MATH 201 - INTRODUCTION TO PROBABILITY AND STATISTICS Combinatorial Analysis Question Bank Set 3 | MATH 201 - Introduction to Probability and Statistics

MATH 201 - INTRODUCTION TO

PROBABILITY AND STATISTICS -

Combinatorial Analysis

Question Bank - Set 3

Liberty University

Question 1

Question

In a group of 10 students, how many ways are there to select a committee of 3

students for a project?

Solution

To find the number of ways to choose a committee of 3 students from a group

of 10, we will use the combination formula.

Step 1: Determine the number of ways to select a committee of 3 students

from a group of 10. This is equivalent to finding the number of combinations of

10 items taken 3 at a time, denoted as (10

3).

Step 2: Use the combination formula (n

k)=n!

k!(n−k)! . Substitute n= 10

and k= 3 into the formula:

(10

3)=10!

3!(10 −3)!

Step 3: Calculate the factorials.

(10

3)=10!

3!7! =10 ×9×8

3×2×1= 120

Step 4: Therefore, there are 120 ways to select a committee of 3 students

from a group of 10 for the project.

Question 2

Question

A committee of 5 people is to be formed from a group of 8 men and 6 women.

If the committee must contain at least 3 men, how many different committees

can be formed?

Solution

Step 1: Find the number of ways to form a committee with exactly 3 men and

2 women.

There are (8

3)ways to choose 3 men from the 8 available men, and (6

2)ways

to choose 2 women from the 6 available women. Therefore, the number of ways

to form a committee with exactly 3 men and 2 women is:

3)×(6

Step 2: Find the number of ways to form a committee with 4 men and 1

woman.

There are (8

4)ways to choose 4 men from the 8 available men, and (6

1)ways

to choose 1 woman from the 6 available women. Therefore, the number of ways

to form a committee with 4 men and 1 woman is:

4)×(6

Step 3: Find the number of ways to form a committee with 5 men.

There are (8

5)ways to choose 5 men from the 8 available men. Since we’ve

exhausted the supply of women, there is only 1 way to form a committee con-

sisting of all 5 men.

Step 4: Add the results from steps 1, 2, and 3 to find the total number of

committees that can be formed.

The total number of committees is the sum of the results from steps 1, 2,

and 3: (8

3)×(6

2)+(8

4)×(6

1)+(8

5)×1

Question 3

Question

In a group of 10 people, how many ways can we choose a committee of 5 people

where one person is designated as the chairperson, one as the vice-chairperson,

and the remaining 3 as regular members?

Solution

Step 1: Choose the chairperson. There are 10 ways to choose the chairperson

out of 10 people.

Step 2: Choose the vice-chairperson. After choosing the chairperson, there

are 9 people left to choose from for the vice-chairperson.

Step 3: Choose the 3 regular members. After choosing the chairperson and

vice-chairperson, there are 8 people left to choose from for the first regular

member, 7 people left for the second regular member, and 6 people left for the

third regular member.

Step 4: Multiply the number of ways in each step. The total number of

ways to choose the committee is the product of the number of ways in each

step: 10 ×9×8×7×6 = 30,240 ways.

Thus, there are 30,240 ways to choose a committee of 5 people where one

is the chairperson, one is the vice-chairperson, and the remaining 3 are regular

members.

Question 4

Question

A committee of 5 people is to be formed from a group of 10 men and 8 women.

1. In how many ways can the committee be formed if it must have 3 men

and 2 women?

2. In how many ways can the committee be formed if at least one man must

be included?

Solution

1. To find the number of ways a committee of 3 men and 2 women can be

formed, we need to consider the number of ways to select 3 men out of 10 and

2 women out of 8.

1. To select 3 men out of 10, we use the combination formula: (10

3)=

10!

3!(10−3)! = 120.

2. To select 2 women out of 8, we use the combination formula: (8

2)=

2!(8−2)! = 28.

So, the total number of ways to form the committee with 3 men and 2 women

is 120 ×28 = 3360.

2. To find the number of ways a committee can be formed if at least one

man must be included, we can consider two cases:

1. Case 1: One man and four women are selected.

• Select 1 man out of 10: (10

1)= 10 ways.

• Select 4 women out of 8: (8

4)= 70 ways.

The total number of ways for this case is 10 ×70 = 700.

2. Case 2: Two men and three women are selected.

• Select 2 men out of 10: (10

2)= 45 ways.

