MATH 350 - DISCRETE MATHEMATICS Permutations and combinations Question Bank Set 2

MATH 350 - DISCRETE

MATHEMATICS - Permutations and

combinations

Question Bank - Set 2

Liberty University

Question 1

Question

How many 6-letter words can be formed using the letters ABBESS if each letter

can be used only once?

Solution

Step 1: Identify the number of letters available.

In this case, we have 6 different letters: A, B, B, E, S, and S.

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

Since the two Bs and two Ss are indistinguishable, we must first calculate the to-

tal number of arrangements and then divide by the permutations of the repeated

letters.

The total number of arrangements of 6 different letters is 6! (6 factorial).

Step 3: Find the permutations of the repeated letters.

The letter B appears twice, and the letter S appears twice. We need to divide

out the arrangements of the repeated letters.

The number of arrangements for the repeated letter B is 2! (2 factorial) and

for the repeated letter S is also 2!.

Step 4: Calculate the total number of unique arrangements.

The total number of unique arrangements is 6!

2! ·2!

Step 5: Perform the calculations.

6! = 720

2! = 2

720

2·2=720

4= 180

Therefore, there are 180 6-letter words that can be formed using the letters

ABBESS where each letter can be used only once.

Question 2

Question

In a group of 10 students, how many ways can a committee of 4 students be

formed if a particular student, Alice, must be included?

Solution

Step 1: Select the particular student (Alice) who must be included in the com-

mittee. There is only 1 way to select Alice.

Step 2: Select the remaining 3 students to complete the committee. Since

we have 9 students left after Alice is selected, we need to choose 3 students from

the remaining 9. This can be done in 9

3ways.

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

number of ways. Total number of ways = (Number of ways in Step 1) ×(Number

of ways in Step 2)

Total number of ways = 1 ×9

3

Total number of ways = 1 ×9!

3!6!

Total number of ways = 9×8×7

3×2×1

Total number of ways = 84

Therefore, there are 84 ways to form a committee of 4 students with Alice

in it.

Question 3

Question

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

many ways can the committee be formed if the committee must include at least

one male student and at least one female student?

Solution

To find the number of ways the committee can be formed, we need to consider

the different cases separately.

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

5 students Since the order in which the students are selected does not matter,

this is a combination problem. We can use the formula for combinations:

Number of ways = n

r=n!

r!(n−r)!

where nis the total number of students (10) and ris the number of students

in the committee (5).

10

5=10!

5!(10 −5)! =10!

5!5! =10 ×9×8×7×6

5×4×3×2×1= 252

So, there are 252 ways to form a committee of 5 students from a group of

10 students.

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

at least one male and at least one female student

Let’s consider the different scenarios:

1. Committee with 1 male and 4 female students 2. Committee with 2

male and 3 female students 3. Committee with 3 male and 2 female students 4.

Committee with 4 male and 1 female student

For each scenario, we will calculate the number of ways to select the students

and then sum those up to find the total number of ways.



Scenario 1: Committee with 1 male and 4 female students

5

1×5

4= 5 ×5 = 25



Scenario 2: Committee with 2 male and 3 female students

5

2×5

3= 10 ×10 = 100



Scenario 3: Committee with 3 male and 2 female students

5

3×5

2= 10 ×10 = 100



Scenario 4: Committee with 4 male and 1 female student

5

4×5

1= 5 ×5 = 25

Adding up the possibilities from each scenario gives us the total number

of ways to form a committee with at least one male and at least one female

student:

25 + 100 + 100 + 25 = 250

Therefore, there are 250 ways to form a committee of 5 students from a

group of 10 students, ensuring that the committee includes at least one male

and at least one female student.

Question 4

Question

You are organizing a bookshelf with 10 history books, 5 science books, and 3

math books. In how many ways can you arrange the books on the shelf if books

of the same subject must be kept together?

Solution

Step 1: First, let’s calculate the number of ways the history books can be

arranged among themselves. There are 10 history books, which can be arranged

among themselves in 10! ways.

Step 2: Next, let’s calculate the number of ways the science books can be

arranged among themselves. There are 5 science books, which can be arranged

among themselves in 5! ways.

