefore, the total number of ways to
choose 2 red balls, 1 blue ball, and 1 yellow ball is (5
2)·(3
1)·(4
1).
Step 3: Calculate the probability. The probability of selecting 2 red balls, 1
blue ball, and 1 yellow ball in any order is given by:
(5
2)·(3
1)·(4
1)
(12
4)=10 ·3·4
495
Therefore, the probability of selecting 2 red balls, 1 blue ball, and 1 yellow
ball in any order is 120
495 =8
33 .
Question 14
Question
In a class of 30 students, there are 12 students who are interested in statistics
and 18 students who are interested in probability. If 8 students are interested in
both statistics and probability, how many students in the class are not interested
in either statistics or probability?
Solution
Step 1: Let’s denote the number of students interested in statistics as S, the
number of students interested in probability as P, and the number of students
interested in both statistics and probability as B. We are given: S= 12 (in-
terested in statistics), P= 18 (interested in probability), B= 8 (interested in
both statistics and probability).
Step 2: We can find the total number of students interested in either statis-
tics or probability (including those interested in both) using the principle of
inclusion-exclusion: S∪P=S+P−B= 12 + 18 −8 = 22.
9
Step 3: Now, we can find the number of students not interested in either
statistics or probability by subtracting the number of students interested in
either from the total number of students: Total students in the class = 30.
Number of students not interested in either = 30 −22 = 8.
Therefore, there are 8students in the class who are not interested in either
statistics or probability.
Question 15
Question
A committee of 5 people is to be formed from a group of 8 men and 6 women.
If the committee must consist of at least 3 men, what is the number of ways the
committee can be formed?
Solution
Step 1: Find the total number of ways to form a committee of 5 people from
14 individuals. Since the committee can consist of any combination of men and
women, we will use the combination formula.
Total ways =(14
5)=14!
5!(14 −5)! =14!
5!9! = 2002
Step 2: Find the number of ways to form a committee with less than 3 men.
The committee can consist of 0, 1, or 2 men.
Ways with 0 men =(8
0)(6
5)= 1 ×6 = 6
Ways with 1 man =(8
1)(6
4)= 8 ×15 = 120
Ways with 2 men =(8
2)(6
3)= 28 ×20 = 560
Total ways with less than 3 men = 6 + 120 + 560 = 686
Step 3: Find the number of ways to form a committee with at least 3 men.
Ways with at least 3 men =Total ways −Ways with less than 3 men
= 2002 −686 = 1316
Therefore, there are 1316 ways to form a committee with at least 3 men from
the group of 8 men and 6 women.
10
Question 16
Question
In a group of 10 students, how many ways can you select a committee of 4
students to represent the class if no student can serve on the committee more
than once?
Solution
Step 1: To solve this problem, we will use the formula for combinations:
(n
k)=n!
k!(n−k)!
where nis the total number of students and kis the number of students we
want to choose for the committee.
Step 2: Substitute n= 10 and k= 4 into the formula:
(10
4)=10!
4!(10 −4)!
Step 3: Simplify the factorial expressions:
(10
4)=10 ×9×8×7
4×3×2×1
Step 4: Perform the multiplication and division:
(10
4)=5040
24
Step 5: Calculate the final result:
(10
4)= 210
Therefore, there are 210 ways to select a committee of 4 students from a
group of 10 students without any student serving on the committee more than
once.
Question 17
Question
A committee of 5 people is to be formed from a group of 9 men and 6 women.
If the committee must include at least 2 men and 2 women, how many different
committees can be formed?
11
Solution
Step 1: Find the number of ways to select 2 men and 3 women. There are (9
2)
ways to select 2 men from 9 men, and (6
3)ways to select 3 women from 6 women.
So, the number of ways to select 2 men and 3 women is (9
2)·(6
3).
Step 2: Find the number of ways to select 3 men and 2 women. There are
(9
3)ways to select 3 men from 9 men, and (6
2)ways to select 2 women from 6
women. So, the number of ways to select 3 men and 2 women is (9
3)·(6
2).
Step 3: Add the results from Step 1 and Step 2 to find the total number of
different committees that can be formed with at least 2 men and 2 women.
Therefore, the total number of different committees that can be formed is:
(9
2)·(6
3)+(9
3)·(6
2)=9!
2!7! ·6!
3!3! +9!
3!6! ·6!
2!4! .
Now, we can simplify this expression to find the final answer.
Question 18
Question
In how many ways can 5 distinct books be arranged on a bookshelf if one
particular pair of books are never to be separated?
Solution
Step 1: Consider the two books that are not to be separated as one entity. Then,
there are 4 entities to arrange on the bookshelf: the pair of books and the other
3 distinct books.
Step 2: Since the two books must always be together, we treat them as one
entity. So, there are 4! ways to arrange the 4 entities.
Step 3: However, within the pair of books, there are 2! ways to arrange the
books themselves.
Step 4: Therefore, the total number of ways to arrange the 5 distinct books
where the particular pair are never separated is 4! ×2! = 48.
Question 19
Question
Suppose a committee of 5 people is to be selected from a group of 10 men and
8 women. In how many ways can the committee be formed if it must have at
least 3 men and 2 women?
12
Solution
Step 1: Calculate the number of ways to select 3 men from 10.
(10
3)=10!
3!(10 −3)! =10 ·9·8
3·2·1= 120
Step 2: Calculate the number of ways to select 2 women from 8.
(8
2)=8!
2!(8 −2)! =8·7
2·1= 28
Step 3: Calculate the number of ways to select the remaining 2 people (could
be any gender) from the remaining pool of 7 people.
(7
2)=7!
2!(7 −2)! =7·6
2·1= 21
Step 4: Multiply the results from Step 1, Step 2, and Step 3 to find the total
number of ways to form the committee. Total ways = 120 ×28 ×21 = 70560
Therefore, the committee can be formed in 70,560 ways if it must have at
least 3 men and 2 women.
Question 20
Question
In a group of 10 friends, how many ways can you choose a committee of 4 people
if there are 2 friends who refuse to be on the same committee together?
Solution
Step 1: Calculate the total number of ways to choose a committee of 4 people
from a group of 10 friends. Step 2: Calculate the number of ways to choose a
committee where the 2 friends who refuse to be on the same committee together
are both included. Step 3: Calculate the number of ways to choose a committee
where the 2 friends who refuse to be on the same committee together are both
excluded. Step 4: Subtract the results from steps 2 and 3 from the result of
step 1 to obtain the final answer.
Step 1: To choose a committee of 4 people from a group of 10 friends, we
use the combination formula:
(10
4)=10!
4!(10 −4)! =10 ×9×8×7
4×3×2×1= 210
Step 2: If the 2 friends who refuse to be on the same committee together are
both included, we have to choose 2 more people from the remaining 8 friends.
This can be done in (8
2)ways.
(8
2)=8!
2!(8 −2)! =8×7
2×1= 28
13
Step 3: If the 2 friends who refuse to be on the same committee together
are both excluded, we can simply choose 4 people from the remaining 8 friends.
This can be done in (8
4)ways.
(8
4)=8!
4!(8 −4)! =8×7×6×5
4×3×2×1= 70
Step 4: The total number of ways to choose a committee of 4 people where
the 2 friends who refuse to be on the same committee together are not included
is: (10
4)−(8
2)−(8
4)= 210 −28 −70 = 112
Therefore, there are 112 ways to choose a committee of 4 people from a group
of 10 friends where 2 friends refuse to be on the same committee together.
Question 21
Question
A committee of 5 people is to be formed from a group of 10 females and 8
males. If the committee must consist of at least 3 females, how many different
committees can be formed?
Solution
Step 1: Find the number of ways to choose exactly 3 females and 2 males for the
committee. Step 2: Find the number of ways to choose exactly 4 females and
1 male for the committee. Step 3: Find the number of ways to choose exactly
5 females for the committee. Step 4: Add the results from Steps 1, 2, and 3 to
find the total number of different committees that can be formed.
Step 1: There are (10
3)ways to choose 3 females from the group of 10 females,
and (8
2)ways to choose 2 males from the group of 8 males. So, the number of
ways to choose exactly 3 females and 2 males is (10
3)×(8
2).
Step 2: Similarly, there are (10
4)ways to choose 4 females and (8
1)ways to
choose 1 male. So, the number of ways to choose exactly 4 females and 1 male
is (10
4)×(8
1).
Step 3: To select all 5 committee members as females, we choose all 5 females
from the group of 10 females. There is only 1 way to choose all 5 females.
Step 4: The total number of different committees that can be formed is
the sum of the results from Steps 1, 2, and 3. Therefore, the total number of
different committees that can be formed is (10
3)×(8
2)+(10
4)×(8
1)+ 1.
14
Question 22
Question
In a group of 10 people, how many different ways can we choose a committee
of 4 people if Alice and Bob refuse to serve on the committee together?
Solution
Step 1: Calculate the total number of ways to choose a committee of 4 people
from a group of 10 people. Step 2: Calculate the number of ways to choose a
committee of 4 people when Alice and Bob are together. Step 3: Subtract the
result from Step 2 from the result in Step 1 to find the number of ways when
Alice and Bob are not together.
Step 1: To choose a committee of 4 people from a group of 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 from the group of
10.
Step 2: When Alice and Bob are serving together, we treat them as one
person. So, we now have 9 people to choose from. We need to choose 3 more
people to complete the committee. Using the combination formula again:
(9
3)=9!
3!(9 −3)! =9×8×7
3×2×1= 84
Therefore, there are 84 ways to choose a committee of 4 people when Alice
and Bob are together.
Step 3: Subtract the number of ways in Step 2 from the total number of
ways in Step 1 to find the number of ways when Alice and Bob are not together:
210 −84 = 126
So, there are 126 ways to choose a committee of 4 people from the group of
10 where Alice and Bob are not both serving on the committee.
Question 23
Question
In a mathematics competition, there are 10 multiple-choice questions with 5
answer choices each. If a student randomly selects an answer for each question,
what is the probability that the student gets at least 8 questions correct?
15
Solution
To solve this problem, we can use the concept of combinations and the binomial
probability formula.
Step 1: Calculate the total number of ways to answer all 10 questions.
Since each question has 5 answer choices, there are 510 total ways the student
can answer all 10 questions.
Step 2: Calculate the number of ways to get exactly 8 questions correct.
To get exactly 8 questions correct, the student must choose 8 correct answers
and 2 incorrect answers out of the 10 questions. The number of ways to do this
is (10
8)=10!
8!(10−8)! ·1·1 = 45.
For each of these ways, the probability of getting exactly 8 questions correct
is (1
5)8×(4
5)2.
Step 3: Calculate the number of ways to get exactly 9 questions correct.
To get exactly 9 questions correct, the student must choose 9 correct answers
and 1 incorrect answer out of the 10 questions. The number of ways to do this
is (10
9)=10!