• Select 3 women out of 8: (8

3)= 56 ways.

The total number of ways for this case is 45 ×56 = 2520.

Therefore, the total number of ways to form the committee with at least one

man is 700 + 2520 = 3220.

Question 5

Question

In a mathematics club with 20 members, a committee of 5 people is to be formed.

If the president and vice president refuse to serve on the committee together,

how many different committees can be formed?

Solution

Step 1: Calculate the total number of ways to form a committee without any

restrictions. Since we are selecting 5 people out of 20, the total number of ways

is given by the combination formula:

(20

5)=20!

5!(20 −5)! =20 ×19 ×18 ×17 ×16

5×4×3×2×1= 15,504.

Step 2: Calculate the number of ways to form a committee when the presi-

dent and vice president are together. Consider the president and vice president

as a single entity. Then, we have 19 people left to choose the remaining 3

committee members, giving us (19

3)ways to form the committee.

Step 3: Calculate the number of ways to form a committee when the pres-

ident and vice president are not together. To find this, we need to subtract

the number of committees formed when the president and vice president are

together from the total number of committees formed.

Number of ways to form a committee =(20

5)−(19

3).

Step 4: Calculate the final answer.

Number of ways to form a committee = 15,504−(19

3)= 15,504−19!

3!(19 −3)! = 15,504−19 ×18 ×17

3×2×1= 15,504−969 = 14,535.

Therefore, there are 14,535 different committees that can be formed in the

mathematics club under the given conditions.

Question 6

Question

How many ways can the letters of the word ”UNIVERSITY” be arranged if the

vowels must always come together?

Solution

Step 1: Consider the word ”UNIVERSITY” as a single entity when the vowels

(U, I, E) are grouped together. Thus, there are 8 entities to arrange: (V, N, R,

S, T, Y) and (U, I, E).

Step 2: Now, within the grouped entity (U, I, E), there are 3 vowels that

can be arranged in 3! = 6 ways.

Step 3: The remaining 6 entities can be arranged in 6! = 720 ways.

Step 4: Finally, the total number of ways the letters of the word ”UNIVER-

SITY” can be arranged if the vowels must always come together is 6! ×3! =

720 ×6 = 4320 ways.

Question 7

Question

In a group of 10 friends, how many ways can we choose a committee of 4 people

to represent the group if two particular friends, John and Mary, must be included

in the committee?

Solution

Step 1: Since John and Mary must be in the committee, we need to choose

2 more people from the remaining 8 friends. Step 2: The number of ways to

choose 2 people from 8 is given by the combination formula C(n, k) = n!

k!(n−k)! .

In this case, we have C(8,2) = 8!

2!(8−2)! . Step 3: Calculate C(8,2):

C(8,2) = 8!

2!6! =8×7

2×1= 28.

Step 4: Now, we need to multiply the number of ways to choose 2 people from

8 with the fixed positions of John and Mary in the committee. Step 5: The

total number of ways to choose a committee of 4 with John and Mary included

is 28 ×1 = 28 ways. Therefore, there are 28 ways to choose a committee of 4

people from a group of 10 friends with John and Mary as members.

Question 8

Question

In how many ways can you arrange the letters in the word ”COMBINATORICS”

such that no two vowels are adjacent?

Solution

To solve this problem, we can treat the word ”COMBINATORICS” as a linear

arrangement of 6 consonants (C, M, B, N, T, R) and 6 vowels (O, I, A, O, I, I).

We will first find the total number of ways to arrange all the letters, and then

subtract the number of ways in which at least 2 vowels are adjacent.

Step 1: Find the total number of ways to arrange all the letters. There are

12 letters in the word ”COMBINATORICS,” with 3 O’s and 3 I’s. Therefore,

the total number of ways to arrange all the letters is 12!

3!3! .

Step 2: Find the number of ways in which at least 2 vowels are adjacent.

Let’s treat the pair of adjacent vowels (O, I) as a single entity. So now we have

6 entities: (O, I), C, M, B, N, T, R. This allows us to arrange these entities in 6!

ways. Within the pair of adjacent vowels (O, I), we have 2 ways to arrange the

vowels. Additionally, we have 2! ways to arrange the remaining vowels. Thus,

the number of ways in which at least 2 vowels are adjacent is 6! ×2×2!.