Step 3: Now, let’s calculate the number of ways the math books can be

arranged among themselves. There are 3 math books, which can be arranged

among themselves in 3! ways.

Step 4: Since the books of each subject must be kept together, we treat each

set of subject books as a single entity. So, we have 3 entities: history books,

science books, and math books. These entities can be arranged in 3! ways.

Step 5: Combining all the permutations calculated in the previous steps, the

total number of ways the books can be arranged on the shelf is:

10! ×5! ×3! ×3!

Step 6: Simplifying, we get:

10!×5!×3!×3! = 10×9×8×7×6×5×4×3×2×1×5×4×3×2×1×3×2×1×3×2×1

Step 7: Finally, calculating the result:

10×9×8×7×6×5×4×3×2×1×5×4×3×2×1×3×2×1×3×2×1 = 3628800

Therefore, the books can be arranged in 3,628,800 ways on the shelf.

Question 5

Question

Consider a group of 10 people, consisting of 4 men and 6 women. If a committee

of 3 people is to be formed, 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 select a committee of 3 people from a

group of 10. Given that there are 10 people in total, to select a committee of 3

people, we calculate the number of ways that can be done using combinations.

Total number of ways to select a committee of 3 people from 10 = 10

3=10!

3!7! = 120

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

woman. There are 4 men and 6 women in the group. To select 2 men from

4, there are 4

2= 6 ways. To select 1 woman from 6, there are 6

1= 6 ways.

Therefore, the number of ways to select a committee with 2 men and 1 woman

is 6 ∗6 = 36.

Step 3: Calculate the probability of selecting a committee with 2 men and 1

woman. The probability is given by the number of favorable outcomes divided

by the total number of outcomes.

Probability = Number of ways to select a committee with 2 men and 1 woman

Total number of ways to select a committee of 3 people from 10

Probability = 36

120 =3

10 = 0.3

Therefore, the probability of forming a committee with 2 men and 1 woman is

0.3.

Question 6

Question

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

the committee must consist of at least 3 women, how many different committees

can be formed?

Solution

Step 1: Calculate the number of committees with exactly 3 women. Choose 3

women from 10: 10

3Choose 2 men from 8: 8

2Total number of committees

with exactly 3 women: 10

3×8

2

Step 2: Calculate the number of committees with exactly 4 women. Choose

4 women from 10: 10

4Choose 1 man from 8: 8

1Total number of committees

with exactly 4 women: 10

4×8

1

Step 3: Calculate the number of committees with exactly 5 women. Choose

5 women from 10: 10

5Total number of committees with exactly 5 women: 10

5

Step 4: Sum up the results from Steps 1, 2, and 3 to find the total number

of committees. Total number of committees = 10

3×8

2+10

4×8

1+10

5

Perform the calculations to find the final answer.

Question 7

Question

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

If at least 2 women must be on the committee, how many different committees

can be formed?

Solution

Step 1: Find the total number of ways to form a committee with no restrictions.

Since we are choosing a committee of 5 people from 8 men and 6 women, we

can calculate the total number of ways without any restrictions using combina-

tions.

Total number of ways = 14

5

Step 2: Find the number of ways to form a committee with less than 2

women.

If less than 2 women are on the committee, that means either 0 women or

1 woman is on the committee. We will calculate the number of ways for each

case separately.

For the case of 0 women:

Number of ways to choose 0 women = 6

0×8

5

For the case of 1 woman:

Number of ways to choose 1 woman = 6

1×8

4

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

Subtracting the total number of ways with less than 2 women from the total

number of ways without restrictions will give us the number of ways to form a

committee with at least 2 women.

Number of ways to form a committee with at least 2 women = Total number of ways−(Number of ways with 0 women + Number of ways with 1 woman)

=14

5−6

0×8

5+6

1×8

4

Now, compute the final answer to find the number of different committees

that can be formed.

Question 8

Question

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

if exactly 2 of the members refuse to work together on the same committee?

Solution

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

from 12 people. The total number of ways to choose a committee of 5 members

from 12 people is given by 12

5=12!

5!(12−5)! = 792.

Step 2: Calculate the number of ways to choose a committee when the 2

members refuse to work together. Since the 2 members refuse to work together,

we have two cases to consider: Case 1: Choose 1 member from the 2 who

refuse to work together and 4 members from the remaining 10. The number

of ways to choose this committee is 2

1×10

4= 2 ×10!