9!(10−9)! ·1·1 = 10.
For each of these ways, the probability of getting exactly 9 questions correct
is (1
5)9×(4
5)1.
Step 4: Calculate the number of ways to get all 10 questions correct. To
get all 10 questions correct, the student must choose all 10 correct answers out
of the 10 questions. There is only 1 way to do this.
The probability of getting all 10 questions correct is (1
5)10.
Step 5: Calculate the total probability of getting at least 8 questions correct.
The total probability is the sum of the probabilities of getting exactly 8, 9, or
10 questions correct.
Total probability =Probability of getting exactly 8 questions correct
+Probability of getting exactly 9 questions correct
+Probability of getting all 10 questions correct
= 45 ×(1
5)8
×(4
5)2
+ 10 ×(1
5)9
×(4
5)1
+(1
5)10
=27735
9765625 +40
9765625 +1
9765625
=27776
9765625
≈0.002841
Therefore, the probability that the student gets at least 8 questions correct
is approximately 0.002841.
16
Question 24
Question
In a group of 10 people, how many ways are there to select a president, vice
president, and treasurer, assuming no person can hold more than one position?
Solution
Step 1: To find the number of ways to select the president, we have 10 choices.
After selecting the president, there remain 9 people for the position of vice
president.
Step 2: After selecting the president and vice president, there remain 8
people for the position of treasurer.
Step 3: Therefore, the total number of ways to select a president, vice pres-
ident, and treasurer is the product of the number of ways for each position:
10 ×9×8 = 720
So, there are 720 ways to select a president, vice president, and treasurer in
a group of 10 people.
Question 25
Question
In a group of 8 friends, how many ways can we select a committee of 3 friends?
Solution
Step 1: To find the number of ways to select a committee of 3 friends from a
group of 8, we will use the combination formula. The formula for combination
is given by:
C(n, r) = n!
r!(n−r)!
where nis the total number of friends and ris the number of friends we
want to choose.
Step 2: Substituting n= 8 and r= 3 into the formula, we have:
C(8,3) = 8!
3!(8 −3)!
Step 3: Calculating the factorials:
C(8,3) = 8×7×6
3×2×1
17
C(8,3) = 336
6
Step 4: Simplifying the fraction:
C(8,3) = 56
Therefore, there are 56 ways to select a committee of 3 friends from a group
of 8.
18
Question 2
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 choose 2 women from 8 women and 3
men from 10 men. Then, calculate the number of ways to choose 3 women from
8 women and 2 men from 10 men.
Step 2: Calculate the number of ways for each combination and sum them
to find the total number of different committees that can be formed.
Let’s calculate the number of ways for each combination.
Step 1: Choosing 2 women from 8 women:
(8
2)=8!
2!(8 −2)! = 28
Choosing 3 men from 10 men:
(10
3)=10!
3!(10 −3)! = 120
Now, we calculate the number of committees with 2 women and 3 men:
28 ×120 = 3360
Step 2: Choosing 3 women from 8 women:
(8
3)=8!
3!(8 −3)! = 56
Choosing 2 men from 10 men:
(10
2)=10!
2!(10 −2)! = 45
Now, we calculate the number of committees with 3 women and 2 men:
56 ×45 = 2520
Therefore, the total number of different committees that can be formed is:
3360 + 2520 = 5880
Thus, there are 5880 different committees that can be formed.
2
Question 3
Question
A committee of 5 people needs to be formed from a group of 10 women and 8
men. If the committee must consist of at least 3 women, how many different
committees can be formed?
Solution
Step 1: Count the number of ways to choose 3 women from 10. Step 2: Count
the number of ways to choose 2 additional members from the remaining 8 people
(men). Step 3: Multiply the results from Step 1 and Step 2 to find the total
number of possible committees.
Step 1: To choose 3 women from 10, we use the combination formula. The
number of ways to choose 3 women from 10 is given by (10
3)=10!
3!(10−3)! =
10×9×8
3×2×1= 120.
Step 2: We need to choose 2 additional members from the remaining 8 people
(men). The number of ways to choose 2 men from 8 is given by (8
2)=8!
2!(8−2)! =
8×7
2×1= 28.
Step 3: To find the total number of different committees that can be formed
consisting of at least 3 women, we multiply the results from Steps 1 and 2:
120 ×28 = 3360.
Therefore, there are 3360 different committees that can be formed.
Question 4
Question
In how many ways can you arrange the letters in the word ”UNIVERSITY”
such that no two vowels are adjacent?
Solution
Step 1: Let’s first count the number of ways to arrange the vowels in the word
”UNIVERSITY”. The word ”UNIVERSITY” has 4 vowels (U, I, E, I) and 6
consonants (N, V, R, S, T, Y). We will treat the vowels as a single entity, so
there are 5 entities to arrange (Vowels, N, V, R, S, T, Y).
Step 2: There are 5! ways to arrange these entities.
Step 3: Within the ”vowels” entity, there are 4! ways to arrange the vowels
themselves.
Step 4: So, the total number of ways to arrange the vowels in the word
”UNIVERSITY” is 5! ×4!.
Step 5: Now, let’s count the number of ways to arrange the 6 consonants in
the word ”UNIVERSITY”. Since there are 6 consonants, we will treat them as
a single entity.
3
Step 6: There are 6! ways to arrange these consonants.
Step 7: So, the total number of ways to arrange the consonants in the word
”UNIVERSITY” is 6!.
Step 8: The total number of ways to arrange the letters in the word ”UNI-
VERSITY” such that no two vowels are adjacent is the product of the number of
ways to arrange the vowels and the number of ways to arrange the consonants.
Step 9: Therefore, the total number of ways to arrange the letters in the word
”UNIVERSITY” such that no two vowels are adjacent is 5! ×4! ×6! = 172,800.
Question 5
Question
In a group of 10 students, how many ways can we choose a committee of 4
students?
Solution
Step 1: To solve this problem, we can use the formula for combinations, which
is given by:
C(n, k) = n!
k!(n−k)!
where nis the total number of items to choose from and kis the number of
items to choose.
Step 2: Plugging in n= 10 and k= 4 into the formula, we get:
C(10,4) = 10!
4!(10 −4)!
Step 3: Calculating the factorials:
C(10,4) = 10 ×9×8×7×6×5×4×3×2×1
(4 ×3×2×1) ×(6 ×5×4×3×2×1)
Step 4: Simplifying the expression:
C(10,4) = 10 ×9×8×7
4×3×2×1= 210
Step 5: Therefore, there are 210 ways to choose a committee of 4 students
from a group of 10.
Question 6
Question
In a group of 10 people, how many ways are there to select a committee of 4
people with exactly 2 men and 2 women if there are 4 men and 6 women in the
group?
4
Solution
Step 1: Calculate the number of ways to choose 2 men from 4 men. There are
(4
2)ways to choose 2 men from the 4 men in the group.
(4
2)=4!
2!(4 −2)! =4×3
2×1= 6
Step 2: Calculate the number of ways to choose 2 women from 6 women.
There are (6
2)ways to choose 2 women from the 6 women in the group.
(6
2)=6!
2!(6 −2)! =6×5
2×1= 15
Step 3: Multiply the number of ways to choose 2 men and 2 women. The
total number of ways to choose 2 men and 2 women from the group of 10 people
is 6×15.
6×15 = 90
Therefore, there are 90 ways to select a committee of 4 people with exactly
2 men and 2 women from a group of 10 people.
Question 7
Question
In a group of 10 students, how many ways can a committee of 4 students be
selected, where one student is the president, one is the vice-president, and the
other two are members?
Solution
To find the number of ways a committee of 4 students can be selected with spe-
cific roles assigned (president, vice-president, members), we can use the concept
of permutations.
Step 1: Choose the president from 10 students. There are 10 choices for
the president.
Step 2: Choose the vice-president from the remaining 9 students. There
are 9 choices for the vice-president.
Step 3: Choose the first member from the remaining 8 students. There are
8 choices for the first member.
Step 4: Choose the second member from the remaining 7 students. There
are 7 choices for the second member.
Therefore, the total number of ways to select a committee of 4 students with
specific roles assigned is given by:
10 ×9×8×7 = 5040
So, there are 5040 ways to select a committee of 4 students with one presi-
dent, one vice-president, and two members from a group of 10 students.
5
Question 8
Question
A committee of 5 people is to be formed from a group of 8 men and 6 women.
How many different committees can be formed if there must be at least 3 men
on the committee?
Solution
Step 1: Find the number of ways to choose 3 men out of 8. There are (8
3)ways
to choose 3 men from 8.
Step 2: Find the number of ways to choose 2 people from the remaining men
and all 6 women. There are (6
2)ways to choose 2 people from the remaining
men and (6
0)ways to choose all 6 women.
Step 3: Calculate the total number of possible committees with at least
3 men. The total number of possible committees with at least 3 men is the
product of the choices in Step 1 and Step 2:
(8
3)×(6
2)×(6
0)= 56 ×15 ×1 = 840
Therefore, there are 840 different committees that can be formed if there
must be at least 3 men on the committee.
Question 9
Question
A committee of 5 people is to be formed from a group of 10 men and 8 women.
How many different committees can be formed if the committee must consist of
at least 2 men and at least 2 women?
Solution
Step 1: Calculate the number of ways to choose 2 men out of 10. There are (10
2)
ways to choose 2 men from a group of 10.
Step 2: Calculate 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.
Step 3: Calculate the number of ways to choose the remaining person (either
a man or a woman). Since the committee must consist of at least 2 men and
at least 2 women, we need to choose 1 more person after selecting 2 men and 2
women. This person can be either a man or a woman, so there are 10+8−4 = 14
people left to choose from.
Step 4: Calculate the total number of committees that can be formed. The
total number of committees that can be formed is the product of the choices in
6
steps 1, 2, and 3:
(10
2)×(8
2)×14
Step 5: Simplify the expression.
(10
2)=10!
2!(10 −2)! =10 ×9
2×1= 45
(8
2)=8!
2!(8 −2)! =8×7
2×1= 28
Therefore, the total number of committees that can be formed is:
45 ×28 ×14 = 17640
So, there are 17,640 different committees that can be formed consisting of
at least 2 men and at least 2 women.
Question 10
Question
Suppose we have a group of 10 people, including Alice and Bob. In how many
ways can we arrange these 10 people in a line such that Alice and Bob are not
next to each other?
Solution
Let’s first find the total number of ways to arrange the 10 people in a line
without any restrictions, and then subtract the number of ways where Alice
and Bob are next to each other.
Step 1: Find the total number of ways to arrange the 10 people in a line.
Since there are 10 people, there are 10! ways to arrange them in a line.
