MATH 350 - DISCRETE MATHEMATICS Permutations and combinations Question Bank Set 1

MATH 350 - DISCRETE

MATHEMATICS - Permutations and

combinations

Question Bank - Set 1

Liberty University

Question 1

Question

A committee of 5 students is to be formed from a group of 10 students. If 3

of the students are math majors and 7 are computer science majors, how many

different committees can be formed if each committee must contain at least 1

math major and at least 1 computer science major?

Solution

To calculate the number of different committees that can be formed, we will use

the principle of combinations.

Step 1: Calculate the total number of committees without any restrictions.

Since each committee must have at least 1 math major and at least 1 com-

puter science major, there are 2 cases to consider:

Case 1: Committee with 1 math major and 4 computer science majors.

There are 3

1ways to choose 1 math major out of 3, and 7

4ways to choose

4 computer science majors out of 7. Using the multiplication principle, the

number of committees with 1 math major and 4 computer science majors is

3

1·7

4.

Case 2: Committee with 2 math majors and 3 computer science majors.

Similarly, there are 3

2ways to choose 2 math majors out of 3, and 7

3ways

to choose 3 computer science majors out of 7. Using the multiplication principle,

the number of committees with 2 math majors and 3 computer science majors

is 3

2·7

3.

Step 2: Add the results from both cases to find the total number of valid

committees.

The total number of valid committees is the sum of the number of com-

mittees with 1 math major and 4 computer science majors, and the number of

committees with 2 math majors and 3 computer science majors. Therefore, the

total number of valid committees is 3

1·7

4+3

2·7

3.

Calculate the values of 3

1,7

4,3

2, and 7

3, and substitute them into the

expression to find the final answer.

Question 2

Question

In a competition, there are 8 finalists, including 3 from the USA, 3 from Canada,

and 2 from Mexico. If the order of the finalists does not matter, how many ways

can the top 3 positions be filled if there must be 1 finalist from each country?

Solution

To solve this problem, we can use the concepts of combinations and permuta-

tions.

Step 1: Select one finalist from each country There are 3 ways to

choose the finalist from the USA, 3 ways to choose the finalist from Canada,

and 2 ways to choose the finalist from Mexico. By the multiplication principle,

there are 3 ×3×2 = 18 ways to select one finalist from each country.

Step 2: Arrange the selected finalists Since the order of the finalists

does not matter, we are interested in combinations rather than permutations.

We want to select 3 finalists from the 8 total finalists, so we have 8

3ways of

arranging the 3 selected finalists.

Step 3: Calculate the total number of ways By multiplying the number

of ways from Step 1 and Step 2, we get the total number of ways to fill the top

3 positions:

18 ×8

3= 18 ×8!

3!(8 −3)! = 18 ×8×7×6

3×2×1= 18 ×56 = 1008.

Therefore, there are 1008 ways to fill the top 3 positions with one finalist

from each country.

Question 3

Question

A committee of 5 members is to be formed from 10 men and 5 women. If the

committee must contain at least 2 women, how many different committees can

be formed?

Solution

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



Choose 2 women from 5: 5

2= 10 ways.



Choose 3 men from 10: 10

3= 120 ways.



Multiply the two possibilities together: 10 ×120 = 1200 ways.

Step 2: Find the number of ways to form a committee with at least 3 women.



Choose 3 women from 5: 5

3= 10 ways.



Choose 2 men from 10: 10

2= 45 ways.



Multiply the two possibilities together: 10 ×45 = 450 ways.

Step 3: Find the total number of ways to form a committee with at least 2

women.



Add the results from Step 1 and Step 2: 1200 + 450 = 1650 ways.

∴There are 1650 different committees that can be formed with at least 2

women.

Question 4

Question

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

that no two identical letters are adjacent?

Solution

To solve this problem, we will use the principle of permutations and inclusion-

exclusion.

Step 1: Break up the word into distinct letters There are 4 ”I”s, 4

”S”s, and 2 ”P”s, so we start by creating a word with distinct letters: ”M, I1,

I2, S1, S2, S3, S4, P1, P2”.

Step 2: Find the total number of arrangements Since all the letters

are distinct, the total number of arrangements of these letters is 9!.

Step 3: Find the number of arrangements with at least one pair of

identical letters together Let’s consider the ”I”s as a single unit (I1I2), the

”S”s as a single unit (S1S2S3S4), and the ”P”s as a single unit (P1P2). Now we

have 6 distinct units in total: M, I1I2, S1S2S3S4, P1P2. We can arrange these

units in 6! ways. Within the units I1I2, S1S2S3S4, and P1P2, we can arrange

the letters in 2!, 4!, and 2! ways respectively. Thus, the number of arrangements

with at least one pair of identical letters together is 6! ×2! ×4! ×2!.

Step 4: Apply inclusion-exclusion principle Now, we can apply the

inclusion-exclusion principle to find the number of arrangements where no two

identical letters are adjacent:

Total number of valid arrangements = Total arrangements - Arrangements

with at least one pair of identical letters together

Total number of valid arrangements = 9! −6! ×2! ×4! ×2!