4!(10−4)! = 420. Case 2:

Exclude both members who refuse to work together and choose all 5 members

from the remaining 10. The number of ways to choose this committee is 10

5=

10!

5!(10−5)! = 252.

Step 3: Calculate the total number of ways to choose a committee of 5

members if exactly 2 refuse to work together. The total number of ways is the

sum of the two cases calculated in Step 2: 420 + 252 = 672.

Therefore, there are 672 ways to choose a committee of 5 members from a

group of 12 people if exactly 2 of the members refuse to work together.

Question 9

Question

In a group of 10 students, how many ways are there to choose a committee of 5

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

together?

Solution

Step 1: Firstly, we find the total number of ways to choose a committee of 5

students from 10 without any restrictions. This is given by the combination

formula n

k=n!

k!(n−k)! .

Total ways = 10

5=10!

5!5! = 252

Step 2: Next, we find the number of ways that Alice and Bob are both on

the committee. This means that we are choosing 3 students from the remaining

8 students. We use the combination formula again.

Ways Alice and Bob are on committee together = 8

3=8!

3!5! = 56

Step 3: Then, we find the number of ways that Alice and Bob are both

not on the committee. This involves choosing 5 students from the remaining 8

students (excluding Alice and Bob).

Ways Alice and Bob are not on committee together = 8

5=8!

5!3! = 56

Step 4: Finally, we find the total number of ways to choose a committee of

5 students with the given restriction.

Number of ways = Total ways−Ways Alice and Bob are on committee together−Ways Alice and Bob are not on committee together

Number of ways = 252 −56 −56 = 140

Therefore, there are 140 ways to choose a committee of 5 students from a

group of 10 students if Alice and Bob refuse to serve on the committee together.

Question 10

Question

A committee of 5 people is to be formed from a group of 7 men and 4 women.

If at least 2 women must be in the committee, how many different committees

can be formed?

Solution

Step 1: Calculate the total number of committees that can be formed with no

restrictions. Since we need to select 5 people from a group of 7 men and 4

women, we have a total of 11 people to choose from. This can be done in 11

5

ways.

Step 2: Calculate the number of committees with only 1 woman. If there is

only 1 woman in the committee, we have 4 choices for the woman and 3 choices

for the remaining 4 members from the 7 men. So the number of committees

with only 1 woman is 4 ×7

4.

Step 3: Calculate the number of committees with no women. If there are

no women in the committee, we need to select all 5 members from the 7 men,

which can be done in 7

5ways.

Step 4: Calculate the total number of committees with at least 2 women.

The total number of committees with at least 2 women is the total number

of committees minus the number of committees with only 1 woman and the

number of committees with no women: 11

5−4×7

4−7

5

Calculating this expression gives us the total number of different committees

that can be formed with at least 2 women.

Question 11

Question

In a group of 10 people, how many ways can we select a committee of 3 people

if 2 of the people are a couple and must serve on the committee together?

Solution

Step 1: First we select the couple to be on the committee together. There is

only 1 way to select the couple.

Step 2: Next, we select the third person to join the couple on the committee.

Since we have already selected 2 people, there are 8 remaining people to choose

from. So, there are 8 ways to select the third person.

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

number of ways. Therefore, the total number of ways to select the committee

is 1 ×8 = 8 .

Question 12

Question

A committee of 5 people is to be formed from a group of 7 men and 6 women.

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

committees can be formed?

Solution

Step 1: Find the total number of ways to form a committee of 5 people. Since

we must choose either 2, 3, 4, or 5 men and the remaining members will be

women, we’ll break this step down into sub-cases.

Case 1: Choosing 2 men and 3 women There are 7

2ways to choose 2 men

from 7 and 6

3ways to choose 3 women from 6. This results in 7

2×6

3ways

to form a committee with 2 men and 3 women.

Case 2: Choosing 3 men and 2 women Similarly, there are 7

3ways to

choose 3 men from 7 and 6

2ways to choose 2 women from 6. This results in

7

3×6

2ways to form a committee with 3 men and 2 women.