Step 3: Calculate the final answer. Subtracting the number of ways in

which at least 2 vowels are adjacent from the total number of ways gives us

the number of ways in which no two vowels are adjacent. Therefore, the final

answer is 12!

3!3! −6! ×2×2!.

Question 9

Question

How many different 10-letter words can be formed using the letters of the word

”UNIVERSITY” if each letter can be used only once?

Solution

Step 1: Count the total number of letters in the word ”UNIVERSITY”. There

are 10 letters in the word ”UNIVERSITY”.

Step 2: Determine the number of ways to arrange the 10 different letters.

Since each letter can be used only once, we can simply use the formula for

permutations of a set of distinct objects. The number of ways to arrange 10

different letters is given by 10!.

Step 3: Account for repeated letters in the word ”UNIVERSITY”. The word

”UNIVERSITY” contains the following repeated letters: - 2 occurrences of the

letter ”I” - 2 occurrences of the letter ”U”

Step 4: Adjust for the repeated letters using factorial division. We divide

by 2! for the 2 occurrences of the letter ”I” and divide by another 2! for the 2

occurrences of the letter ”U”.

Step 5: Calculate the total number of different 10-letter words. The total

number of different 10-letter words is: 10!

2!·2! =10×9×8×7×6×5×4×3×2×1

2×1×2×1= 453600.

Therefore, there are 453600 different 10-letter words that can be formed

using the letters of the word ”UNIVERSITY” if each letter can be used only

once.

Question 10

Question

A committee of 5 people is to be formed from a group of 9 men and 6 women.

a) How many ways can the committee be formed if it consists of 3 men and 2

women?

b) How many ways can the committee be formed if there must be at least 1

man and 1 woman?

Solution

a) To find the number of ways the committee can be formed with 3 men and 2

women, we will first choose the men and then the women.

Step 1: Choose 3 men out of 9

3)=9!

3!(9 −3)! =9×8×7

3×2×1= 84

Step 2: Choose 2 women out of 6

2)=6!

2!(6 −2)! =6×5

2×1= 15

Step 3: Multiply the results from Step 1 and Step 2 to find the

total number of ways

84 ×15 = 1260

So, there are 1260 ways to form a committee with 3 men and 2 women.

b) To find the number of ways the committee can be formed with at least 1

man and 1 woman, we will consider two cases:

Case 1: 1 man and 4 women

Step 1: Choose 1 man out of 9

1)= 9

Step 2: Choose 4 women out of 6

4)=6!

4!(6 −4)! =6×5

2×1= 15

Step 3: Multiply the results from Step 1 and Step 2 to find the

total number of ways for case 1

9×15 = 135

Question 11

Question

A committee of 6 people is to be formed from a group of 10 men and 8 women.

If the committee must consist of 3 men and 3 women, how many different

committees can be formed?

Solution

Step 1: Calculate the number of ways to choose 3 men from 10. There are (10

ways to choose 3 men from a group of 10.

(10

3)=10!

3!(10 −3)! =10 ×9×8

3×2×1= 120

Step 2: Calculate the number of ways to choose 3 women from 8. There are

3)ways to choose 3 women from a group of 8.

3)=8!

3!(8 −3)! =8×7×6

3×2×1= 56

Step 3: Multiply the results from Step 1 and Step 2. The total number

of ways to form a committee with 3 men and 3 women is the product of the

number of ways to choose 3 men and 3 women.

120 ×56 = 6720

Thus, there are 6720 different committees that can be formed with 3 men

and 3 women.

Question 12

Question

A committee of 4 people is to be formed from a group of 8 men and 5 women.

How many different committees can be formed if the committee must consist of

at least 2 men and 2 women?

Solution

Step 1: Calculate the number of committees with exactly 2 men and 2 women.

• Select 2 men out of 8: (8

2)= 28 ways

• Select 2 women out of 5: (5

2)= 10 ways

Therefore, the total number of committees with exactly 2 men and 2 women

is 28 ×10 = 280 ways.

Step 2: Calculate the number of committees with 3 men and 1 woman.

• Select 3 men out of 8: (8

3)= 56 ways

• Select 1 woman out of 5: (5

1)= 5 ways

Therefore, the total number of committees with 3 men and 1 woman is

56 ×5 = 280 ways.

Step 3: Calculate the number of committees with 4 men and 0 women.

• Select 4 men out of 8: (8

4)= 70 ways

Therefore, the total number of committees with 4 men and 0 women is 70

ways.