= 362,880 −6×2×24 ×24 ×2

= 362,880 −55,296

= 307,584

Question 5

Question

In a bookstore, there are 5 different math textbooks and 3 different computer

science textbooks on a shelf. If a student can only choose 2 books to purchase,

how many different pairs of textbooks can the student select?

Solution

To find the number of different pairs of textbooks that a student can select, we

need to use the concept of combinations.

Step 1: Find the number of ways to choose 2 math textbooks Since

there are 5 different math textbooks, the number of ways to choose 2 of them

is given by 5

2.

Step 2: Find the number of ways to choose 2 computer science

textbooks Similarly, with 3 different computer science textbooks, the number

of ways to choose 2 of them is 3

2.

Step 3: Calculate the total number of pairs The total number of

different pairs of textbooks that can be selected is the product of the number

of ways to choose math textbooks and the number of ways to choose computer

science textbooks. Therefore, the total number of pairs is 5

2×3

2.

Step 4: Calculate the total number of pairs

5

2=5!

2!(5 −2)! =5×4

2×1= 10

3

2=3!

2!(3 −2)! =3×2

2×1= 3

So, the total number of pairs = 10 ×3 = 30.

Therefore, there are 30 different pairs of textbooks that the student can

select.

Question 6

Question

A committee of 5 people is to be formed from a group of 10 professors and 8

graduate students. If the committee must consist of 3 professors and 2 graduate

students, how many different committees can be formed?

Solution

Step 1: Calculate the number of ways to choose 3 professors from 10 professors.

Step 2: Calculate the number of ways to choose 2 graduate students from 8

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

the total number of different committees that can be formed.

Step 1: The number of ways to choose 3 professors from 10 professors is

given by the combination formula:

10

3=10!

3!(10 −3)! =10 ×9×8

3×2×1= 120

Step 2: The number of ways to choose 2 graduate students from 8 graduate

students is given by the combination formula:

8

2=8!

2!(8 −2)! =8×7

2×1= 28

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

of different committees that can be formed:

120 ×28 = 3360

Therefore, there are 3360 different committees that can be formed consisting

of 3 professors and 2 graduate students.

Question 7

Question

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

If the committee must consist of at least 3 men and at least 2 women, how many

different committees can be formed?

Solution

Step 1: Calculate the number of ways to choose 3 men from 10. Step 2: Calculate

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

number of ways to form the committee with the given restrictions.

Step 1: The number of ways to choose 3 men from 10 is given by the

combination formula C(10,3):

C(10,3) = 10!

3!(10 −3)! =10 ×9×8

3×2×1= 120

Step 2: The number of ways to choose 2 women from 8 is given by the

combination formula C(8,2):

C(8,2) = 8!

2!(8 −2)! =8×7

2×1= 28

Step 3: To form the committee with the given restrictions, we multiply the

results from step 1 and step 2:

Total number of ways = 120 ×28 = 3360

Therefore, there are 3360 different committees that can be formed with at

least 3 men and at least 2 women.

Question 8

Question

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

the committee must consist of at least 2 men and 2 women, how many different

committees can be formed?

Solution

Step 1: Calculate the number of ways to choose 2 men from 8 men. There are

8

2ways to choose 2 men from 8, which is equal to 8!

2!(8−2)! =8×7

2×1= 28.

Step 2: Calculate the number of ways to choose 2 women from 6 women.

There are 6

2ways to choose 2 women from 6, which is equal to 6!

2!(6−2)! =

6×5

2×1= 15.

Step 3: Calculate the number of ways to choose the remaining person (either

a man or a woman) for the committee. There are 8 men and 6 women left to

choose from. Since the committee must consist of at least 2 men and 2 women,

the remaining person can be a man or a woman. Therefore, there are 8 + 6 - 2

= 12 options for the remaining person.

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

the results from Step 1, Step 2, and Step 3 to get the total number of ways to

form the committee: Total = Number of ways to choose 2 men ×Number of

ways to choose 2 women ×Number of ways to choose the remaining person

Total = 28 ×15 ×12 Total = 5040

Therefore, there are 5040 different committees that can be formed with at

least 2 men and 2 women.

Question 9

Question

A company wants to form a committee of 5 members from a pool of 10 employ-

ees, including 4 men and 6 women. If the committee must consist of at least

2 men and at least 2 women, how many different ways can the committee be

formed?

Solution

Let’s break this problem down into cases: - Case 1: 2 men and 3 women on the

committee - Case 2: 3 men and 2 women on the committee - Case 3: 4 men and

1 woman on the committee

We will calculate the number of ways for each case and then add them up

to get the total number of ways the committee can be formed.

Case 1: 2 men and 3 women on the committee Step 1: Choose 2

men out of 4 men in 4

2ways. Step 2: Choose 3 women out of 6 women in 6

3

ways. Step 3: Multiply the number of ways in Step 1 and Step 2 to get the total

number of ways for Case 1. So, the total number of ways for Case 1 is 4

2·6

3.

Case 2: 3 men and 2 women on the committee Step 1: Choose 3

men out of 4 men in 4

3ways. Step 2: Choose 2 women out of 6 women in 6

2

ways. Step 3: Multiply the number of ways in Step 1 and Step 2 to get the total

number of ways for Case 2. So, the total number of ways for Case 2 is 4

3·6

2.