Case 3: Choosing 4 men and 1 woman There are 7

4ways to choose 4 men

from 7 and 6

1ways to choose 1 woman from 6. This results in 7

4×6

1ways

to form a committee with 4 men and 1 woman.

Case 4: Choosing 5 men There is only 7

5way to choose all 5 men from 7,

as there are no women left. This results in 7

5ways to form a committee with

5 men.

Step 2: Add up the results from all the cases to find the total number

of different committees that can be formed. Total number of committees =

7

2×6

3+7

3×6

2+7

4×6

1+7

5

Calculating the values:

7

2= 21,6

3= 20,7

3= 35,6

2= 15,7

4= 35,6

1= 6,7

5= 21

Total number of committees = 21 ×20 + 35 ×15 + 35 ×6 + 21 = 420 + 525 +

210 + 21 = 1176

Therefore, there are 1176 different committees that can be formed.

Question 13

Question

A committee of 5 people is to be formed from a group of 8 students and 4

professors. If at least one student must be on the committee, how many different

committees can be formed?

Solution

Step 1: Find the total number of ways to form a committee with no restrictions.

The total number of ways to form a committee with 5 people from 8 students

and 4 professors is given by the combination formula:

8+4

5=12

5=12!

5!7! = 792

Step 2: Find the number of ways to form a committee with only professors.

The number of ways to select 5 professors from 4 professors is given by the

combination formula: 4

5= 0

Step 3: Find the number of ways to form a committee with only students.

The number of ways to select 5 students from 8 students is given by the combi-

nation formula: 8

5= 56

Step 4: Find the number of ways to form a committee with at least one

student. The total number of ways to form a committee with 5 people with

at least one student is given by the total number of ways minus the number of

ways with only professors:

792 −0 = 792

Therefore, there are 792 different committees that can be formed with at

least one student.

Question 14

Question

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

of 4 students if exactly 2 of the students, Alice and Bob, refuse to be on the

committee together?

Solution

Step 1: First, we find the total number of ways to select a committee of 4 stu-

dents from a group of 10. The total number of ways is given by the combination

formula: 10

4=10!

4!(10 −4)! =10 ×9×8×7

4×3×2×1= 210

Step 2: Next, we find the number of ways where Alice and Bob are both on

the committee. The number of ways to select the remaining 2 students from

the remaining 8 students is:

8

2=8!

2!(8 −2)! =8×7

2×1= 28

Step 3: Finally, we subtract the number of ways both Alice and Bob are on

the committee from the total number of ways.

210 −28 = 182

Therefore, there are 182 ways to select a committee of 4 students from a

group of 10 where Alice and Bob do not want to be on the committee together.

Question 15

Question

In how many ways can 5 red, 4 blue, and 3 green balls be arranged in a row

such that no two green balls are adjacent?

Solution

Step 1: Find the total number of ways to arrange all the balls without restric-

tions. In this case, we have a total of 5 + 4 + 3 = 12 balls to arrange. The total

number of ways to arrange them is 12!.

Step 2: Find the number of ways the green balls can be arranged where they

are adjacent. Since no two green balls can be adjacent, let’s treat the 3 green

balls as a single entity. This means we have 12 −3 + 1 = 10 entities to arrange:

the 3 green balls as a group and the other 9 non-green balls. The total number

of ways to arrange these 10 entities is 10!. However, within the green group,

the balls can be arranged in 3! ways. So the total number of ways to arrange

all balls with the greens adjacent is 10! ×3!.

Step 3: Find the number of ways the green balls can be arranged with the

constraint that they are not adjacent. Subtract the number of arrangements

with the green balls adjacent from the total number of arrangements:

12! −10! ×3!

Therefore, the number of ways the 5 red, 4 blue, and 3 green balls can be

arranged in a row such that no two green balls are adjacent is 12! −10! ×3!.

Question 16

Question

In how many ways can a committee of 4 people be selected from a group of

8 women and 6 men if the committee must consist of at least 2 women and 2

men?

Solution

Step 1: Calculate the number of ways to select 2 women from 8. There are 8

2

ways to select 2 women from 8.

Step 2: Calculate the number of ways to select 2 men from 6. Similarly,

there are 6

2ways to select 2 men from 6.