Step 4: Add the results from Step 1, Step 2, and Step 3 to find the total

number of committees with at least 2 men and 2 women. Total number of

committees = 280 + 280 + 70 = 630 ways.

Question 13

Question

How many ways are there to select a committee of 5 people from a group of 10

people if 3 of them are good friends and want to be on the committee together?

Solution

Step 1: Choose the 3 friends to be on the committee. There is only 1 way to

choose the 3 friends to be on the committee (since they all want to be together).

Step 2: Choose the remaining 2 people to be on the committee. Since the 3

friends have already been chosen, we need to choose 2 people from the remaining

7 people. This can be done in (7

2)ways.

Step 3: Multiply the number of ways for each step to find the total number

of ways. The total number of ways to select the committee is: 1×(7

2)=

1×7!

2!(7−2)! = 1 ×7×6

2×1= 21

Therefore, there are 21 ways to select a committee of 5 people from a group

of 10 people if 3 of them are good friends and want to be on the committee

together.

Question 14

Question

In a group of 8 people, how many ways can we arrange 3 individuals to sit in a

row for a photo?

Solution

Step 1: Identify the total number of ways to arrange the 3 individuals. Step 2:

Calculate the total number of ways to arrange the remaining 5 individuals.

Step 1: To arrange 3 individuals in a row, we use the permutation formula:

P(n, r) = n!

(n−r)!

where n= 8 (total number of people) and r= 3 (number of individuals to

arrange).

Therefore, the total number of ways to arrange 3 individuals in a row is:

P(8,3) = 8!

(8 −3)! =8×7×6

1= 336

Step 2: After arranging the 3 individuals, there are 5 individuals remaining

to be arranged. To arrange the remaining 5 individuals in the remaining 5 seats,

we use the permutation formula again:

P(n, r) = n!

(n−r)!

where n= 5 (remaining number of individuals) and r= 5 (remaining seats to

arrange them).

Therefore, the total number of ways to arrange the remaining 5 individuals

in the remaining 5 seats is:

P(5,5) = 5!

(5 −5)! =5!

0! = 5! = 5 ×4×3×2×1 = 120

Final Answer: The total number of ways to arrange 3 individuals to sit in

a row for a photo in a group of 8 people is:

336 ×120 = 40320

Therefore, there are 40320 ways to arrange 3 individuals in a row for the photo.

Question 15

Question

A committee of 5 people is to be formed from a group of 9 mathematicians and

7 computer scientists. If the committee must consist of 3 mathematicians and

2 computer scientists, how many different committees can be formed?

Solution

Step 1: Calculate the number of ways to choose 3 mathematicians from 9.

3)=9!

3!(9 −3)! =9×8×7

3×2×1= 84

Step 2: Calculate the number of ways to choose 2 computer scientists from

7. (7

2)=7!

2!(7 −2)! =7×6

2×1= 21

Step 3: Multiply the number of ways to choose mathematicians and com-

puter scientists to form the committee. Total number of committees = 84×21 =

1764

Therefore, there are 1764 different committees that can be formed with 3

mathematicians and 2 computer scientists from the given group.

Question 16

Question

In a group of 12 people, there are 5 men and 7 women. A committee of 3 people

is to be selected at random. What is the probability that the committee will

consist of 2 men and 1 woman?

Solution

Step 1: Find the total number of ways to choose a committee of 3 people from

the group of 12 people. Step 2: Find the number of ways to choose 2 men from

the 5 men. Step 3: Find the number of ways to choose 1 woman from the 7

women. Step 4: Calculate the probability of selecting 2 men and 1 woman for

the committee.

Step 1: The total number of ways to choose a committee of 3 people from

12 is given by the combination formula:

(12

3)=12!

3!(12 −3)! =12 ×11 ×10

3×2×1= 220

Step 2: The number of ways to choose 2 men from the 5 men is given by:

2)=5!

2!(5 −2)! =5×4

2×1= 10

Step 3: The number of ways to choose 1 woman from the 7 women is given

by:

1)= 7

Step 4: The total number of ways to select 2 men and 1 woman for the

committee is the product of the number of ways from Step 2 and Step 3:

10 ×7 = 70

Therefore, the probability of selecting 2 men and 1 woman for the committee

is: 70

220 =7

22 ≈0.3182

Question 17

Question

A group of 10 people are taking turns to ride a rollercoaster. The rollercoaster

has 5 seats. In how many ways can the 10 people be seated on the rollercoaster?