Case 3: 4 men and 1 woman on the committee Step 1: Choose 4

men out of 4 men in 4

4ways. Step 2: Choose 1 woman out of 6 women in 6

1

ways. Step 3: Multiply the number of ways in Step 1 and Step 2 to get the total

number of ways for Case 3. So, the total number of ways for Case 3 is 4

4·6

1.

Now, we add the number of ways for each case to get the total number of ways

the committee can be formed: Total number of ways = 4

2·6

3+4

3·6

2+4

4·6

1.

Question 10

Question

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

the committee must consist of at least 2 men and 2 women, how many different

committees can be formed?

Solution

To find the number of different committees that can be formed, we need to

consider the number of ways we can select 2 men and 3 women, 3 men and 2

women, or 4 men and 1 woman.

Step 1: Selecting 2 men and 3 women To select 2 men out of 8, we use

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

k!(n−k)! :

C(8,2) = 8!

2!(8 −2)! =8×7

2×1= 28

To select 3 women out of 6, we use the combination formula again:

C(6,3) = 6!

3!(6 −3)! =6×5×4

3×2×1= 20

Therefore, the number of ways to select 2 men and 3 women is 28 ×20.

Step 2: Selecting 3 men and 2 women To select 3 men out of 8, we

have:

C(8,3) = 8!

3!(8 −3)! =8×7×6

3×2×1= 56

To select 2 women out of 6, we have:

C(6,2) = 6!

2!(6 −2)! =6×5

2×1= 15

Therefore, the number of ways to select 3 men and 2 women is 56 ×15.

Step 3: Selecting 4 men and 1 woman To select 4 men out of 8, we

have:

C(8,4) = 8!

4!(8 −4)! =8×7×6×5

4×3×2×1= 70

To select 1 woman out of 6, we have:

C(6,1) = 6!

1!(6 −1)! =6

1= 6

Therefore, the number of ways to select 4 men and 1 woman is 70 ×6.

Step 4: Total number of committees Finally, we add up the number of

committees from each scenario to find the total number of different committees

that can be formed:

28 ×20 + 56 ×15 + 70 ×6 = 560 + 840 + 420 = 1820

Therefore, there are 1820 different committees that can be formed.

Question 11

Question

A company is selecting a team of 5 employees to attend a conference. If there

are 10 qualified employees to choose from, how many different teams can be

formed if:

1. The order in which the employees are selected does not matter?

2. The order in which the employees are selected does matter?

Solution

1. When the order does not matter, we are dealing with combinations.

1. We need to calculate the number of ways to choose 5 employees out of 10.

This is denoted by 10

5.

2. Applying the formula for combinations, we get:

10

5=10!

5!(10 −5)! =10 ×9×8×7×6

5×4×3×2×1= 252

2. When the order does matter, we are dealing with permutations.

1. We need to calculate the number of ways to arrange 5 employees out of

10. This is denoted by P(10,5).

2. Applying the formula for permutations, we get:

P(10,5) = 10!

(10 −5)! =10 ×9×8×7×6

5×4×3×2×1= 30,240

Question 12

Question

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

If the committee must have at least 2 men and 2 women, how many different

committees can be formed?

Solution

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

3 women. Since we need at least 2 men and 2 women, we must consider all

possible combinations of 2 men and 3 women. The number of ways to select

2 men from 9 is 9

2= 36. The number of ways to select 3 women from 6 is

6

3= 20. Therefore, the number of ways to form a committee with exactly 2

men and 3 women is 36 ×20 = 720.

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

women. The number of ways to select 3 men from 9 is 9

3= 84. The number

of ways to select 2 women from 6 is 6

2= 15. Therefore, the number of ways to

form a committee with 3 men and 2 women is 84 ×15 = 1260.

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

ways to form the committee. Total number of ways = 720 + 1260 = 1980.

Thus, there are 1980 different committees that can be formed with at least

2 men and 2 women.

Question 13

Question

How many ways are there to choose a committee of 3 students from a group of

10 students if one of the students, Alice, must be on the committee?

Solution

To find the number of ways to choose a committee of 3 students with Alice

included, we need to choose 2 more students from the remaining 9 students.



Step 1: Calculate the number of ways to choose 2 students from the

remaining 9 students.

Since the order does not matter, we are dealing with combinations. The

number of ways to choose 2 students from 9 can be calculated using the

combination formula n

r=n!

r!(n−r)! , where nis the total number of options

and ris the number of choices. Here, n= 9 and r= 2.

9

2=9!

2!(9 −2)! =9!

2!7! =9×8

2×1= 36

So, there are 36 ways to choose 2 students from the remaining 9.



Step 2: Multiply the result from Step 1 by 1 (since Alice is already

chosen).

We have already chosen Alice to be on the committee, so we just need

to choose 2 students from the remaining 9 students. This means there is

only 1 way to choose Alice.



Step 3: Calculate the total number of ways to form the committee.

By the multiplication principle, the total number of ways to choose a

committee with 3 students, including Alice, is the product of the number

of ways to choose 2 students from the remaining 9 and the number of ways

to choose Alice.