Step 3: Calculate the number of ways to select the remaining 1 woman and

1 man. There are 6 remaining women and 4 remaining men, so there are 6

1·4

1

ways to select the remaining 1 woman and 1 man.

Step 4: Calculate the total number of ways to select 2 women and 2 men.

Multiplying the results from Step 1, Step 2, and Step 3, we have: 8

2·6

1·

4

1= 28 ·15 ·6·4 = 16800.

Therefore, there are 16800 ways to select a committee of 4 people consisting

of at least 2 women and 2 men from a group of 8 women and 6 men.

Question 17

Question

How many ways are there to arrange the letters in the word ”MISSISSIPPI”

such that no two consecutive letters are the same?

Solution

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

SIPPI”. Let’s denote the word ”MISSISSIPPI” as M1, I4, S4, P2. The total

number of ways to arrange these letters is given by:

11!

1!4!4!2! = 34,650

Step 2: Calculate the total number of ways with consecutive letters the

same. To calculate the number of ways with consecutive letters the same, we

treat each set of consecutive identical letters as a single entity. Consider the

sets M, IIII, SSSS, PP. We treat each set as a single entity to calculate the total

number of ways:

7! = 5,040

Step 3: Calculate the total number of ways with no consecutive letters the

same. The total number of ways with no consecutive letters the same is the

difference between the total number of arrangements and the number of ar-

rangements with consecutive letters the same:

34,650 −5,040 = 29,610

Therefore, there are 29,610 ways to arrange the letters in ”MISSISSIPPI”

such that no two consecutive letters are the same.

Question 18

Question

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

committee must consist of 4 members, with at least one woman and one man.

How many different committees can be formed?

Solution

Step 1: Calculate the total number of committees that can be formed. Since

the committee must consist of 4 members, we need to find the number of ways

to choose 4 people from the total group of 18 (10 women and 8 men). This can

be calculated using combinations:

Total number of committees = 18

4

Step 2: Calculate the number of committees without any women. If the

committee has no women, then all 4 members must be men. We need to find

the number of ways to choose 4 men from the group of 8 men:

Number of committees without any women = 8

4

Step 3: Calculate the number of committees without any men. If the com-

mittee has no men, then all 4 members must be women. We need to find the

number of ways to choose 4 women from the group of 10 women:

Number of committees without any men = 10

4

Step 4: Calculate the number of committees with at least one woman and

one man. To find the number of committees with at least one woman and one

man, we can subtract the number of committees without any women or without

any men from the total number of committees:

Number of committees with at least one woman and one man = Total number of committees−Number of committees without any women−Number of committees without any men

=18

4−8

4−10

4

Question 19

Question

In how many ways can 5 mathematics books, 4 physics books, and 3 chemistry

books be arranged on a bookshelf if books on the same subject must be kept

together, but the subjects can be in any order?

Solution

Step 1: Calculate the number of ways to arrange the mathematics books. There

are 5 mathematics books, and they must be kept together. They can be arranged

among themselves in 5! ways.

Step 2: Calculate the number of ways to arrange the physics books. Simi-

larly, there are 4 physics books, and they must be kept together. They can be

arranged among themselves in 4! ways.

Step 3: Calculate the number of ways to arrange the chemistry books. There

are 3 chemistry books, and they must be kept together. They can be arranged

among themselves in 3! ways.

Step 4: Calculate the number of ways the subjects can be arranged on the

bookshelf. Now, we have 3 groups of books (math, physics, and chemistry)

which can be arranged on the bookshelf. Since the subjects can be in any order,

these 3 groups can be arranged among themselves in 3! ways.

Step 5: Multiply the number of ways from each step to find the total number

of arrangements. The total number of arrangements is given by: 5!×4!×3!×3! =

120 ×24 ×6×6 = 51840

Therefore, there are 51,840 ways to arrange the books on the bookshelf.

Question 20

Question

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

of the students are math majors and the other 7 are computer science majors,

determine the number of ways to form the committee if at least one math major

must be included.

Solution

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

out of 10. Step 2: Determine the number of ways to form a committee with no

math majors. Step 3: Subtract the result from Step 2 from the total to find the

number of ways to form a committee with at least one math major.

Step 1: The total number of ways to form a committee of 5 students out of

10 is given by the combination formula:

10

5=10!