(Assume that the order of seating does not matter)

Solution

Step 1: Since the order of seating does not matter, we are dealing with combi-

nations. We are selecting 5 people out of 10 to sit on the rollercoaster.

Step 2: The number of ways to choose 5 people out of 10 can be calculated

using the combination formula C(n, k) = n!

k!(n−k)!

Step 3: Substitute n= 10 and k= 5 into the formula: C(10,5) = 10!

5!(10−5)!

Step 4: Calculate the factorials: C(10,5) = 10×9×8×7×6

5×4×3×2×1

Step 5: Simplify the expression: C(10,5) = 252

Therefore, there are 252 ways for the 10 people to be seated on the roller-

coaster.

Question 18

Question

A committee of 5 people is to be formed from a group of 10 men and 8 women.

In how many ways can the committee be formed if it must consist of at least 3

men and at least 2 women?

Solution

Step 1: Calculate the number of ways to form the committee with exactly 3

men and 2 women. Given that there are 10 men and 8 women to choose from,

the number of ways to select 3 men from 10 men is (10

3), and the number of

ways to select 2 women from 8 women is (8

2). Therefore, the number of ways to

form the committee with exactly 3 men and 2 women is:

(10

3)×(8

Step 2: Calculate the number of ways to form the committee with 4 men

and 1 woman. The number of ways to select 4 men from 10 men is (10

4), and the

number of ways to select 1 woman from 8 women is (8

1). Therefore, the number

of ways to form the committee with 4 men and 1 woman is:

(10

4)×(8

Step 3: Calculate the number of ways to form the committee with 5 men.

The number of ways to select 5 men from 10 men is (10

5). Therefore, the number

of ways to form the committee with 5 men is:

(10

Step 4: Add the results from Step 1, Step 2, and Step 3 to find the total

number of ways to form the committee with at least 3 men and at least 2 women.

(10

3)×(8

2)+(10

4)×(8

1)+(10

Question 19

Question

A committee of 6 people is to be formed from a group of 10 individuals. If 4 of

the 10 individuals are women and 6 are men, and the committee must consist

of at least 2 women, how many different committees can be formed?

Solution

Step 1: Find the number of ways to form a committee with 2 women and 4 men.

Since we must have at least 2 women on the committee, there are 4 ways to

choose the 2 women from the 4 women, and 6 ways to choose the 4 men from

the 6 men. Therefore, the number of ways to form a committee with 2 women

and 4 men is given by:

4×6 = 24

Step 2: Find the number of ways to form a committee with 3 women and 3

men.

There are 4 ways to choose the 3 women from the 4 women, and 6 ways to

choose the 3 men from the 6 men. Therefore, the number of ways to form a

committee with 3 women and 3 men is given by:

4×6 = 24

Step 3: Find the number of ways to form a committee with 4 women and 2

men.

Since there are only 4 women in the group, there is only 1 way to choose all

4 women. There are 6 ways to choose the 2 men from the 6 men. Thus, the

number of ways to form a committee with 4 women and 2 men is:

1×6 = 6

Step 4: Add the number of committees from each case to get the total

number of different committees that can be formed.

24 + 24 + 6 = 54

Therefore, there are 54 different committees that can be formed from the

group of 10 individuals.

Question 20

Question

A committee of 5 members is to be formed from 8 men and 6 women. In how

many ways can the committee be formed if it must contain at least 2 men and

at least 2 women?

Solution

Step 1: Calculate the number of ways to choose exactly 2 men and 3 women:

There are (8

2)ways to choose 2 men from the 8 available men, and (6

3)ways to

choose 3 women from the 6 available women. Therefore, the number of ways to

choose exactly 2 men and 3 women is (8

2)×(6

3).

Step 2: Calculate the number of ways to choose 3 men and 2 women: Sim-

ilarly, there are (8

3)ways to choose 3 men from the 8 available men, and (6

ways to choose 2 women from the 6 available women. Therefore, the number of

ways to choose 3 men and 2 women is (8

3)×(6

2).

Step 3: Calculate the number of ways to choose 4 men and 1 woman: There

are (8

4)ways to choose 4 men from the 8 available men, and (6

1)ways to choose

1 woman from the 6 available women. Therefore, the number of ways to choose

4 men and 1 woman is (8

4)×(6

1).