Total ways = 36 ×1 = 36

Therefore, there are 36 ways to choose a committee of 3 students from a

group of 10 students with Alice on the committee.

Question 14

Question

How many ways are there to seat 6 people at a round table if 3 of them refuse

to sit next to each other?

Solution

Step 1: First consider the total number of ways to seat 6 people at a round table

without any restrictions. Let’s seat the first person arbitrarily. After that, there

are 5 remaining seats for the 5 other people. Thus, there are 5! ways to seat the

6 people in total without any restrictions.

Step 2: Next, consider the number of ways to seat the 6 people so that 3

specific people refuse to sit next to each other. Let’s treat the 3 people as a

single entity. Then, we have 4 entities to seat: the group of 3 people who cannot

sit next to each other, and the 3 remaining individuals. This can be done in 4!

ways.

Step 3: Consider how the 3 individuals in the group can be arranged within

the group. The 3 individuals can be arranged among themselves in 3! ways.

Step 4: Put all the pieces together to find the total number of ways to seat

the 6 people with the given restriction. The total number of ways is 5! −4! ·3! =

120 −24 ·6 = 120 −144 = −24 ways.

Therefore, there are 24 ways to seat the 6 people at a round table such that

3 specific people refuse to sit next to each other.

Question 15

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 3 men and

2 women?

Solution

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

10

3ways to select 3 men from 10.

Step 2: Calculate the number of ways to select 2 women from 8 women.

There are 8

2ways to select 2 women from 8.

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

of ways to form the committee. Total number of ways = 10

3×8

2.

Step 4: Calculate the values of 10

3and 8

2.10

3=10!

3!(10−3)! =10×9×8

3×2×1=

120.

8

2=8!

2!(8−2)! =8×7

2×1= 28.

Step 5: Substitute the values back into the formula. Total number of ways

= 120 ×28 = 3360.

Therefore, the committee can be formed in 3360 different ways.

Question 16

Question

In a class of 20 students, a group of 7 students is to be chosen to form a team.

However, 2 particular students, Alice and Bob, refuse to be on the same team.

How many ways can the team be formed?

Solution

Step 1: Calculate the total number of ways to choose a team of 7 students from

20. Step 2: Calculate the number of ways in which Alice and Bob can be on

the same team. Step 3: Subtract the number of ways in Step 2 from Step 1 to

find the total number of ways without Alice and Bob on the same team.

Step 1: The total number of ways to choose 7 students out of 20 is given

by the combination formula:

20

7=20!

7!(20 −7)! = 77520

Step 2: Now, consider Alice and Bob as a single entity (since they must al-

ways be together). Choosing this ”entity” along with the other 5 team members

from the remaining 18 students can be done in:

18

5=18!

5!(18 −5)! = 8568

However, within this entity, Alice and Bob can be arranged in 2 different ways.

So, the total number of ways Alice and Bob can be on the same team is 8568×2 =

17136.

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

team from the total number of ways to choose 7 students:

77520 −17136 = 60384

Therefore, the total number of ways to form a team of 7 students without

Alice and Bob on the same team is 60,384 ways.

Question 17

Question

In how many ways can the letters of the word ”UNIVERSITY” be arranged

such that no two vowels are adjacent?

Solution

Step 1: First, we need to determine the total number of ways to arrange all the

letters in the word ”UNIVERSITY”. Since there are 10 letters in ”UNIVER-

SITY”, we have 10 positions to fill.

Step 2: There are 3 vowels (U, I, E) and 7 consonants (N, V, R, S, T, Y) in

the word ”UNIVERSITY”. Let’s consider the vowels as a group (e.g., Vowels:

UIE) and the consonants as another group (e.g., Consonants: NVRSTY).

Step 3: Since no two vowels can be adjacent, we must place the consonants

in between the vowels. Thus, we have 4 places to put the consonant group

(either before the first vowel, between the vowels, after the last vowel).

Step 4: The consonant group (NVRSTY) can be arranged among themselves

in 7! ways.

Step 5: The vowel group (UIE) can be arranged among themselves in 3!

ways.

Step 6: Finally, the total number of ways to arrange the letters of the word

”UNIVERSITY” such that no two vowels are adjacent is given by:

4×7! ×3! = 4 ×5040 ×6 = 120960

Therefore, there are 120,960 ways to arrange the letters of the word ”UNI-

VERSITY” such that no two vowels are adjacent.

Question 18

Question

A committee of 5 people is to be formed from a group of 8 men and 7 women.

If the committee must consist of 3 men and 2 women, how many different

committees can be formed?

Solution

Step 1: Calculate the number of ways to choose 3 men from 8 men. Step

2: Calculate the number of ways to choose 2 women from 7 women. Step 3:

Multiply the results from step 1 and step 2 to find the total number of different

committees that can be formed.

Step 1: Calculate the number of ways to choose 3 men from 8 men. Since

order does not matter, we will use combinations. The number of ways to choose

3 men from 8 men is given by 8

3=8!

3!(8−3)! = 56.

Step 2: Calculate the number of ways to choose 2 women from 7 women.

Similarly, the number of ways to choose 2 women from 7 women is 7

2=

2!(7−2)! = 21.