5!(10 −5)! = 252.

Step 2: To find the number of ways to form a committee with no math

majors, we need to choose 5 students from the 7 computer science majors:

7

5=7!

5!(7 −5)! = 21.

Step 3: To find the number of ways to form a committee with at least one

math major, we subtract the result from Step 2 from the total:

252 −21 = 231.

Therefore, there are 231 ways to form a committee of 5 students from a

group of 10 students with at least one math major.

Question 21

Question

In how many ways can 5 different math books, 4 different history books, and

3 different science books be arranged on a bookshelf if the books within each

subject must remain together?

Solution

Step 1: Calculate the number of ways to arrange the math books. Since there

are 5 different math books, the number of ways to arrange them is 5!.

Step 2: Calculate the number of ways to arrange the history books. Since

there are 4 different history books, the number of ways to arrange them is 4!.

Step 3: Calculate the number of ways to arrange the science books. Since

there are 3 different science books, the number of ways to arrange them is 3!.

Step 4: Calculate the total number of ways to arrange all the books. To find

the total number of ways to arrange all the books, we multiply the number of

ways to arrange the books within each subject together. Thus, the total number

of ways is 5! ×4! ×3!. Therefore, the total number of ways to arrange the books

is 5! ×4! ×3! = 1440.

Question 22

Question

In how many ways can a committee of 5 people be formed from a group of 10

women and 8 men if the committee must have at least 2 women and at least 2

men?

Solution

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

and 3 men.

From a group of 10 women, the number of ways to choose 2 women is denoted

by 10

2.

From a group of 8 men, the number of ways to choose 3 men is denoted by

8

3.

Therefore, the number of ways to form a committee with exactly 2 women

and 3 men is: 10

2×8

3

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

and 2 men.

From a group of 10 women, the number of ways to choose 3 women is denoted

by 10

3.

From a group of 8 men, the number of ways to choose 2 men is denoted by

8

2.

Therefore, the number of ways to form a committee with 3 women and 2

men is: 10

3×8

2

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

least 2 women and at least 2 men.

Add the results from Step 1 and Step 2:

10

2×8

3+10

3×8

2

Calculating the values:

10

2= 45,8

3= 56,10

3= 120,8

2= 28

Therefore, the total number of ways to form a committee with at least 2

women and at least 2 men is:

45 ×56 + 120 ×28 = 2520 + 3360 = 5880

Question 23

Question

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

If the committee is to consist of 3 women and 2 men, how many different ways

can the committee be formed?

Solution

Step 1: Calculate the number of ways to choose 3 women from 10. Step 2: Cal-

culate the number of ways to choose 2 men from 5. Step 3: Multiply the results

from steps 1 and 2 to find the total number of ways to form the committee.

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

combination formula:

10

3=10!

3!(10 −3)! =10!

3!7! =10 ×9×8

3×2×1= 120

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

bination formula:

5

2=5!

2!(5 −2)! =5!

2!3! =5×4

2×1= 10

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

ways to form the committee:

120 ×10 = 1200

Therefore, there are 1200 different ways to form the committee consisting of

3 women and 2 men.

Question 24

Question

In how many ways can 5 identical red roses, 4 identical yellow roses, and 3

identical white roses be distributed among 4 different vases if each vase must

contain at least one rose of each color?

Solution

Step 1: Calculate the total number of ways to distribute the roses without re-

strictions. Let nbe the total number of ways to distribute the roses without

restrictions. We can use the stars and bars method to represent this distribu-

tion. There are 12 roses in total, and 4 vases, so we need to place 3 dividers

(representing the vases) among the 12 roses. Thus, n=12+4−1

4−1=15

3= 455.

Step 2: Calculate the number of ways to distribute the roses such that at

least one rose of each color is in each vase. Let mbe the number of ways to

distribute the roses such that at least one rose of each color is in each vase. To

ensure each vase has at least one rose of each color, we can first distribute one

red, one yellow, and one white rose to each vase. This leaves us with 2 red roses,

1 yellow rose, and 1 white rose left to distribute among the vases.

Step 3: Distribute the remaining roses. Since the red roses are identical,

there is only 1 way to distribute the 2 red roses among the 4 vases. Similarly,

there is only 1 way to distribute the remaining yellow and white roses.