Step 2: Find the number of ways to arrange the 10 people in a line such
that Alice and Bob are next to each other. Consider Alice and Bob as a single
entity. Then there are 9 entities to arrange (the group of Alice and Bob counted
as one, and the remaining 8 people). Within the group of Alice and Bob, there
are 2 ways to arrange them. Hence, there are 9! ×2! ways to arrange the 10
people in a line with Alice and Bob next to each other.
Step 3: Subtract the number of ways to arrange the 10 people with Alice
and Bob next to each other from the total number of ways to arrange the 10
people. The number of ways to arrange the 10 people in a line such that Alice
and Bob are not next to each other is 10! −9! ×2!.
Therefore, there are 10! −9! ×2! = 3,026,880 −725,760 = 2,301,120 ways
to arrange the 10 people in a line such that Alice and Bob are not next to each
other.
7
Question 11
Question
In a group of 12 people, how many ways can we choose a committee of 5 people
if there are 3 specific people that must be on the committee?
Solution
Step 1: First, we choose the 3 specific people that must be on the committee.
Since these 3 people are already chosen, we only need to choose 2 more people
from the remaining 9 people.
(9
2)=9!
2!(9 −2)! =9×8
2×1= 36
Step 2: So, there are 36 ways to choose 2 more people from the remaining 9
people. Therefore, there are a total of 3×36 = 108 ways to choose a committee
of 5 people with 3 specific people included.
Question 12
Question
In how many ways can 5 people be seated in a row with restrictions that two of
them, A and B, must not sit together?
Solution
Step 1: First we calculate the total number of ways the 5 people can be seated
without any restrictions.
The number of ways to seat 5 people in a row is 5! = 5 ×4×3×2×1 = 120.
Step 2: Next, we calculate the number of ways when A and B sit together.
Since A and B must sit together, consider them as one group. Then, there
are 4 entities (AB, C, D, E) that can be seated in 4! ways. However, A and B can
switch positions among themselves in 2 ways, so the final count is 4! ×2 = 48.
Step 3: Finally, we subtract the number of ways A and B can sit together
from the total number of ways to get the number of ways they must not sit
together.
Therefore, the number of ways the 5 people can be seated with A and B not
sitting together is 120 −48 = 72.
8
Question 13
Question
Suppose you have 5 red balls, 3 blue balls, and 4 yellow balls. If you randomly
select 4 balls without replacement, what is the probability of selecting 2 red
balls, 1 blue ball, and 1 yellow ball in any order?
Solution
Step 1: Find the total number of ways to choose 4 balls out of 12. Since order
does not matter, we use combinations. The total number of ways to choose 4
balls out of 12 is given by (12
4).
Step 2: Find the number of ways to choose 2 red balls, 1 blue ball, and 1
yellow ball. The number of ways to choose 2 red balls out of 5 is (5
2). The
number of ways to choose 1 blue ball out of 3 is (3
1). The number of ways to
choose 1 yellow ball out of 4 is (4
1). Therefore, the total number of ways to
choose 2 red balls, 1 blue ball, and 1 yellow ball is (5
2)·(3
1)·(4
1).
Step 3: Calculate the probability. The probability of selecting 2 red balls, 1
blue ball, and 1 yellow ball in any order is given by:
(5
2)·(3
1)·(4
1)
(12
4)=10 ·3·4
495
Therefore, the probability of selecting 2 red balls, 1 blue ball, and 1 yellow
ball in any order is 120
495 =8
33 .
Question 14
Question
In a class of 30 students, there are 12 students who are interested in statistics
and 18 students who are interested in probability. If 8 students are interested in
both statistics and probability, how many students in the class are not interested
in either statistics or probability?
Solution
Step 1: Let’s denote the number of students interested in statistics as S, the
number of students interested in probability as P, and the number of students
interested in both statistics and probability as B. We are given: S= 12 (in-
terested in statistics), P= 18 (interested in probability), B= 8 (interested in
both statistics and probability).
Step 2: We can find the total number of students interested in either statis-
tics or probability (including those interested in both) using the principle of
inclusion-exclusion: S∪P=S+P−B= 12 + 18 −8 = 22.
9
Step 3: Now, we can find the number of students not interested in either
statistics or probability by subtracting the number of students interested in
either from the total number of students: Total students in the class = 30.
Number of students not interested in either = 30 −22 = 8.
Therefore, there are 8students in the class who are not interested in either
statistics or probability.
Question 15
Question
A committee of 5 people is to be formed from a group of 8 men and 6 women.
If the committee must consist of at least 3 men, what is the number of ways the
committee can be formed?
Solution
Step 1: Find the total number of ways to form a committee of 5 people from
14 individuals. Since the committee can consist of any combination of men and
women, we will use the combination formula.
Total ways =(14
5)=14!
5!(14 −5)! =14!
5!9! = 2002
Step 2: Find the number of ways to form a committee with less than 3 men.
The committee can consist of 0, 1, or 2 men.
Ways with 0 men =(8
0)(6
5)= 1 ×6 = 6
Ways with 1 man =(8
1)(6
4)= 8 ×15 = 120
Ways with 2 men =(8
2)(6
3)= 28 ×20 = 560
Total ways with less than 3 men = 6 + 120 + 560 = 686
Step 3: Find the number of ways to form a committee with at least 3 men.
Ways with at least 3 men =Total ways −Ways with less than 3 men
= 2002 −686 = 1316
Therefore, there are 1316 ways to form a committee with at least 3 men from
the group of 8 men and 6 women.
10
Question 16
Question
In a group of 10 students, how many ways can you select a committee of 4
students to represent the class if no student can serve on the committee more
than once?
Solution
Step 1: To solve this problem, we will use the formula for combinations:
(n
k)=n!
k!(n−k)!
where nis the total number of students and kis the number of students we
want to choose for the committee.
Step 2: Substitute n= 10 and k= 4 into the formula:
(10
4)=10!
4!(10 −4)!
Step 3: Simplify the factorial expressions:
(10
4)=10 ×9×8×7
4×3×2×1
Step 4: Perform the multiplication and division:
(10
4)=5040
24
Step 5: Calculate the final result:
(10
4)= 210
Therefore, there are 210 ways to select a committee of 4 students from a
group of 10 students without any student serving on the committee more than
once.
Question 17
Question
A committee of 5 people is to be formed from a group of 9 men and 6 women.
If the committee must include at least 2 men and 2 women, how many different
committees can be formed?
11
Solution
Step 1: Find the number of ways to select 2 men and 3 women. There are (9
2)
ways to select 2 men from 9 men, and (6
3)ways to select 3 women from 6 women.
So, the number of ways to select 2 men and 3 women is (9
2)·(6
3).
Step 2: Find the number of ways to select 3 men and 2 women. There are
(9
3)ways to select 3 men from 9 men, and (6
2)ways to select 2 women from 6
women. So, the number of ways to select 3 men and 2 women is (9
3)·(6
2).
Step 3: Add the results from Step 1 and Step 2 to find the total number of
different committees that can be formed with at least 2 men and 2 women.
Therefore, the total number of different committees that can be formed is:
(9
2)·(6
3)+(9
3)·(6
2)=9!
2!7! ·6!
3!3! +9!
3!6! ·6!
2!4! .
Now, we can simplify this expression to find the final answer.
Question 18
Question
In how many ways can 5 distinct books be arranged on a bookshelf if one
particular pair of books are never to be separated?
Solution
Step 1: Consider the two books that are not to be separated as one entity. Then,
there are 4 entities to arrange on the bookshelf: the pair of books and the other
3 distinct books.
Step 2: Since the two books must always be together, we treat them as one
entity. So, there are 4! ways to arrange the 4 entities.
Step 3: However, within the pair of books, there are 2! ways to arrange the
books themselves.
Step 4: Therefore, the total number of ways to arrange the 5 distinct books
where the particular pair are never separated is 4! ×2! = 48.
Question 19
Question
Suppose a committee of 5 people is to be selected from a group of 10 men and
8 women. In how many ways can the committee be formed if it must have at
least 3 men and 2 women?
12
Solution
Step 1: Calculate the number of ways to select 3 men from 10.
(10
3)=10!
3!(10 −3)! =10 ·9·8
3·2·1= 120
Step 2: Calculate the number of ways to select 2 women from 8.
(8
2)=8!
2!(8 −2)! =8·7
2·1= 28
Step 3: Calculate the number of ways to select the remaining 2 people (could
be any gender) from the remaining pool of 7 people.
(7
2)=7!
2!(7 −2)! =7·6
2·1= 21
Step 4: Multiply the results from Step 1, Step 2, and Step 3 to find the total
number of ways to form the committee. Total ways = 120 ×28 ×21 = 70560
Therefore, the committee can be formed in 70,560 ways if it must have at
least 3 men and 2 women.
Question 20
Question
In a group of 10 friends, how many ways can you choose a committee of 4 people
if there are 2 friends who refuse to be on the same committee together?
Solution
Step 1: Calculate the total number of ways to choose a committee of 4 people
from a group of 10 friends. Step 2: Calculate the number of ways to choose a
committee where the 2 friends who refuse to be on the same committee together
are both included. Step 3: Calculate the number of ways to choose a committee
where the 2 friends who refuse to be on the same committee together are both
excluded. Step 4: Subtract the results from steps 2 and 3 from the result of
step 1 to obtain the final answer.
Step 1: To choose a committee of 4 people from a group of 10 friends, we
use the combination formula:
(10
4)=10!
4!(10 −4)! =10 ×9×8×7
4×3×2×1= 210
Step 2: If the 2 friends who refuse to be on the same committee together are
both included, we have to choose 2 more people from the remaining 8 friends.
This can be done in (8
2)ways.
(8
2)=8!
2!(8 −2)! =8×7
2×1= 28
13
Step 3: If the 2 friends who refuse to be on the same committee together
are both excluded, we can simply choose 4 people from the remaining 8 friends.
This can be done in (8
4)ways.
(8
4)=8!
4!(8 −4)! =8×7×6×5
4×3×2×1= 70
Step 4: The total number of ways to choose a committee of 4 people where
the 2 friends who refuse to be on the same committee together are not included
is: (10
4)−(8
2)−(8
4)= 210 −28 −70 = 112
Therefore, there are 112 ways to choose a committee of 4 people from a group
of 10 friends where 2 friends refuse to be on the same committee together.
Question 21
Question
A committee of 5 people is to be formed from a group of 10 females and 8
males. If the committee must consist of at least 3 females, how many different
committees can be formed?
Solution
Step 1: Find the number of ways to choose exactly 3 females and 2 males for the
committee. Step 2: Find the number of ways to choose exactly 4 females and
1 male for the committee. Step 3: Find the number of ways to choose exactly
5 females for the committee. Step 4: Add the results from Steps 1, 2, and 3 to
find the total number of different committees that can be formed.