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

different committees that can be formed with 3 men and 2 women is 56 ×21 =

1176.

Therefore, there are 1176 different committees that can be formed with 3

men and 2 women from the group.

Question 19

Question

In a group of 12 people, how many ways are there to choose a committee of 5

people if two particular people, Emily and Alex, refuse to serve on the committee

together?

Solution

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

the 12 available. Step 2: Find the number of ways to choose a committee of

5 people when Emily and Alex are both on the committee. Step 3: Find the

number of ways to choose a committee of 5 people when Emily and Alex are

both excluded from the committee. Step 4: Subtract the result from Step 2

from the result in Step 1 to get the final answer.

Step 1: The total number of ways to choose a committee of 5 people from

12 is given by the combination formula 12

5.

12

5=12!

5!(12 −5)! =12 ×11 ×10 ×9×8

5×4×3×2×1= 792

Step 2: The number of ways to choose a committee of 5 people when Emily

and Alex are both on the committee is the number of ways to choose 3 people

from the remaining 10 people. This is given by 10

3.

10

3=10!

3!(10 −3)! =10 ×9×8

3×2×1= 120

Step 3: The number of ways to choose a committee of 5 people when Emily

and Alex are both excluded is the total number of ways to choose a committee

of 5 people from the remaining 10 people, excluding Emily and Alex. This is

10

5.

10

5=10!

5!(10 −5)! =10 ×9×8×7×6

5×4×3×2×1= 252

Step 4: The final answer is obtained by subtracting the result from Step 2

from Step 1.

792 −120 = 672

Thus, there are 672 ways to choose a committee of 5 people from the group

of 12 people if Emily and Alex refuse to serve on the committee together.

Question 20

Question

A committee of 5 people is to be formed from a group of 12 students. If 4 of the

students must be on the committee, how many different ways can the committee

be formed?

Solution

Step 1: Determine the number of ways to choose the required 4 students out of

12. This is a combination problem.

Number of ways to choose 4 students from 12 = 12

4

Step 2: Calculate 12

4.

12

4=12!

4!(12 −4)!

12

4=12 ×11 ×10 ×9

4×3×2×1

12

4= 495

Step 3: Once the 4 students are selected, there is only 1 way to select the

5th member for the committee.

Step 4: Multiply the number of ways in step 2 and step 3 to get the total

number of ways to form the committee.

Total number of ways to form the committee = 12

4×1

Total number of ways to form the committee = 495 ×1

Total number of ways to form the committee = 495

Therefore, there are 495 different ways the committee can be formed.

Question 21

Question

A committee of 5 people is to be formed from a group of 9 women and 7 men.

In how many ways can this committee be formed if there must be at least 2

women and 2 men on the committee?

Solution

To find the total number of ways a committee of 5 people can be formed with

at least 2 women and 2 men, we will consider the different cases separately.

Step 1: Choose 2 women and 3 men We can choose 2 women from the 9

available women in 9

2ways, and we can choose 3 men from the 7 available men

in 7

3ways.

Total number of ways to choose 2 women and 3 men: 9

2×7

3= 36 ×35 =

1260 ways

Step 2: Choose 3 women and 2 men We can choose 3 women from the 9

available women in 9

3ways, and we can choose 2 men from the 7 available men

in 7

2ways.

Total number of ways to choose 3 women and 2 men: 9

3×7

2= 84 ×21 =

1764 ways

Step 3: Choose 4 women and 1 man We can choose 4 women from the 9

available women in 9

4ways, and we can choose 1 man from the 7 available men

in 7

1ways.

Total number of ways to choose 4 women and 1 man: 9

4×7

1= 126 ×7 =

882 ways

Step 4: Choose 5 women We can choose all 5 members to be women from

the 9 available women in 9

5ways.

Total number of ways to choose 5 women: 9

5= 126 ways

Step 5: Add the results from all cases Adding the number of ways found in

each step gives us the total number of ways to form a committee with at least

2 women and 2 men: 1260 + 1764 + 882 + 126 = 4032 ways

Therefore, there are 4032 ways to form a committee of 5 people with at least

2 women and 2 men.

Question 22

Question

A bakery sells 8 different types of cupcakes. If a customer wants to buy a box

of 5 cupcakes with at least one of each type, how many different choices does

the customer have?

Solution

Step 1: First, we can calculate the total number of ways to choose 5 cupcakes

from 8 types without any restrictions. This is a combination problem and can

be solved using the formula for combinations:

n

k=n!

k!(n−k)!

where nis the total number of items to choose from and kis the number of

items to choose. So, the total number of ways to choose 5 cupcakes from 8

types is:

8

5=8!

5!(8 −5)! =8×7×6

3×2×1= 56

Step 2: Next, we need to consider the ways in which the customer can choose

5 cupcakes with at least one of each type. This can be done by subtracting the

number of ways to choose 5 cupcakes without any restrictions from the total

number of ways to choose 5 cupcakes. Which is:

Total choices −Choices without any restrictions = 56 −8 = 48

Therefore, the customer has 48 different choices to buy a box of 5 cupcakes

with at least one of each type.