Step 4: Calculate the total number of ways to form the committee with

at least 2 men and at least 2 women: The total number of ways is the sum

of the ways calculated in Steps 1, 2, and 3. So, the total number of ways =

2)×(6

3)+(8

3)×(6

2)+(8

4)×(6

1). Calculate this sum to find the final answer.

Question 21

Question

In a group of 12 students, how many ways can we choose a committee of 5

students if 2 of the students, Alice and Bob, refuse to be on the committee

together?

Solution

Step 1: Find the total number of ways to choose a committee of 5 students from

the group of 12. Step 2: Subtract the number of ways in which Alice and Bob

are on the committee together.

Step 1: To choose a committee of 5 students from 12, we use the combination

formula: (12

5)=12!

5!(12 −5)! =12 ×11 ×10 ×9×8

5×4×3×2×1= 792

So, there are 792 ways to choose a committee of 5 students from the group of

12.

Step 2: Now, we need to find the number of ways in which Alice and Bob

are on the committee together. This can be done by considering Alice and Bob

as a single entity and choosing the remaining 3 students from the remaining 10

students. This can be done in the following ways:

(10

3)=10!

3!(10 −3)! =10 ×9×8

3×2×1= 120

Therefore, there are 120 ways in which Alice and Bob are on the committee

together.

Step 3: Subtracting the number of ways in which Alice and Bob are on the

committee together from the total number of ways to choose a committee of 5

students:

792 −120 = 672

So, there are 672 ways to choose a committee of 5 students from the group

of 12 if Alice and Bob refuse to be on the committee together.

Question 22

Question

A committee of 5 people is to be formed from a group of 10 men and 8 women.

How many ways can this committee be formed if there must be at least 2 men

and 2 women on the committee?

Solution

Step 1: Calculate the number of ways to choose 2 men and 3 women for the

committee. Step 2: Calculate the number of ways to choose 3 men and 2 women

for the committee. Step 3: Add the results from Step 1 and Step 2 to find the

total number of ways to form the committee.

Step 1: To choose 2 men out of 10, we use the combination formula:

(10

2)=10!

2!(10 −2)! = 45

Similarly, to choose 3 women out of 8, we have:

3)=8!

3!(8 −3)! = 56

Therefore, the total number of ways to choose 2 men and 3 women is 45 ×56 =

2520.

Step 2: To choose 3 men out of 10, we have:

(10

3)=10!

3!(10 −3)! = 120

And to choose 2 women out of 8, we get:

2)=8!

2!(8 −2)! = 28

Hence, the total number of ways to choose 3 men and 2 women is 120×28 = 3360.

Step 3: Finally, we add the results from Step 1 and Step 2 to get the total

number of ways to form the committee: 2520 + 3360 = 5880.

Therefore, there are 5880 ways to form a committee of 5 people with at least

2 men and 2 women.

Question 23

Question

A committee of 5 people is to be formed from a group of 10 men and 12 women.

How many different committees can be formed if each committee must have at

least 1 man and 1 woman?

Solution

Step 1: Calculate the number of committees with 1 man and 1 woman. There

are (10

1)ways to choose 1 man and (12

1)ways to choose 1 woman. Therefore, the

number of committees with 1 man and 1 woman is (10

1)·(12

1)= 10 ·12 = 120.

Step 2: Calculate the number of committees with 2 men and 3 women. There

are (10

2)ways to choose 2 men and (12

3)ways to choose 3 women. Therefore, the

number of committees with 2 men and 3 women is (10

2)·(12

3)= 45 ·220 = 9900.

Step 3: Calculate the number of committees with 3 men and 2 women. There

are (10

3)ways to choose 3 men and (12

2)ways to choose 2 women. Therefore, the

number of committees with 3 men and 2 women is (10

3)·(12

2)= 120 ·66 = 7920.

Step 4: Calculate the total number of valid committees. The total number

of valid committees is the sum of the committees from Step 1, Step 2, and Step

3. So, the total number of valid committees is 120 + 9900 + 7920 = 17940.

Therefore, there are 17940 different committees that can be formed with

at least 1 man and 1 woman from the group of 10 men and 12 women.

Question 24

Question

You are organizing a team of 4 members from a group of 10 individuals, including

5 women and 5 men. In how many ways can you select the team if at least two

women must be on the team?

Solution

Step 1: Calculate the number of ways to select a team with exactly two women.