Step 1: There are (10
3)ways to choose 3 females from the group of 10 females,
and (8
2)ways to choose 2 males from the group of 8 males. So, the number of
ways to choose exactly 3 females and 2 males is (10
3)×(8
2).
Step 2: Similarly, there are (10
4)ways to choose 4 females and (8
1)ways to
choose 1 male. So, the number of ways to choose exactly 4 females and 1 male
is (10
4)×(8
1).
Step 3: To select all 5 committee members as females, we choose all 5 females
from the group of 10 females. There is only 1 way to choose all 5 females.
Step 4: The total number of different committees that can be formed is
the sum of the results from Steps 1, 2, and 3. Therefore, the total number of
different committees that can be formed is (10
3)×(8
2)+(10
4)×(8
1)+ 1.
14
Question 22
Question
In a group of 10 people, how many different ways can we choose a committee
of 4 people if Alice and Bob refuse to serve on the committee together?
Solution
Step 1: Calculate the total number of ways to choose a committee of 4 people
from a group of 10 people. Step 2: Calculate the number of ways to choose a
committee of 4 people when Alice and Bob are together. Step 3: Subtract the
result from Step 2 from the result in Step 1 to find the number of ways when
Alice and Bob are not together.
Step 1: To choose a committee of 4 people from a group of 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 from the group of
10.
Step 2: When Alice and Bob are serving together, we treat them as one
person. So, we now have 9 people to choose from. We need to choose 3 more
people to complete the committee. Using the combination formula again:
(9
3)=9!
3!(9 −3)! =9×8×7
3×2×1= 84
Therefore, there are 84 ways to choose a committee of 4 people when Alice
and Bob are together.
Step 3: Subtract the number of ways in Step 2 from the total number of
ways in Step 1 to find the number of ways when Alice and Bob are not together:
210 −84 = 126
So, there are 126 ways to choose a committee of 4 people from the group of
10 where Alice and Bob are not both serving on the committee.
Question 23
Question
In a mathematics competition, there are 10 multiple-choice questions with 5
answer choices each. If a student randomly selects an answer for each question,
what is the probability that the student gets at least 8 questions correct?
15
Solution
To solve this problem, we can use the concept of combinations and the binomial
probability formula.
Step 1: Calculate the total number of ways to answer all 10 questions.
Since each question has 5 answer choices, there are 510 total ways the student
can answer all 10 questions.
Step 2: Calculate the number of ways to get exactly 8 questions correct.
To get exactly 8 questions correct, the student must choose 8 correct answers
and 2 incorrect answers out of the 10 questions. The number of ways to do this
is (10
8)=10!
8!(10−8)! ·1·1 = 45.
For each of these ways, the probability of getting exactly 8 questions correct
is (1
5)8×(4
5)2.
Step 3: Calculate the number of ways to get exactly 9 questions correct.
To get exactly 9 questions correct, the student must choose 9 correct answers
and 1 incorrect answer out of the 10 questions. The number of ways to do this
is (10
9)=10!
9!(10−9)! ·1·1 = 10.
For each of these ways, the probability of getting exactly 9 questions correct
is (1
5)9×(4
5)1.
Step 4: Calculate the number of ways to get all 10 questions correct. To
get all 10 questions correct, the student must choose all 10 correct answers out
of the 10 questions. There is only 1 way to do this.
The probability of getting all 10 questions correct is (1
5)10.
Step 5: Calculate the total probability of getting at least 8 questions correct.
The total probability is the sum of the probabilities of getting exactly 8, 9, or
10 questions correct.
Total probability =Probability of getting exactly 8 questions correct
+Probability of getting exactly 9 questions correct
+Probability of getting all 10 questions correct
= 45 ×(1
5)8
×(4
5)2
+ 10 ×(1
5)9
×(4
5)1
+(1
5)10
=27735
9765625 +40
9765625 +1
9765625
=27776
9765625
≈0.002841
Therefore, the probability that the student gets at least 8 questions correct
is approximately 0.002841.
16
Question 24
Question
In a group of 10 people, how many ways are there to select a president, vice
president, and treasurer, assuming no person can hold more than one position?
Solution
Step 1: To find the number of ways to select the president, we have 10 choices.
After selecting the president, there remain 9 people for the position of vice
president.
Step 2: After selecting the president and vice president, there remain 8
people for the position of treasurer.
Step 3: Therefore, the total number of ways to select a president, vice pres-
ident, and treasurer is the product of the number of ways for each position:
10 ×9×8 = 720
So, there are 720 ways to select a president, vice president, and treasurer in
a group of 10 people.
Question 25
Question
In a group of 8 friends, how many ways can we select a committee of 3 friends?
Solution
Step 1: To find the number of ways to select a committee of 3 friends from a
group of 8, we will use the combination formula. The formula for combination
is given by:
C(n, r) = n!
r!(n−r)!
where nis the total number of friends and ris the number of friends we
want to choose.
Step 2: Substituting n= 8 and r= 3 into the formula, we have:
C(8,3) = 8!
3!(8 −3)!
Step 3: Calculating the factorials:
C(8,3) = 8×7×6
3×2×1
17
C(8,3) = 336
6
Step 4: Simplifying the fraction:
C(8,3) = 56
Therefore, there are 56 ways to select a committee of 3 friends from a group
of 8.
18
Question 2
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 choose 2 women from 8 women and 3
men from 10 men. Then, calculate the number of ways to choose 3 women from
8 women and 2 men from 10 men.
Step 2: Calculate the number of ways for each combination and sum them
to find the total number of different committees that can be formed.
Let’s calculate the number of ways for each combination.
Step 1: Choosing 2 women from 8 women:
(8
2)=8!
2!(8 −2)! = 28
Choosing 3 men from 10 men:
(10
3)=10!
3!(10 −3)! = 120
Now, we calculate the number of committees with 2 women and 3 men:
28 ×120 = 3360
Step 2: Choosing 3 women from 8 women:
(8
3)=8!
3!(8 −3)! = 56
Choosing 2 men from 10 men:
(10
2)=10!
2!(10 −2)! = 45
Now, we calculate the number of committees with 3 women and 2 men:
56 ×45 = 2520
Therefore, the total number of different committees that can be formed is:
3360 + 2520 = 5880
Thus, there are 5880 different committees that can be formed.
2
Question 3
Question
A committee of 5 people needs to be formed from a group of 10 women and 8
men. If the committee must consist of at least 3 women, how many different
committees can be formed?
Solution
Step 1: Count the number of ways to choose 3 women from 10. Step 2: Count
the number of ways to choose 2 additional members from the remaining 8 people
(men). Step 3: Multiply the results from Step 1 and Step 2 to find the total
number of possible committees.
Step 1: To choose 3 women from 10, we use the combination formula. The
number of ways to choose 3 women from 10 is given by (10
3)=10!
3!(10−3)! =
10×9×8
3×2×1= 120.
Step 2: We need to choose 2 additional members from the remaining 8 people
(men). The number of ways to choose 2 men from 8 is given by (8
2)=8!
2!(8−2)! =
8×7
2×1= 28.
Step 3: To find the total number of different committees that can be formed
consisting of at least 3 women, we multiply the results from Steps 1 and 2:
120 ×28 = 3360.
Therefore, there are 3360 different committees that can be formed.
Question 4
Question
In how many ways can you arrange the letters in the word ”UNIVERSITY”
such that no two vowels are adjacent?
Solution
Step 1: Let’s first count the number of ways to arrange the vowels in the word
”UNIVERSITY”. The word ”UNIVERSITY” has 4 vowels (U, I, E, I) and 6
consonants (N, V, R, S, T, Y). We will treat the vowels as a single entity, so
there are 5 entities to arrange (Vowels, N, V, R, S, T, Y).
Step 2: There are 5! ways to arrange these entities.
Step 3: Within the ”vowels” entity, there are 4! ways to arrange the vowels
themselves.
Step 4: So, the total number of ways to arrange the vowels in the word
”UNIVERSITY” is 5! ×4!.
Step 5: Now, let’s count the number of ways to arrange the 6 consonants in
the word ”UNIVERSITY”. Since there are 6 consonants, we will treat them as
a single entity.
3
Step 6: There are 6! ways to arrange these consonants.
Step 7: So, the total number of ways to arrange the consonants in the word
”UNIVERSITY” is 6!.
Step 8: The total number of ways to arrange the letters in the word ”UNI-
VERSITY” such that no two vowels are adjacent is the product of the number of
ways to arrange the vowels and the number of ways to arrange the consonants.
Step 9: Therefore, the total number of ways to arrange the letters in the word
”UNIVERSITY” such that no two vowels are adjacent is 5! ×4! ×6! = 172,800.
Question 5
Question
In a group of 10 students, how many ways can we choose a committee of 4
students?
Solution
Step 1: To solve this problem, we can use the formula for combinations, which
is given by:
C(n, k) = n!
k!(n−k)!
where nis the total number of items to choose from and kis the number of
items to choose.
Step 2: Plugging in n= 10 and k= 4 into the formula, we get:
C(10,4) = 10!
4!(10 −4)!
Step 3: Calculating the factorials:
C(10,4) = 10 ×9×8×7×6×5×4×3×2×1
(4 ×3×2×1) ×(6 ×5×4×3×2×1)
Step 4: Simplifying the expression:
C(10,4) = 10 ×9×8×7
4×3×2×1= 210
Step 5: Therefore, there are 210 ways to choose a committee of 4 students
from a group of 10.
Question 6
Question
In a group of 10 people, how many ways are there to select a committee of 4
people with exactly 2 men and 2 women if there are 4 men and 6 women in the
group?
4
Solution
Step 1: Calculate the number of ways to choose 2 men from 4 men. There are
(4
2)ways to choose 2 men from the 4 men in the group.
(4
2)=4!
2!(4 −2)! =4×3
2×1= 6
Step 2: Calculate the number of ways to choose 2 women from 6 women.
There are (6
2)ways to choose 2 women from the 6 women in the group.
(6
2)=6!
2!(6 −2)! =6×5
2×1= 15
Step 3: Multiply the number of ways to choose 2 men and 2 women. The
total number of ways to choose 2 men and 2 women from the group of 10 people
is 6×15.
6×15 = 90
Therefore, there are 90 ways to select a committee of 4 people with exactly
2 men and 2 women from a group of 10 people.
Question 7
Question
In a group of 10 students, how many ways can a committee of 4 students be
selected, where one student is the president, one is the vice-president, and the
other two are members?
Solution
To find the number of ways a committee of 4 students can be selected with spe-
cific roles assigned (president, vice-president, members), we can use the concept
of permutations.
Step 1: Choose the president from 10 students. There are 10 choices for
the president.
Step 2: Choose the vice-president from the remaining 9 students. There
are 9 choices for the vice-president.
Step 3: Choose the first member from the remaining 8 students. There are
8 choices for the first member.