Question 23

Question

A student has 8 different textbooks on mathematics and wants to choose 4

of them to bring on a study trip. However, the student only wants to bring at

most 2 textbooks that focus on the same topic. How many ways can the student

choose the textbooks to bring on the study trip?

Solution

To solve this problem, we can break it down into cases based on the number of

textbooks on the same topic that the student can bring.

Case 1: Bringing 4 textbooks on different topics In this case, the

student must choose 4 textbooks from the 8 different textbooks available. This

can be done in 8

4ways.

Case 2: Bringing 3 textbooks on one topic and 1 textbook on a

different topic In this case, the student must choose 3 textbooks out of the

textbooks on the same topic and 1 textbook out of the remaining textbooks.

There are 8

3ways to choose 3 textbooks on the same topic and 5

1ways to

choose 1 textbook on a different topic.

Therefore, the total number of ways for this case is 8

3·5

1.

Case 3: Bringing 2 textbooks on one topic and 2 textbooks on

another topic In this case, the student must choose 2 textbooks out of the

textbooks on one topic and 2 textbooks out of the textbooks on another topic.

There are 4

2ways to choose 2 textbooks on one topic and 4

2ways to choose

2 textbooks on another topic.

Therefore, the total number of ways for this case is 4

2·4

2.

Now, we can find the total number of ways by adding up the number of ways

from each case: 8

4+8

3·5

1+4

2·4

2

Calculating these combinations gives us the final answer.

Question 24

Question

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

no two S’s are adjacent?

Solution

Step 1: Calculate the total number of ways to arrange the letters in ”MISSIS-

SIPPI” without any restrictions.

There are 11 letters in the word ”MISSISSIPPI”, with the following counts:

- 4 S’s - 4 I’s - 2 P’s - 1 M

The total number of ways to arrange these letters without any restrictions

is given by the formula for permutations of a multiset:

11!

4! ·4! ·2! ·1! = 34,650

Step 2: Calculate the number of ways that the S’s are adjacent.

Treat the 4 S’s as a single letter (denoted as ”SSSS”) and find the total

number of ways to arrange the letters now, which will include the cases when

the S’s are adjacent.

Now we have 8 letters (SSSS, M, II, PPP) to arrange, with the following

counts: - 1 SSSS - 1 M - 2 I’s - 3 P’s

The number of ways to arrange these letters is:

3! ·2! = 3360

Step 3: Calculate the number of ways that the S’s are not adjacent.

To find the number of ways the S’s are not adjacent, subtract the number

of ways that the S’s are adjacent from the total number of ways without any

restrictions:

34,650 −3360 = 31,290

Therefore, there are 31,290 ways to arrange the letters in the word ”MIS-

SISSIPPI” if no two S’s are adjacent.

Question 25

Question

In how many ways can we arrange the letters of the word ”UNIVERSITY” such

that no two vowels are adjacent?

Solution

Step 1: Calculate the total number of ways to arrange the letters in the word

”UNIVERSITY” without any restrictions. The word ”UNIVERSITY” has 10

letters, including 3 vowels (U, I, and I) and 7 consonants (N, V, R, S, T, and

Y). The total number of ways to arrange the 10 letters is 10!.

Step 2: Calculate the number of ways that the vowels can be arranged such

that no two vowels are adjacent. First, we can arrange the 4 blocks containing

the 3 vowels (U, I, and I) and 1 block containing the 7 consonants (N, V, R, S,

T, and Y) in 4! ways. Within the vowel blocks, the vowels can be arranged in

3! ways (since there are 3 vowels). Therefore, the number of ways the vowels

can be arranged is 4! ×3!.

Step 3: Calculate the number of ways to arrange the letters such that no

two vowels are adjacent. Using the principle of multiplication, we multiply the

results from Step 2 and Step 3. Thus, the number of ways to arrange the letters

of ”UNIVERSITY” such that no two vowels are adjacent is 10!

4!×3! .

Question 26

Question

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

If the committee must consist of 3 men and 2 women, how many different

committees can be formed?

Solution

Step 1: Find the number of ways to choose 3 men out of 10.

There are 10 men to choose from, and we need to select 3 of them. This can

be done using a combination, denoted as C(n, k), where nis the total number

of items and kis the number of items to choose.

C(10,3) = 10!

3!(10 −3)! =10 ×9×8

3×2×1= 120

So, there are 120 ways to choose 3 men out of 10.

Step 2: Find the number of ways to choose 2 women out of 8.

Similarly, we can find the number of ways to choose 2 women out of 8 using

a combination.

C(8,2) = 8!

2!(8 −2)! =8×7

2×1= 28

Thus, there are 28 ways to choose 2 women out of 8.

Step 3: Find the total number of different committees.

To find the total number of different committees that can be formed with

3 men and 2 women, we multiply the number of ways to choose 3 men and 2

women.

Total number of committees = Number of ways to choose 3 men * Number

of ways to choose 2 women Total number of committees = 120 * 28 = 3360

Therefore, there are 3360 different committees that can be formed with 3

men and 2 women.

Question 27

Question

In how many ways can a committee of 4 be chosen from a group of 10 people,

where 2 of the people refuse to serve on the committee together?