• Choose 2 women from 5: (5

2)= 10 ways.

• Choose 2 men from 5: (5

2)= 10 ways.

So, there are 10 ×10 = 100 ways to select a team with exactly two women.

Step 2: Calculate the number of ways to select a team with exactly three

women.

• Choose 3 women from 5: (5

3)= 10 ways.

• Choose 1 man from 5: (5

1)= 5 ways.

So, there are 10 ×5 = 50 ways to select a team with exactly three women.

Step 3: Calculate the number of ways to select a team with all four members

being women.

• Choose 4 women from 5: (5

4)= 5 ways.

Therefore, there are 5 ways to select a team with all four members being women.

Step 4: Total number of ways to select a team with at least two women.

Add the results from Steps 1, 2, and 3: 100 + 50 + 5 = 155

So, there are 155 ways to select a team of 4 members with at least two

women.

Question 25

Question

A committee of 5 people is to be selected from a group of 10 women and 7 men.

If the committee must have at least 3 women, how many different committees

can be formed?

Solution

Step 1: Calculate the number of ways to choose a committee with exactly 3

women and 2 men. To choose a committee with exactly 3 women and 2 men,

we can choose 3 women from the 10 available women and 2 men from the 7

available men. The number of ways to choose 3 women from 10 women is

given by (10

3)and the number of ways to choose 2 men from 7 men is given by

2). Therefore, the total number of ways to choose a committee with exactly 3

women and 2 men is (10

3)×(7

2).

Step 2: Calculate the number of ways to choose a committee with 4 women

and 1 man. To choose a committee with 4 women and 1 man, we can choose

4 women from the 10 available women and 1 man from the 7 available men.

The number of ways to choose 4 women from 10 women is given by (10

4)and

the number of ways to choose 1 man from 7 men is given by (7

1). Therefore,

the total number of ways to choose a committee with 4 women and 1 man is

(10

4)×(7

1).

Step 3: Calculate the total number of ways to choose a committee with at

least 3 women. The total number of ways to choose a committee with at least 3

women is the sum of the number of ways to choose a committee with exactly 3

women and 2 men and the number of ways to choose a committee with 4 women

and 1 man. Therefore, the total number of ways to choose a committee with at

least 3 women is (10

3)×(7

2)+(10

4)×(7

1).

Now, we can calculate the total number of ways to form the committee by

computing the expression above.

Question 2

Question

A committee of 5 people is to be formed from a group of 8 men and 6 women.

If the committee must contain at least 3 men, how many different committees

can be formed?

Solution

Step 1: Find the number of ways to form a committee with exactly 3 men and

2 women.

There are (8

3)ways to choose 3 men from the 8 available men, and (6

2)ways

to choose 2 women from the 6 available women. Therefore, the number of ways

to form a committee with exactly 3 men and 2 women is:

3)×(6

Step 2: Find the number of ways to form a committee with 4 men and 1

woman.

There are (8

4)ways to choose 4 men from the 8 available men, and (6

1)ways

to choose 1 woman from the 6 available women. Therefore, the number of ways

to form a committee with 4 men and 1 woman is:

4)×(6

Step 3: Find the number of ways to form a committee with 5 men.

There are (8

5)ways to choose 5 men from the 8 available men. Since we’ve

exhausted the supply of women, there is only 1 way to form a committee con-

sisting of all 5 men.

Step 4: Add the results from steps 1, 2, and 3 to find the total number of

committees that can be formed.

The total number of committees is the sum of the results from steps 1, 2,

and 3: (8

3)×(6

2)+(8

4)×(6

1)+(8

5)×1

Question 3

Question

In a group of 10 people, how many ways can we choose a committee of 5 people

where one person is designated as the chairperson, one as the vice-chairperson,

and the remaining 3 as regular members?

Solution

Step 1: Choose the chairperson. There are 10 ways to choose the chairperson

out of 10 people.

Step 2: Choose the vice-chairperson. After choosing the chairperson, there

are 9 people left to choose from for the vice-chairperson.

Step 3: Choose the 3 regular members. After choosing the chairperson and

vice-chairperson, there are 8 people left to choose from for the first regular

member, 7 people left for the second regular member, and 6 people left for the

third regular member.

Step 4: Multiply the number of ways in each step. The total number of

ways to choose the committee is the product of the number of ways in each

step: 10 ×9×8×7×6 = 30,240 ways.