Step 4: Choose the second member from the remaining 7 students. There
are 7 choices for the second member.
Therefore, the total number of ways to select a committee of 4 students with
specific roles assigned is given by:
10 ×9×8×7 = 5040
So, there are 5040 ways to select a committee of 4 students with one presi-
dent, one vice-president, and two members from a group of 10 students.
5
Question 8
Question
A committee of 5 people is to be formed from a group of 8 men and 6 women.
How many different committees can be formed if there must be at least 3 men
on the committee?
Solution
Step 1: Find the number of ways to choose 3 men out of 8. There are (8
3)ways
to choose 3 men from 8.
Step 2: Find the number of ways to choose 2 people from the remaining men
and all 6 women. There are (6
2)ways to choose 2 people from the remaining
men and (6
0)ways to choose all 6 women.
Step 3: Calculate the total number of possible committees with at least
3 men. The total number of possible committees with at least 3 men is the
product of the choices in Step 1 and Step 2:
(8
3)×(6
2)×(6
0)= 56 ×15 ×1 = 840
Therefore, there are 840 different committees that can be formed if there
must be at least 3 men on the committee.
Question 9
Question
A committee of 5 people is to be formed from a group of 10 men and 8 women.
How many different committees can be formed if the committee must consist of
at least 2 men and at least 2 women?
Solution
Step 1: Calculate the number of ways to choose 2 men out of 10. There are (10
2)
ways to choose 2 men from a group of 10.
Step 2: Calculate 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.
Step 3: Calculate the number of ways to choose the remaining person (either
a man or a woman). Since the committee must consist of at least 2 men and
at least 2 women, we need to choose 1 more person after selecting 2 men and 2
women. This person can be either a man or a woman, so there are 10+8−4 = 14
people left to choose from.
Step 4: Calculate the total number of committees that can be formed. The
total number of committees that can be formed is the product of the choices in
6
steps 1, 2, and 3:
(10
2)×(8
2)×14
Step 5: Simplify the expression.
(10
2)=10!
2!(10 −2)! =10 ×9
2×1= 45
(8
2)=8!
2!(8 −2)! =8×7
2×1= 28
Therefore, the total number of committees that can be formed is:
45 ×28 ×14 = 17640
So, there are 17,640 different committees that can be formed consisting of
at least 2 men and at least 2 women.
Question 10
Question
Suppose we have a group of 10 people, including Alice and Bob. In how many
ways can we arrange these 10 people in a line such that Alice and Bob are not
next to each other?
Solution
Let’s first find the total number of ways to arrange the 10 people in a line
without any restrictions, and then subtract the number of ways where Alice
and Bob are next to each other.
Step 1: Find the total number of ways to arrange the 10 people in a line.
Since there are 10 people, there are 10! ways to arrange them in a line.
Step 2: Find the number of ways to arrange the 10 people in a line such
that Alice and Bob are next to each other. Consider Alice and Bob as a single
entity. Then there are 9 entities to arrange (the group of Alice and Bob counted
as one, and the remaining 8 people). Within the group of Alice and Bob, there
are 2 ways to arrange them. Hence, there are 9! ×2! ways to arrange the 10
people in a line with Alice and Bob next to each other.
Step 3: Subtract the number of ways to arrange the 10 people with Alice
and Bob next to each other from the total number of ways to arrange the 10
people. The number of ways to arrange the 10 people in a line such that Alice
and Bob are not next to each other is 10! −9! ×2!.
Therefore, there are 10! −9! ×2! = 3,026,880 −725,760 = 2,301,120 ways
to arrange the 10 people in a line such that Alice and Bob are not next to each
other.
7
Question 11
Question
In a group of 12 people, how many ways can we choose a committee of 5 people
if there are 3 specific people that must be on the committee?
Solution
Step 1: First, we choose the 3 specific people that must be on the committee.
Since these 3 people are already chosen, we only need to choose 2 more people
from the remaining 9 people.
(9
2)=9!
2!(9 −2)! =9×8
2×1= 36
Step 2: So, there are 36 ways to choose 2 more people from the remaining 9
people. Therefore, there are a total of 3×36 = 108 ways to choose a committee
of 5 people with 3 specific people included.
Question 12
Question
In how many ways can 5 people be seated in a row with restrictions that two of
them, A and B, must not sit together?
Solution
Step 1: First we calculate the total number of ways the 5 people can be seated
without any restrictions.
The number of ways to seat 5 people in a row is 5! = 5 ×4×3×2×1 = 120.
Step 2: Next, we calculate the number of ways when A and B sit together.
Since A and B must sit together, consider them as one group. Then, there
are 4 entities (AB, C, D, E) that can be seated in 4! ways. However, A and B can
switch positions among themselves in 2 ways, so the final count is 4! ×2 = 48.
Step 3: Finally, we subtract the number of ways A and B can sit together
from the total number of ways to get the number of ways they must not sit
together.
Therefore, the number of ways the 5 people can be seated with A and B not
sitting together is 120 −48 = 72.
8
Question 13
Question
Suppose you have 5 red balls, 3 blue balls, and 4 yellow balls. If you randomly
select 4 balls without replacement, what is the probability of selecting 2 red
balls, 1 blue ball, and 1 yellow ball in any order?
Solution
Step 1: Find the total number of ways to choose 4 balls out of 12. Since order
does not matter, we use combinations. The total number of ways to choose 4
balls out of 12 is given by (12
4).
Step 2: Find the number of ways to choose 2 red balls, 1 blue ball, and 1
yellow ball. The number of ways to choose 2 red balls out of 5 is (5
2). The
number of ways to choose 1 blue ball out of 3 is (3
1). The number of ways to
choose 1 yellow ball out of 4 is (4
1). Therefore, the total number of ways to
choose 2 red balls, 1 blue ball, and 1 yellow ball is (5
2)·(3
1)·(4
1).
Step 3: Calculate the probability. The probability of selecting 2 red balls, 1
blue ball, and 1 yellow ball in any order is given by:
(5
2)·(3
1)·(4
1)
(12
4)=10 ·3·4
495
Therefore, the probability of selecting 2 red balls, 1 blue ball, and 1 yellow
ball in any order is 120
495 =8
33 .
Question 14
Question
In a class of 30 students, there are 12 students who are interested in statistics
and 18 students who are interested in probability. If 8 students are interested in
both statistics and probability, how many students in the class are not interested
in either statistics or probability?
Solution
Step 1: Let’s denote the number of students interested in statistics as S, the
number of students interested in probability as P, and the number of students
interested in both statistics and probability as B. We are given: S= 12 (in-
terested in statistics), P= 18 (interested in probability), B= 8 (interested in
both statistics and probability).
Step 2: We can find the total number of students interested in either statis-
tics or probability (including those interested in both) using the principle of
inclusion-exclusion: S∪P=S+P−B= 12 + 18 −8 = 22.
9
Step 3: Now, we can find the number of students not interested in either
statistics or probability by subtracting the number of students interested in
either from the total number of students: Total students in the class = 30.
Number of students not interested in either = 30 −22 = 8.
Therefore, there are 8students in the class who are not interested in either
statistics or probability.
Question 15
Question
A committee of 5 people is to be formed from a group of 8 men and 6 women.
If the committee must consist of at least 3 men, what is the number of ways the
committee can be formed?
Solution
Step 1: Find the total number of ways to form a committee of 5 people from
14 individuals. Since the committee can consist of any combination of men and
women, we will use the combination formula.
Total ways =(14
5)=14!
5!(14 −5)! =14!
5!9! = 2002
Step 2: Find the number of ways to form a committee with less than 3 men.
The committee can consist of 0, 1, or 2 men.
Ways with 0 men =(8
0)(6
5)= 1 ×6 = 6
Ways with 1 man =(8
1)(6
4)= 8 ×15 = 120
Ways with 2 men =(8
2)(6
3)= 28 ×20 = 560
Total ways with less than 3 men = 6 + 120 + 560 = 686
Step 3: Find the number of ways to form a committee with at least 3 men.
Ways with at least 3 men =Total ways −Ways with less than 3 men
= 2002 −686 = 1316
Therefore, there are 1316 ways to form a committee with at least 3 men from
the group of 8 men and 6 women.
10
Question 16
Question
In a group of 10 students, how many ways can you select a committee of 4
students to represent the class if no student can serve on the committee more
than once?
Solution
Step 1: To solve this problem, we will use the formula for combinations:
(n
k)=n!
k!(n−k)!
where nis the total number of students and kis the number of students we
want to choose for the committee.
Step 2: Substitute n= 10 and k= 4 into the formula:
(10
4)=10!
4!(10 −4)!
Step 3: Simplify the factorial expressions:
(10
4)=10 ×9×8×7
4×3×2×1
Step 4: Perform the multiplication and division:
(10
4)=5040
24
Step 5: Calculate the final result:
(10
4)= 210
Therefore, there are 210 ways to select a committee of 4 students from a
group of 10 students without any student serving on the committee more than
once.
Question 17
Question
A committee of 5 people is to be formed from a group of 9 men and 6 women.
If the committee must include at least 2 men and 2 women, how many different
committees can be formed?
11
Solution
Step 1: Find the number of ways to select 2 men and 3 women. There are (9
2)
ways to select 2 men from 9 men, and (6
3)ways to select 3 women from 6 women.
So, the number of ways to select 2 men and 3 women is (9
2)·(6
3).
Step 2: Find the number of ways to select 3 men and 2 women. There are
(9
3)ways to select 3 men from 9 men, and (6
2)ways to select 2 women from 6
women. So, the number of ways to select 3 men and 2 women is (9
3)·(6
2).
Step 3: Add the results from Step 1 and Step 2 to find the total number of
different committees that can be formed with at least 2 men and 2 women.
Therefore, the total number of different committees that can be formed is:
(9
2)·(6
3)+(9
3)·(6
2)=9!
2!7! ·6!
3!3! +9!
3!6! ·6!
2!4! .
Now, we can simplify this expression to find the final answer.
Question 18
Question
In how many ways can 5 distinct books be arranged on a bookshelf if one
particular pair of books are never to be separated?
Solution
Step 1: Consider the two books that are not to be separated as one entity. Then,
there are 4 entities to arrange on the bookshelf: the pair of books and the other
3 distinct books.
Step 2: Since the two books must always be together, we treat them as one
entity. So, there are 4! ways to arrange the 4 entities.
Step 3: However, within the pair of books, there are 2! ways to arrange the
books themselves.
Step 4: Therefore, the total number of ways to arrange the 5 distinct books
where the particular pair are never separated is 4! ×2! = 48.
Question 19
Question
Suppose a committee of 5 people is to be selected from a group of 10 men and
8 women. In how many ways can the committee be formed if it must have at
least 3 men and 2 women?
12
Solution
Step 1: Calculate the number of ways to select 3 men from 10.