Solution

Step 1: We first calculate the total number of ways to choose a committee of 4

from 10 people. This is given by the combination formula C(n, k) = n!

k!(n−k)! ,

where nis the total number of people and kis the number of people on the

committee.

Total ways to choose a committee of 4 from 10 people = C(10,4) = 10!

4!6! = 210

Step 2: Next, we calculate the number of ways the two people who refuse

to serve on the committee can be chosen. Since they cannot serve together, we

have 2 choices for the first person and 7 choices for the second person.

Number of ways to choose the two people who refuse to serve together = 2×7 = 14

Step 3: Now, we subtract the number of committees that include both of

the people who refuse to serve from the total number of committees to get the

final answer.

Number of committees that include both of the people who refuse to serve = Total ways−Ways with both refusing = 210−14 = 196

Therefore, there are 196 ways to choose a committee of 4 from a group of

10 people, where 2 of the people refuse to serve on the committee together.

Question 28

Question

In a computer science class, there are 7 students: Alex, Ben, Chris, Dana, Eve,

Fran, and Gabe. The professor wants to form a team consisting of a president,

a vice president, and a treasurer. How many different ways can the professor

choose the team if:

1. Students can hold more than one position?

2. Each student can only hold one position?

Solution

1. In this case, since students can hold more than one position, we can simply

count the number of ways each position can be filled.



For the president position, there are 7 choices.



For the vice president position, there are still 7 choices.



For the treasurer position, there are again 7 choices.

Therefore, the total number of ways the team can be chosen is 7 ×7×7 = 343

ways.

2. Now, if each student can only hold one position, we will consider the

positions one at a time.



For the president position, there are 7 choices.



For the vice president position, after choosing the president, there are 6

choices remaining.



For the treasurer position, after choosing the president and vice president,

there are 5 choices remaining.

Therefore, the total number of ways the team can be chosen with each student

holding one position is 7 ×6×5 = 210 ways.

Question 29

Question

Suppose a car rental company offers 5 different models of cars. A customer

wishes to rent 3 cars for a road trip. How many ways can the customer choose

3 cars to rent if the order in which the cars are chosen does not matter?

Solution

Let’s solve this problem using combinations (since the order in which the cars

are chosen does not matter). We can use the formula for combinations, which

is denoted as n

k=n!

k!(n−k)!, where nis the total number of options and kis

the number of choices.

Step 1: Calculate the number of ways to choose 3 cars out of 5. The

formula for combinations is 5

3=5!

3!(5 −3)!. Calculating the factorials, we get:

5

3=120

6·2= 10.

Step 2: Determine the number of ways the customer can choose 3 cars to

rent. There are 10 ways for the customer to choose 3 cars out of the 5 available

models.

Therefore, there are 10 ways for the customer to choose 3 cars to rent from

the 5 different models offered by the car rental company.

Question 30

Question

In how many ways can 5 boys and 5 girls sit in a row such that no two boys or

two girls are sitting next to each other?

Solution

Step 1: First, we arrange the boys and girls separately. Since no two boys or two

girls can sit next to each other, we treat the boys and girls as distinct entities.

Step 2: Arrange the 5 boys in a row. There are 5 boys and 5 positions, so

there are 5! ways to arrange the boys.

Step 3: Arrange the 5 girls in a row. There are 5 girls and 6 positions (the

5 spaces between boys and the ends), so there are 6P5 = 6!

(6−5)! = 6! ways to

arrange the girls.

Step 4: Multiply the number of ways to arrange the boys and girls to get

the total number of ways to arrange all 10 people.

Therefore, the total number of ways the 5 boys and 5 girls can sit in a row

such that no two boys or two girls are sitting next to each other is

5! ×6! = 120 ×720 = 86400

Question 31

Question

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

people, where 2 of the people on the committee must be a married couple that

always come together?

Solution

Step 1: First we choose the married couple to be on the committee. Since they

always come together, we can treat them as one entity. There are 2

2= 1 way

to choose the married couple.

Step 2: Now we need to choose the remaining 3 people to be on the committee

from the remaining 8 people (10 people total - 2 people of the married couple).

This can be done in 8

3ways.

Step 3: To find the total number of ways to form the committee with the

specified conditions, we multiply the results from Step 1 and Step 2:

1×8

3= 1 ×8!

3!(8−3)! =8×7×6

3×2×1= 56

Therefore, there are 56 ways to choose a committee of 5 people from a group

of 10 people, where 2 of the people on the committee must be a married couple

that always come together.

Question 32

Question

A committee of 5 people is to be formed from a group of 9 men and 7 women.

In how many ways can the committee be formed if the committee must have at

least 3 women?

Solution

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

the committee. To choose 3 women from 7, we use combinations: 7

3= 35. To

choose 2 men from 9, we use combinations: 9

2= 36. So, the number of ways

to choose exactly 3 women and 2 men is (35)(36) = 1260.

Step 2: Count the number of ways to choose exactly 4 women and 1 man for

the committee. To choose 4 women from 7, we use combinations: 7

4= 35. To

choose 1 man from 9, we use combinations: 9

1= 9. So, the number of ways to

choose exactly 4 women and 1 man is (35)(9) = 315.