Step 4: Calculate the total number of ways. Therefore, the total number of

ways to distribute the roses such that each vase has at least one rose of each

color is m= 1 ×1×1 = 1.

Step 5: Calculate the final answer. The number of ways to distribute the

roses among the vases with the given restrictions is m= 1 out of n= 455 total

ways. Therefore, the final answer is 1 way to distribute the roses among the

vases with the given restrictions.

Question 25

Question

A student club has 12 members, and they need to choose a president, vice

president, and treasurer from among the members. How many different ways

can they fill these positions if no person can hold more than one position?

Solution

Let’s solve this problem by using the concepts of permutations and combina-

tions.

Step 1: To choose the president, we have 12 choices. There are 12 possible

candidates for the president position.

Step 2: After the president is chosen, there are 11 remaining members to

choose from for the vice president position. There are 11 candidates left for the

vice president position.

Step 3: Finally, after the president and vice president are chosen, there are

10 remaining members to choose from for the treasurer position. There are 10

candidates left for the treasurer position.

Step 4: To find the total number of ways to fill the three positions, we

multiply the number of choices at each step. Total number of ways = 12×11×10

Therefore, the total number of ways the club can fill the positions is 12 ×

11 ×10 = 1320.

Question 26

Question

In a mathematics competition, there are 10 multiple-choice questions. Each

question has 4 answer choices: A, B, C, and D. How many ways can a contestant

answer the 10 questions such that exactly 6 of the answers are correct and the

remaining 4 are incorrect?

Solution

Step 1: To solve this problem, we need to consider the number of ways the

contestant can answer 6 questions correctly and 4 questions incorrectly.

Step 2: There are 10

6ways to choose which 6 questions the contestant

answers correctly.

Step 3: For each of the 6 correct questions, there is 1 way to answer correctly

and 3 ways to answer incorrectly. Similarly, for each of the 4 incorrect questions,

there are 3 ways to answer incorrectly.

Step 4: Therefore, the total number of ways a contestant can answer the

questions as required is:

10

6×16×34= 210 ×81 = 17010

So, there are 17,010 ways for a contestant to answer the 10 questions such

that exactly 6 of the answers are correct and the remaining 4 are incorrect.

Question 27

Question

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

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

can be formed?

Solution

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

and 2 men.

There are 6

3ways to choose 3 women from the 6 available women. Similarly,

there are 8

2ways to choose 2 men from the 8 available men.

Therefore, the number of ways to form a committee with exactly 3 women

and 2 men is: 6

3×8

2

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

and 1 man.

There are 6

4ways to choose 4 women from the 6 available women. Similarly,

there are 8

1ways to choose 1 man from the 8 available men.

Therefore, the number of ways to form a committee with exactly 4 women

and 1 man is: 6

4×8

1

Step 3: Find the number of ways to form a committee with all 5 members

being women.

There is only 1 way to choose all 5 members from the 6 available women.

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

number of different committees that can be formed.

The total number of different committees that can be formed is:

6

3×8

2+6

4×8

1+ 1

Question 28

Question

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

if two particular people refuse to be on the committee together?

Solution

To solve this problem, we can consider two cases: when the two particular people

are both on the committee, and when they are not both on the committee.

Case 1: Both particular people are on the committee

Step 1: Choose the two particular people who will be on the committee.

There are 2

2= 1 way to choose them.

Step 2: Choose the remaining 2 people from the remaining 8 people. There

are 8

2ways to do this.

Case 2: Both particular people are not on the committee

Step 3: Choose the committee of 4 people from the remaining 8 people

(excluding the two particular people). There are 8

4ways to do this.

Step 4: Add the results from Case 1 and Case 2 to find the total number

of ways to choose the committee.

Therefore, the total number of ways to choose a committee of 4 people from

a group of 10 people where two particular people refuse to be on the committee

together is

2

2×8

2+8

4.

Question 29

Question

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

if there are 3 specific people that must be on the committee?