(10
3)=10!
3!(10 −3)! =10 ·9·8
3·2·1= 120
Step 2: Calculate the number of ways to select 2 women from 8.
(8
2)=8!
2!(8 −2)! =8·7
2·1= 28
Step 3: Calculate the number of ways to select the remaining 2 people (could
be any gender) from the remaining pool of 7 people.
(7
2)=7!
2!(7 −2)! =7·6
2·1= 21
Step 4: Multiply the results from Step 1, Step 2, and Step 3 to find the total
number of ways to form the committee. Total ways = 120 ×28 ×21 = 70560
Therefore, the committee can be formed in 70,560 ways if it must have at
least 3 men and 2 women.
Question 20
Question
In a group of 10 friends, how many ways can you choose a committee of 4 people
if there are 2 friends who refuse to be on the same committee together?
Solution
Step 1: Calculate the total number of ways to choose a committee of 4 people
from a group of 10 friends. Step 2: Calculate the number of ways to choose a
committee where the 2 friends who refuse to be on the same committee together
are both included. Step 3: Calculate the number of ways to choose a committee
where the 2 friends who refuse to be on the same committee together are both
excluded. Step 4: Subtract the results from steps 2 and 3 from the result of
step 1 to obtain the final answer.
Step 1: To choose a committee of 4 people from a group of 10 friends, we
use the combination formula:
(10
4)=10!
4!(10 −4)! =10 ×9×8×7
4×3×2×1= 210
Step 2: If the 2 friends who refuse to be on the same committee together are
both included, we have to choose 2 more people from the remaining 8 friends.
This can be done in (8
2)ways.
(8
2)=8!
2!(8 −2)! =8×7
2×1= 28
13
Step 3: If the 2 friends who refuse to be on the same committee together
are both excluded, we can simply choose 4 people from the remaining 8 friends.
This can be done in (8
4)ways.
(8
4)=8!
4!(8 −4)! =8×7×6×5
4×3×2×1= 70
Step 4: The total number of ways to choose a committee of 4 people where
the 2 friends who refuse to be on the same committee together are not included
is: (10
4)−(8
2)−(8
4)= 210 −28 −70 = 112
Therefore, there are 112 ways to choose a committee of 4 people from a group
of 10 friends where 2 friends refuse to be on the same committee together.
Question 21
Question
A committee of 5 people is to be formed from a group of 10 females and 8
males. If the committee must consist of at least 3 females, how many different
committees can be formed?
Solution
Step 1: Find the number of ways to choose exactly 3 females and 2 males for the
committee. Step 2: Find the number of ways to choose exactly 4 females and
1 male for the committee. Step 3: Find the number of ways to choose exactly
5 females for the committee. Step 4: Add the results from Steps 1, 2, and 3 to
find the total number of different committees that can be formed.
Step 1: There are (10
3)ways to choose 3 females from the group of 10 females,
and (8
2)ways to choose 2 males from the group of 8 males. So, the number of
ways to choose exactly 3 females and 2 males is (10
3)×(8
2).
Step 2: Similarly, there are (10
4)ways to choose 4 females and (8
1)ways to
choose 1 male. So, the number of ways to choose exactly 4 females and 1 male
is (10
4)×(8
1).
Step 3: To select all 5 committee members as females, we choose all 5 females
from the group of 10 females. There is only 1 way to choose all 5 females.
Step 4: The total number of different committees that can be formed is
the sum of the results from Steps 1, 2, and 3. Therefore, the total number of
different committees that can be formed is (10
3)×(8
2)+(10
4)×(8
1)+ 1.
14
Question 22
Question
In a group of 10 people, how many different ways can we choose a committee
of 4 people if Alice and Bob refuse to serve on the committee together?
Solution
Step 1: Calculate the total number of ways to choose a committee of 4 people
from a group of 10 people. Step 2: Calculate the number of ways to choose a
committee of 4 people when Alice and Bob are together. Step 3: Subtract the
result from Step 2 from the result in Step 1 to find the number of ways when
Alice and Bob are not together.
Step 1: To choose a committee of 4 people from a group of 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 from the group of
10.
Step 2: When Alice and Bob are serving together, we treat them as one
person. So, we now have 9 people to choose from. We need to choose 3 more
people to complete the committee. Using the combination formula again:
(9
3)=9!
3!(9 −3)! =9×8×7
3×2×1= 84
Therefore, there are 84 ways to choose a committee of 4 people when Alice
and Bob are together.
Step 3: Subtract the number of ways in Step 2 from the total number of
ways in Step 1 to find the number of ways when Alice and Bob are not together:
210 −84 = 126
So, there are 126 ways to choose a committee of 4 people from the group of
10 where Alice and Bob are not both serving on the committee.
Question 23
Question
In a mathematics competition, there are 10 multiple-choice questions with 5
answer choices each. If a student randomly selects an answer for each question,
what is the probability that the student gets at least 8 questions correct?
15
Solution
To solve this problem, we can use the concept of combinations and the binomial
probability formula.
Step 1: Calculate the total number of ways to answer all 10 questions.
Since each question has 5 answer choices, there are 510 total ways the student
can answer all 10 questions.
Step 2: Calculate the number of ways to get exactly 8 questions correct.
To get exactly 8 questions correct, the student must choose 8 correct answers
and 2 incorrect answers out of the 10 questions. The number of ways to do this
is (10
8)=10!
8!(10−8)! ·1·1 = 45.
For each of these ways, the probability of getting exactly 8 questions correct
is (1
5)8×(4
5)2.
Step 3: Calculate the number of ways to get exactly 9 questions correct.
To get exactly 9 questions correct, the student must choose 9 correct answers
and 1 incorrect answer out of the 10 questions. The number of ways to do this
is (10
9)=10!
9!(10−9)! ·1·1 = 10.
For each of these ways, the probability of getting exactly 9 questions correct
is (1
5)9×(4
5)1.
Step 4: Calculate the number of ways to get all 10 questions correct. To
get all 10 questions correct, the student must choose all 10 correct answers out
of the 10 questions. There is only 1 way to do this.
The probability of getting all 10 questions correct is (1
5)10.
Step 5: Calculate the total probability of getting at least 8 questions correct.
The total probability is the sum of the probabilities of getting exactly 8, 9, or
10 questions correct.
Total probability =Probability of getting exactly 8 questions correct
+Probability of getting exactly 9 questions correct
+Probability of getting all 10 questions correct
= 45 ×(1
5)8
×(4
5)2
+ 10 ×(1
5)9
×(4
5)1
+(1
5)10
=27735
9765625 +40
9765625 +1
9765625
=27776
9765625
≈0.002841
Therefore, the probability that the student gets at least 8 questions correct
is approximately 0.002841.
16
Question 24
Question
In a group of 10 people, how many ways are there to select a president, vice
president, and treasurer, assuming no person can hold more than one position?
Solution
Step 1: To find the number of ways to select the president, we have 10 choices.
After selecting the president, there remain 9 people for the position of vice
president.
Step 2: After selecting the president and vice president, there remain 8
people for the position of treasurer.
Step 3: Therefore, the total number of ways to select a president, vice pres-
ident, and treasurer is the product of the number of ways for each position:
10 ×9×8 = 720
So, there are 720 ways to select a president, vice president, and treasurer in
a group of 10 people.
Question 25
Question
In a group of 8 friends, how many ways can we select a committee of 3 friends?
Solution
Step 1: To find the number of ways to select a committee of 3 friends from a
group of 8, we will use the combination formula. The formula for combination
is given by:
C(n, r) = n!
r!(n−r)!
where nis the total number of friends and ris the number of friends we
want to choose.
Step 2: Substituting n= 8 and r= 3 into the formula, we have:
C(8,3) = 8!
3!(8 −3)!
Step 3: Calculating the factorials:
C(8,3) = 8×7×6
3×2×1
17
C(8,3) = 336
6
Step 4: Simplifying the fraction:
C(8,3) = 56
Therefore, there are 56 ways to select a committee of 3 friends from a group
of 8.
18
Question 2
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 choose 2 women from 8 women and 3
men from 10 men. Then, calculate the number of ways to choose 3 women from
8 women and 2 men from 10 men.
Step 2: Calculate the number of ways for each combination and sum them
to find the total number of different committees that can be formed.
Let’s calculate the number of ways for each combination.
Step 1: Choosing 2 women from 8 women:
(8
2)=8!
2!(8 −2)! = 28
Choosing 3 men from 10 men:
(10
3)=10!
3!(10 −3)! = 120
Now, we calculate the number of committees with 2 women and 3 men:
28 ×120 = 3360
Step 2: Choosing 3 women from 8 women:
(8
3)=8!
3!(8 −3)! = 56
Choosing 2 men from 10 men:
(10
2)=10!
2!(10 −2)! = 45
Now, we calculate the number of committees with 3 women and 2 men:
56 ×45 = 2520
Therefore, the total number of different committees that can be formed is:
3360 + 2520 = 5880
Thus, there are 5880 different committees that can be formed.
2
Question 3
Question
A committee of 5 people needs to be formed from a group of 10 women and 8
men. If the committee must consist of at least 3 women, how many different
committees can be formed?
Solution
Step 1: Count the number of ways to choose 3 women from 10. Step 2: Count
the number of ways to choose 2 additional members from the remaining 8 people
(men). Step 3: Multiply the results from Step 1 and Step 2 to find the total
number of possible committees.
Step 1: To choose 3 women from 10, we use the combination formula. The
number of ways to choose 3 women from 10 is given by (10
3)=10!
3!(10−3)! =
10×9×8
3×2×1= 120.
Step 2: We need to choose 2 additional members from the remaining 8 people
(men). The number of ways to choose 2 men from 8 is given by (8
2)=8!
2!(8−2)! =
8×7
2×1= 28.
Step 3: To find the total number of different committees that can be formed
consisting of at least 3 women, we multiply the results from Steps 1 and 2:
120 ×28 = 3360.
Therefore, there are 3360 different committees that can be formed.
Question 4
Question
In how many ways can you arrange the letters in the word ”UNIVERSITY”
such that no two vowels are adjacent?
Solution
Step 1: Let’s first count the number of ways to arrange the vowels in the word
”UNIVERSITY”. The word ”UNIVERSITY” has 4 vowels (U, I, E, I) and 6
consonants (N, V, R, S, T, Y). We will treat the vowels as a single entity, so
there are 5 entities to arrange (Vowels, N, V, R, S, T, Y).
Step 2: There are 5! ways to arrange these entities.
Step 3: Within the ”vowels” entity, there are 4! ways to arrange the vowels
themselves.
Step 4: So, the total number of ways to arrange the vowels in the word
”UNIVERSITY” is 5! ×4!.
Step 5: Now, let’s count the number of ways to arrange the 6 consonants in
the word ”UNIVERSITY”. Since there are 6 consonants, we will treat them as
a single entity.