Thus, there are 30,240 ways to choose a committee of 5 people where one

is the chairperson, one is the vice-chairperson, and the remaining 3 are regular

members.

Question 4

Question

A committee of 5 people is to be formed from a group of 10 men and 8 women.

1. In how many ways can the committee be formed if it must have 3 men

and 2 women?

2. In how many ways can the committee be formed if at least one man must

be included?

Solution

1. To find the number of ways a committee of 3 men and 2 women can be

formed, we need to consider the number of ways to select 3 men out of 10 and

2 women out of 8.

1. To select 3 men out of 10, we use the combination formula: (10

3)=

10!

3!(10−3)! = 120.

2. To select 2 women out of 8, we use the combination formula: (8

2)=

2!(8−2)! = 28.

So, the total number of ways to form the committee with 3 men and 2 women

is 120 ×28 = 3360.

2. To find the number of ways a committee can be formed if at least one

man must be included, we can consider two cases:

1. Case 1: One man and four women are selected.

• Select 1 man out of 10: (10

1)= 10 ways.

• Select 4 women out of 8: (8

4)= 70 ways.

The total number of ways for this case is 10 ×70 = 700.

2. Case 2: Two men and three women are selected.

• Select 2 men out of 10: (10

2)= 45 ways.

• Select 3 women out of 8: (8

3)= 56 ways.

The total number of ways for this case is 45 ×56 = 2520.

Therefore, the total number of ways to form the committee with at least one

man is 700 + 2520 = 3220.

Question 5

Question

In a mathematics club with 20 members, a committee of 5 people is to be formed.

If the president and vice president refuse to serve on the committee together,

how many different committees can be formed?

Solution

Step 1: Calculate the total number of ways to form a committee without any

restrictions. Since we are selecting 5 people out of 20, the total number of ways

is given by the combination formula:

(20

5)=20!

5!(20 −5)! =20 ×19 ×18 ×17 ×16

5×4×3×2×1= 15,504.

Step 2: Calculate the number of ways to form a committee when the presi-

dent and vice president are together. Consider the president and vice president

as a single entity. Then, we have 19 people left to choose the remaining 3

committee members, giving us (19

3)ways to form the committee.

Step 3: Calculate the number of ways to form a committee when the pres-

ident and vice president are not together. To find this, we need to subtract

the number of committees formed when the president and vice president are

together from the total number of committees formed.

Number of ways to form a committee =(20

5)−(19

3).

Step 4: Calculate the final answer.

Number of ways to form a committee = 15,504−(19

3)= 15,504−19!

3!(19 −3)! = 15,504−19 ×18 ×17

3×2×1= 15,504−969 = 14,535.

Therefore, there are 14,535 different committees that can be formed in the

mathematics club under the given conditions.

Question 6

Question

How many ways can the letters of the word ”UNIVERSITY” be arranged if the

vowels must always come together?

Solution

Step 1: Consider the word ”UNIVERSITY” as a single entity when the vowels

(U, I, E) are grouped together. Thus, there are 8 entities to arrange: (V, N, R,

S, T, Y) and (U, I, E).

Step 2: Now, within the grouped entity (U, I, E), there are 3 vowels that

can be arranged in 3! = 6 ways.

Step 3: The remaining 6 entities can be arranged in 6! = 720 ways.

Step 4: Finally, the total number of ways the letters of the word ”UNIVER-

SITY” can be arranged if the vowels must always come together is 6! ×3! =

720 ×6 = 4320 ways.

Question 7

Question

In a group of 10 friends, how many ways can we choose a committee of 4 people

to represent the group if two particular friends, John and Mary, must be included

in the committee?

Solution

Step 1: Since John and Mary must be in the committee, we need to choose

2 more people from the remaining 8 friends. Step 2: The number of ways to

choose 2 people from 8 is given by the combination formula C(n, k) = n!

k!(n−k)! .

In this case, we have C(8,2) = 8!

2!(8−2)! . Step 3: Calculate C(8,2):

C(8,2) = 8!

2!6! =8×7

2×1= 28.

Step 4: Now, we need to multiply the number of ways to choose 2 people from

8 with the fixed positions of John and Mary in the committee. Step 5: The

total number of ways to choose a committee of 4 with John and Mary included

is 28 ×1 = 28 ways. Therefore, there are 28 ways to choose a committee of 4

people from a group of 10 friends with John and Mary as members.