Step 3: Count the number of ways to choose exactly 5 women for the com-

mittee. To choose 5 women from 7, we use combinations: 7

5= 21. Since we

have to choose all women, the number of ways is 1.

Step 4: Add up the total number of ways by summing the results from all

cases. Total number of ways = 1260 + 315 + 1 = 1576.

Therefore, the committee can be formed in 1576 different ways if it must

have at least 3 women.

Question 33

Question

A committee of 5 people is to be formed from a group of 10 students and 6

professors. In how many ways can the committee be formed if it must consist

of 3 students and 2 professors?

Solution

Let’s break down the problem step by step:

Step 1: Find the number of ways to select 3 students from 10. This

is a combination problem where we need to choose 3 students from a group of

10 students. 10

3=10!

3!7! = 120

Step 2: Find the number of ways to select 2 professors from 6.

Similarly, this is a combination problem where we need to choose 2 professors

from a group of 6 professors.

6

2=6!

2!4! = 15

Step 3: Find the total number of ways to form the committee. To

find the total number of ways to form the committee, multiply the number of

ways to choose 3 students and 2 professors.

120 ×15 = 1800

So, there are 1800 ways to form the committee consisting of 3 students and

2 professors.

Question 34

Question

A committee of 5 members is to be formed from a group of 10 people. If 3

of the 10 people refuse to serve together on the committee because of personal

conflicts, how many different committees can be formed?

Solution

Step 1: Calculate the total number of ways to form a committee of 5 members

from 10 people. Step 2: Calculate the number of ways which include the 3 people

who refuse to serve together. Step 3: Subtract the number of committees with

the 3 people from the total number to get the final answer.

Step 1: The total number of ways to choose a committee of 5 members

from 10 people is given by the binomial coefficient 10

5, which is calculated as:

10

5=10!

5! ·(10 −5)! =10 ·9·8·7·6

5·4·3·2·1= 252

Step 2: To calculate the number of ways to form a committee including

the 3 people who refuse to serve together, we choose 2 more members from the

remaining 7 people. This is given by 7

2, which is calculated as:

7

2=7!

2! ·(7 −2)! =7·6

2·1= 21

Step 3: Finally, subtract the number of committees with all 3 people from

the total number of committees:

252 −21 = 231

Therefore, there are 231 different committees that can be formed.

Question 35

Question

A committee of 5 people is to be formed from 8 men and 7 women. In how

many ways can the committee be formed if it 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 8. Step 2: Calculate

the number of ways to choose 2 women out of 7. Step 3: Calculate the number

of ways to choose the remaining person (man or woman) for the committee.

Step 4: Multiply the results from steps 1-3 to find the total number of ways the

committee can be formed.

Step 1: The number of ways to choose 2 men out of 8 is given by the

combination formula 8

2:

8

2=8!

2!(8 −2)! =8×7

2×1= 28

Step 2: The number of ways to choose 2 women out of 7 is given by the

combination formula 7

2:

7

2=7!

2!(7 −2)! =7×6

2×1= 21

Step 3: To form a committee of 5 people with at least 2 men and at least

2 women, we need to choose the remaining person as either a man or a woman.

There are 6 remaining people to choose from (6 men and 5 women).

Step 4: Multiply the results from steps 1-3 to find the total number of ways

the committee can be formed:

28 ×21 ×6 = 3528

Therefore, there are 3528 ways to form a committee of 5 people with at least

2 men and at least 2 women from 8 men and 7 women.

The total number of valid committees is the sum of the number of com-

mittees with 1 math major and 4 computer science majors, and the number of

committees with 2 math majors and 3 computer science majors. Therefore, the

total number of valid committees is 3

1·7

4+3

2·7

3.

Calculate the values of 3

1,7

4,3

2, and 7

3, and substitute them into the

expression to find the final answer.

Question 2

Question

In a competition, there are 8 finalists, including 3 from the USA, 3 from Canada,

and 2 from Mexico. If the order of the finalists does not matter, how many ways

can the top 3 positions be filled if there must be 1 finalist from each country?

Solution

To solve this problem, we can use the concepts of combinations and permuta-

tions.

Step 1: Select one finalist from each country There are 3 ways to

choose the finalist from the USA, 3 ways to choose the finalist from Canada,

and 2 ways to choose the finalist from Mexico. By the multiplication principle,

there are 3 ×3×2 = 18 ways to select one finalist from each country.

Step 2: Arrange the selected finalists Since the order of the finalists

does not matter, we are interested in combinations rather than permutations.

We want to select 3 finalists from the 8 total finalists, so we have 8

3ways of

arranging the 3 selected finalists.

Step 3: Calculate the total number of ways By multiplying the number

of ways from Step 1 and Step 2, we get the total number of ways to fill the top

3 positions:

18 ×8

3= 18 ×8!

3!(8 −3)! = 18 ×8×7×6

3×2×1= 18 ×56 = 1008.

Therefore, there are 1008 ways to fill the top 3 positions with one finalist

from each country.

Question 3

Question

A committee of 5 members is to be formed from 10 men and 5 women. If the

committee must contain at least 2 women, how many different committees can

be formed?