3
Step 6: There are 6! ways to arrange these consonants.
Step 7: So, the total number of ways to arrange the consonants in the word
”UNIVERSITY” is 6!.
Step 8: The total number of ways to arrange the letters in the word ”UNI-
VERSITY” such that no two vowels are adjacent is the product of the number of
ways to arrange the vowels and the number of ways to arrange the consonants.
Step 9: Therefore, the total number of ways to arrange the letters in the word
”UNIVERSITY” such that no two vowels are adjacent is 5! ×4! ×6! = 172,800.
Question 5
Question
In a group of 10 students, how many ways can we choose a committee of 4
students?
Solution
Step 1: To solve this problem, we can use the formula for combinations, which
is given by:
C(n, k) = n!
k!(n−k)!
where nis the total number of items to choose from and kis the number of
items to choose.
Step 2: Plugging in n= 10 and k= 4 into the formula, we get:
C(10,4) = 10!
4!(10 −4)!
Step 3: Calculating the factorials:
C(10,4) = 10 ×9×8×7×6×5×4×3×2×1
(4 ×3×2×1) ×(6 ×5×4×3×2×1)
Step 4: Simplifying the expression:
C(10,4) = 10 ×9×8×7
4×3×2×1= 210
Step 5: Therefore, there are 210 ways to choose a committee of 4 students
from a group of 10.
Question 6
Question
In a group of 10 people, how many ways are there to select a committee of 4
people with exactly 2 men and 2 women if there are 4 men and 6 women in the
group?
4
Solution
Step 1: Calculate the number of ways to choose 2 men from 4 men. There are
(4
2)ways to choose 2 men from the 4 men in the group.
(4
2)=4!
2!(4 −2)! =4×3
2×1= 6
Step 2: Calculate the number of ways to choose 2 women from 6 women.
There are (6
2)ways to choose 2 women from the 6 women in the group.
(6
2)=6!
2!(6 −2)! =6×5
2×1= 15
Step 3: Multiply the number of ways to choose 2 men and 2 women. The
total number of ways to choose 2 men and 2 women from the group of 10 people
is 6×15.
6×15 = 90
Therefore, there are 90 ways to select a committee of 4 people with exactly
2 men and 2 women from a group of 10 people.
Question 7
Question
In a group of 10 students, how many ways can a committee of 4 students be
selected, where one student is the president, one is the vice-president, and the
other two are members?
Solution
To find the number of ways a committee of 4 students can be selected with spe-
cific roles assigned (president, vice-president, members), we can use the concept
of permutations.
Step 1: Choose the president from 10 students. There are 10 choices for
the president.
Step 2: Choose the vice-president from the remaining 9 students. There
are 9 choices for the vice-president.
Step 3: Choose the first member from the remaining 8 students. There are
8 choices for the first member.
Step 4: Choose the second member from the remaining 7 students. There
are 7 choices for the second member.
Therefore, the total number of ways to select a committee of 4 students with
specific roles assigned is given by:
10 ×9×8×7 = 5040
So, there are 5040 ways to select a committee of 4 students with one presi-
dent, one vice-president, and two members from a group of 10 students.
5
Question 8
Question
A committee of 5 people is to be formed from a group of 8 men and 6 women.
How many different committees can be formed if there must be at least 3 men
on the committee?
Solution
Step 1: Find the number of ways to choose 3 men out of 8. There are (8
3)ways
to choose 3 men from 8.
Step 2: Find the number of ways to choose 2 people from the remaining men
and all 6 women. There are (6
2)ways to choose 2 people from the remaining
men and (6
0)ways to choose all 6 women.
Step 3: Calculate the total number of possible committees with at least
3 men. The total number of possible committees with at least 3 men is the
product of the choices in Step 1 and Step 2:
(8
3)×(6
2)×(6
0)= 56 ×15 ×1 = 840
Therefore, there are 840 different committees that can be formed if there
must be at least 3 men on the committee.
Question 9
Question
A committee of 5 people is to be formed from a group of 10 men and 8 women.
How many different committees can be formed if the committee must consist of
at least 2 men and at least 2 women?
Solution
Step 1: Calculate the number of ways to choose 2 men out of 10. There are (10
2)
ways to choose 2 men from a group of 10.
Step 2: Calculate 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.
Step 3: Calculate the number of ways to choose the remaining person (either
a man or a woman). Since the committee must consist of at least 2 men and
at least 2 women, we need to choose 1 more person after selecting 2 men and 2
women. This person can be either a man or a woman, so there are 10+8−4 = 14
people left to choose from.
Step 4: Calculate the total number of committees that can be formed. The
total number of committees that can be formed is the product of the choices in
6
steps 1, 2, and 3:
(10
2)×(8
2)×14
Step 5: Simplify the expression.
(10
2)=10!
2!(10 −2)! =10 ×9
2×1= 45
(8
2)=8!
2!(8 −2)! =8×7
2×1= 28
Therefore, the total number of committees that can be formed is:
45 ×28 ×14 = 17640
So, there are 17,640 different committees that can be formed consisting of
at least 2 men and at least 2 women.
Question 10
Question
Suppose we have a group of 10 people, including Alice and Bob. In how many
ways can we arrange these 10 people in a line such that Alice and Bob are not
next to each other?
Solution
Let’s first find the total number of ways to arrange the 10 people in a line
without any restrictions, and then subtract the number of ways where Alice
and Bob are next to each other.
Step 1: Find the total number of ways to arrange the 10 people in a line.
Since there are 10 people, there are 10! ways to arrange them in a line.
Step 2: Find the number of ways to arrange the 10 people in a line such
that Alice and Bob are next to each other. Consider Alice and Bob as a single
entity. Then there are 9 entities to arrange (the group of Alice and Bob counted
as one, and the remaining 8 people). Within the group of Alice and Bob, there
are 2 ways to arrange them. Hence, there are 9! ×2! ways to arrange the 10
people in a line with Alice and Bob next to each other.
Step 3: Subtract the number of ways to arrange the 10 people with Alice
and Bob next to each other from the total number of ways to arrange the 10
people. The number of ways to arrange the 10 people in a line such that Alice
and Bob are not next to each other is 10! −9! ×2!.
Therefore, there are 10! −9! ×2! = 3,026,880 −725,760 = 2,301,120 ways
to arrange the 10 people in a line such that Alice and Bob are not next to each
other.
7
Question 11
Question
In a group of 12 people, how many ways can we choose a committee of 5 people
if there are 3 specific people that must be on the committee?
Solution
Step 1: First, we choose the 3 specific people that must be on the committee.
Since these 3 people are already chosen, we only need to choose 2 more people
from the remaining 9 people.
(9
2)=9!
2!(9 −2)! =9×8
2×1= 36
Step 2: So, there are 36 ways to choose 2 more people from the remaining 9
people. Therefore, there are a total of 3×36 = 108 ways to choose a committee
of 5 people with 3 specific people included.
Question 12
Question
In how many ways can 5 people be seated in a row with restrictions that two of
them, A and B, must not sit together?
Solution
Step 1: First we calculate the total number of ways the 5 people can be seated
without any restrictions.
The number of ways to seat 5 people in a row is 5! = 5 ×4×3×2×1 = 120.
Step 2: Next, we calculate the number of ways when A and B sit together.
Since A and B must sit together, consider them as one group. Then, there
are 4 entities (AB, C, D, E) that can be seated in 4! ways. However, A and B can
switch positions among themselves in 2 ways, so the final count is 4! ×2 = 48.
Step 3: Finally, we subtract the number of ways A and B can sit together
from the total number of ways to get the number of ways they must not sit
together.
Therefore, the number of ways the 5 people can be seated with A and B not
sitting together is 120 −48 = 72.
8
Question 13
Question
Suppose you have 5 red balls, 3 blue balls, and 4 yellow balls. If you randomly
select 4 balls without replacement, what is the probability of selecting 2 red
balls, 1 blue ball, and 1 yellow ball in any order?
Solution
Step 1: Find the total number of ways to choose 4 balls out of 12. Since order
does not matter, we use combinations. The total number of ways to choose 4
balls out of 12 is given by (12
4).
Step 2: Find the number of ways to choose 2 red balls, 1 blue ball, and 1
yellow ball. The number of ways to choose 2 red balls out of 5 is (5
2). The
number of ways to choose 1 blue ball out of 3 is (3
1). The number of ways to
choose 1 yellow ball out of 4 is (4
1). Therefore, the total number of ways to
choose 2 red balls, 1 blue ball, and 1 yellow ball is (5
2)·(3
1)·(4
1).
Step 3: Calculate the probability. The probability of selecting 2 red balls, 1
blue ball, and 1 yellow ball in any order is given by:
(5
2)·(3
1)·(4
1)
(12
4)=10 ·3·4
495
Therefore, the probability of selecting 2 red balls, 1 blue ball, and 1 yellow
ball in any order is 120
495 =8
33 .
Question 14
Question
In a class of 30 students, there are 12 students who are interested in statistics
and 18 students who are interested in probability. If 8 students are interested in
both statistics and probability, how many students in the class are not interested
in either statistics or probability?
Solution
Step 1: Let’s denote the number of students interested in statistics as S, the
number of students interested in probability as P, and the number of students
interested in both statistics and probability as B. We are given: S= 12 (in-
terested in statistics), P= 18 (interested in probability), B= 8 (interested in
both statistics and probability).
Step 2: We can find the total number of students interested in either statis-
tics or probability (including those interested in both) using the principle of
inclusion-exclusion: S∪P=S+P−B= 12 + 18 −8 = 22.
9
Step 3: Now, we can find the number of students not interested in either
statistics or probability by subtracting the number of students interested in
either from the total number of students: Total students in the class = 30.
Number of students not interested in either = 30 −22 = 8.
Therefore, there are 8students in the class who are not interested in either
statistics or probability.
Question 15
Question
A committee of 5 people is to be formed from a group of 8 men and 6 women.
If the committee must consist of at least 3 men, what is the number of ways the
committee can be formed?
Solution
Step 1: Find the total number of ways to form a committee of 5 people from
14 individuals. Since the committee can consist of any combination of men and
women, we will use the combination formula.
Total ways =(14
5)=14!
5!(14 −5)! =14!
5!9! = 2002
Step 2: Find the number of ways to form a committee with less than 3 men.
The committee can consist of 0, 1, or 2 men.
Ways with 0 men =(8
0)(6
5)= 1 ×6 = 6
Ways with 1 man =(8
1)(6
4)= 8 ×15 = 120
Ways with 2 men =(8
2)(6
3)= 28 ×20 = 560
Total ways with less than 3 men = 6 + 120 + 560 = 686
Step 3: Find the number of ways to form a committee with at least 3 men.
Ways with at least 3 men =Total ways −Ways with less than 3 men
= 2002 −686 = 1316
Therefore, there are 1316 ways to form a committee with at least 3 men from
the group of 8 men and 6 women.
10
Question 16
Question
In a group of 10 students, how many ways can you select a committee of 4
students to represent the class if no student can serve on the committee more
than once?
Solution
Step 1: To solve this problem, we will use the formula for combinations:
(n
k)=n!
k!(n−k)!
where nis the total number of students and kis the number of students we
want to choose for the committee.
Step 2: Substitute n= 10 and k= 4 into the formula:
(10
4)=10!
4!(10 −4)!
Step 3: Simplify the factorial expressions:
(10
4)=10 ×9×8×7
4×3×2×1
Step 4: Perform the multiplication and division:
(10
4)=5040
24
Step 5: Calculate the final result:
(10
4)= 210
Therefore, there are 210 ways to select a committee of 4 students from a
group of 10 students without any student serving on the committee more than
once.
Question 17
Question
A committee of 5 people is to be formed from a group of 9 men and 6 women.
If the committee must include at least 2 men and 2 women, how many different
committees can be formed?
11
Solution
Step 1: Find the number of ways to select 2 men and 3 women. There are (9
2)
ways to select 2 men from 9 men, and (6
3)ways to select 3 women from 6 women.
So, the number of ways to select 2 men and 3 women is (9
2)·(6
3).
Step 2: Find the number of ways to select 3 men and 2 women. There are
(9
3)ways to select 3 men from 9 men, and (6
2)ways to select 2 women from 6
women. So, the number of ways to select 3 men and 2 women is (9
3)·(6
2).
Step 3: Add the results from Step 1 and Step 2 to find the total number of
different committees that can be formed with at least 2 men and 2 women.
Therefore, the total number of different committees that can be formed is:
(9
2)·(6
3)+(9
3)·(6
2)=9!
2!7! ·6!
3!3! +9!
3!6! ·6!
2!4! .
Now, we can simplify this expression to find the final answer.
Question 18
Question
In how many ways can 5 distinct books be arranged on a bookshelf if one
particular pair of books are never to be separated?
Solution
Step 1: Consider the two books that are not to be separated as one entity. Then,
there are 4 entities to arrange on the bookshelf: the pair of books and the other
3 distinct books.
Step 2: Since the two books must always be together, we treat them as one
entity. So, there are 4! ways to arrange the 4 entities.
Step 3: However, within the pair of books, there are 2! ways to arrange the
books themselves.
Step 4: Therefore, the total number of ways to arrange the 5 distinct books
where the particular pair are never separated is 4! ×2! = 48.
Question 19
Question
Suppose a committee of 5 people is to be selected from a group of 10 men and
8 women. In how many ways can the committee be formed if it must have at
least 3 men and 2 women?
12
Solution
Step 1: Calculate the number of ways to select 3 men from 10.
(10
3)=10!
3!(10 −3)! =10 ·9·8
3·2·1= 120
Step 2: Calculate the number of ways to select 2 women from 8.
(8
2)=8!
2!(8 −2)! =8·7
2·1= 28
Step 3: Calculate the number of ways to select the remaining 2 people (could
be any gender) from the remaining pool of 7 people.
(7
2)=7!
2!(7 −2)! =7·6
2·1= 21
Step 4: Multiply the results from Step 1, Step 2, and Step 3 to find the total
number of ways to form the committee. Total ways = 120 ×28 ×21 = 70560
Therefore, the committee can be formed in 70,560 ways if it must have at
least 3 men and 2 women.
Question 20
Question
In a group of 10 friends, how many ways can you choose a committee of 4 people
if there are 2 friends who refuse to be on the same committee together?
Solution
Step 1: Calculate the total number of ways to choose a committee of 4 people
from a group of 10 friends. Step 2: Calculate the number of ways to choose a
committee where the 2 friends who refuse to be on the same committee together
are both included. Step 3: Calculate the number of ways to choose a committee
where the 2 friends who refuse to be on the same committee together are both
excluded. Step 4: Subtract the results from steps 2 and 3 from the result of
step 1 to obtain the final answer.
Step 1: To choose a committee of 4 people from a group of 10 friends, we
use the combination formula:
(10
4)=10!
4!(10 −4)! =10 ×9×8×7
4×3×2×1= 210
Step 2: If the 2 friends who refuse to be on the same committee together are
both included, we have to choose 2 more people from the remaining 8 friends.
This can be done in (8
2)ways.
(8
2)=8!
2!(8 −2)! =8×7
2×1= 28
13
Step 3: If the 2 friends who refuse to be on the same committee together
are both excluded, we can simply choose 4 people from the remaining 8 friends.
This can be done in (8
4)ways.
(8
4)=8!
4!(8 −4)! =8×7×6×5
4×3×2×1= 70
Step 4: The total number of ways to choose a committee of 4 people where
the 2 friends who refuse to be on the same committee together are not included
is: (10
4)−(8
2)−(8
4)= 210 −28 −70 = 112
Therefore, there are 112 ways to choose a committee of 4 people from a group
of 10 friends where 2 friends refuse to be on the same committee together.
Question 21
Question
A committee of 5 people is to be formed from a group of 10 females and 8
males. If the committee must consist of at least 3 females, how many different
committees can be formed?
Solution
Step 1: Find the number of ways to choose exactly 3 females and 2 males for the
committee. Step 2: Find the number of ways to choose exactly 4 females and
1 male for the committee. Step 3: Find the number of ways to choose exactly
5 females for the committee. Step 4: Add the results from Steps 1, 2, and 3 to
find the total number of different committees that can be formed.
Step 1: There are (10
3)ways to choose 3 females from the group of 10 females,
and (8
2)ways to choose 2 males from the group of 8 males. So, the number of
ways to choose exactly 3 females and 2 males is (10
3)×(8
2).
Step 2: Similarly, there are (10
4)ways to choose 4 females and (8
1)ways to
choose 1 male. So, the number of ways to choose exactly 4 females and 1 male
is (10
4)×(8
1).
Step 3: To select all 5 committee members as females, we choose all 5 females
from the group of 10 females. There is only 1 way to choose all 5 females.
Step 4: The total number of different committees that can be formed is
the sum of the results from Steps 1, 2, and 3. Therefore, the total number of
different committees that can be formed is (10
3)×(8
2)+(10
4)×(8
1)+ 1.
14
Question 22
Question
In a group of 10 people, how many different ways can we choose a committee
of 4 people if Alice and Bob refuse to serve on the committee together?
Solution
Step 1: Calculate the total number of ways to choose a committee of 4 people
from a group of 10 people. Step 2: Calculate the number of ways to choose a
committee of 4 people when Alice and Bob are together. Step 3: Subtract the
result from Step 2 from the result in Step 1 to find the number of ways when
Alice and Bob are not together.
Step 1: To choose a committee of 4 people from a group of 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 from the group of
10.
Step 2: When Alice and Bob are serving together, we treat them as one
person. So, we now have 9 people to choose from. We need to choose 3 more
people to complete the committee. Using the combination formula again:
(9
3)=9!
3!(9 −3)! =9×8×7
3×2×1= 84
Therefore, there are 84 ways to choose a committee of 4 people when Alice
and Bob are together.
Step 3: Subtract the number of ways in Step 2 from the total number of
ways in Step 1 to find the number of ways when Alice and Bob are not together:
210 −84 = 126
So, there are 126 ways to choose a committee of 4 people from the group of
10 where Alice and Bob are not both serving on the committee.
Question 23
Question
In a mathematics competition, there are 10 multiple-choice questions with 5
answer choices each. If a student randomly selects an answer for each question,
what is the probability that the student gets at least 8 questions correct?
15
Solution
To solve this problem, we can use the concept of combinations and the binomial
probability formula.
Step 1: Calculate the total number of ways to answer all 10 questions.
Since each question has 5 answer choices, there are 510 total ways the student
can answer all 10 questions.
Step 2: Calculate the number of ways to get exactly 8 questions correct.
To get exactly 8 questions correct, the student must choose 8 correct answers
and 2 incorrect answers out of the 10 questions. The number of ways to do this
is (10
8)=10!
8!(10−8)! ·1·1 = 45.
For each of these ways, the probability of getting exactly 8 questions correct
is (1
5)8×(4
5)2.
Step 3: Calculate the number of ways to get exactly 9 questions correct.
To get exactly 9 questions correct, the student must choose 9 correct answers
and 1 incorrect answer out of the 10 questions. The number of ways to do this
is (10
9)=10!
9!(10−9)! ·1·1 = 10.
For each of these ways, the probability of getting exactly 9 questions correct
is (1
5)9×(4
5)1.
Step 4: Calculate the number of ways to get all 10 questions correct. To
get all 10 questions correct, the student must choose all 10 correct answers out
of the 10 questions. There is only 1 way to do this.
The probability of getting all 10 questions correct is (1
5)10.
Step 5: Calculate the total probability of getting at least 8 questions correct.
The total probability is the sum of the probabilities of getting exactly 8, 9, or
10 questions correct.
Total probability =Probability of getting exactly 8 questions correct
+Probability of getting exactly 9 questions correct
+Probability of getting all 10 questions correct
= 45 ×(1
5)8
×(4
5)2
+ 10 ×(1
5)9
×(4
5)1
+(1
5)10
=27735
9765625 +40
9765625 +1
9765625
=27776
9765625
≈0.002841
Therefore, the probability that the student gets at least 8 questions correct
is approximately 0.002841.
16
Question 24
Question
In a group of 10 people, how many ways are there to select a president, vice
president, and treasurer, assuming no person can hold more than one position?
Solution
Step 1: To find the number of ways to select the president, we have 10 choices.
After selecting the president, there remain 9 people for the position of vice
president.
Step 2: After selecting the president and vice president, there remain 8
people for the position of treasurer.
Step 3: Therefore, the total number of ways to select a president, vice pres-
ident, and treasurer is the product of the number of ways for each position:
10 ×9×8 = 720
So, there are 720 ways to select a president, vice president, and treasurer in
a group of 10 people.
Question 25
Question
In a group of 8 friends, how many ways can we select a committee of 3 friends?
Solution
Step 1: To find the number of ways to select a committee of 3 friends from a
group of 8, we will use the combination formula. The formula for combination
is given by:
C(n, r) = n!
r!(n−r)!
where nis the total number of friends and ris the number of friends we
want to choose.
Step 2: Substituting n= 8 and r= 3 into the formula, we have:
C(8,3) = 8!
3!(8 −3)!
Step 3: Calculating the factorials:
C(8,3) = 8×7×6
3×2×1
17
C(8,3) = 336
6
Step 4: Simplifying the fraction:
C(8,3) = 56
Therefore, there are 56 ways to select a committee of 3 friends from a group
of 8.
18