Solution

Step 1: Since there are 3 specific people that must be on the committee, we

need to choose the remaining 1 person from the remaining 7 people. Step 2:

Calculate the number of ways to choose the 1 person from 7:

7

1= 7

Step 3: Therefore, the total number of ways to form the committee is the

product of the number of ways to choose the 3 specific people and the 1 person

from 7:

3

3·7

1= 1 ·7=7

Step 4: So, there are 7 ways to choose a committee of 4 people with 3

specific people on it from a group of 10 people.

Question 30

Question

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

professors. If at least 2 professors must be in the committee, how many different

committees can be formed?

Solution

Step 1: Calculate the number of ways to choose 2 professors from 5. Step 2:

Calculate the number of ways to choose 3 other members from the remaining

10 students and 3 professors. 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: There are 5

2ways to choose 2 professors out of 5.

5

2=5!

2!(5 −2)! =5×4

2×1= 10

Step 2: Since at least 2 professors must be in the committee, we have 3

spots left. We can choose those 3 spots from the remaining 3 professors and 10

students. There are 8

3ways to choose 3 members out of 8.

8

3=8!

3!(8 −3)! =8×7×6

3×2×1= 56

Step 3: To find the total number of different committees that can be formed,

we multiply the results from Step 1 and Step 2. Total number of committees =

10 ×56 = 560

Therefore, there are 560 different committees that can be formed when at

least 2 professors must be in the committee.

Question 31

Question

A committee of 4 people is to be formed from a group of 8 women and 6 men.

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

committees can be formed?

Solution

Step 1: Calculate the number of ways to choose 2 women out of 8. Step 2:

Calculate the number of ways to choose 2 men out of 6. 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: Choosing 2 women out of 8 can be done using a combination,

denoted as n

k=n!

k!(n−k)! .

The number of ways to choose 2 women out of 8 is 8

2=8!

2!(8−2)! =8×7

2×1= 28.

Step 2: Similarly, choosing 2 men out of 6 can be done using a combination.

The number of ways to choose 2 men out of 6 is 6

2=6!

2!(6−2)! =6×5

2×1= 15.

Step 3: To find the total number of different committees that can be formed

with 2 women and 2 men, multiply the results from Step 1 and Step 2.

Total number of different committees = 28 ×15 = 420.

Therefore, there are 420 different committees that can be formed with 2

women and 2 men from the group of 8 women and 6 men.

Question 32

Question

A committee of 5 people is to be formed from a group of 8 students and 4

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

of at least 2 professors?

Solution

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

students for the committee. Since there are 4 professors and 8 students available,

the number of ways to choose 2 professors from 4 is 4

2and the number of ways

to choose 3 students from 8 is 8

3. Therefore, the number of ways to choose

exactly 2 professors and 3 students is 4

2×8

3.

Step 2: Calculate the number of ways to choose exactly 3 professors and 2

students for the committee. Similarly, the number of ways to choose 3 profes-

sors from 4 is 4

3and the number of ways to choose 2 students from 8 is 8

2.

Therefore, the number of ways to choose exactly 3 professors and 2 students is

4

3×8

2.

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

least 2 professors. To find the total number of ways to form the committee with

at least 2 professors, we sum the results from Step 1 and Step 2: Total number

of ways = 4

2×8

3+4

3×8

2.

Question 33

Question

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

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

committees can be formed?

Solution

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

Let’s choose 2 women out of 10 and 2 men out of 5.

There are 10

2= 45 ways to choose 2 women and 5

2= 10 ways to choose 2

men.

Therefore, there are 45 ×10 = 450 ways to choose 2 women and 2 men.

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

Let’s choose 3 women out of 10 and 2 men out of 5.

There are 10

3= 120 ways to choose 3 women and 5

2= 10 ways to choose

2 men.

Therefore, there are 120 ×10 = 1200 ways to choose 3 women and 2 men.

Step 3: Calculate the number of committees with 4 women and 1 man.

Let’s choose 4 women out of 10 and 1 man out of 5.

There are 10

4= 210 ways to choose 4 women and 5

1= 5 ways to choose 1

man.

Therefore, there are 210 ×5 = 1050 ways to choose 4 women and 1 man.

Step 4: Add the number of committees with 2 women and 2 men, 3 women

and 2 men, and 4 women and 1 man together to find the total number of

committees that can be formed.

Total number of committees = 450 + 1200 + 1050 = 2700.

Therefore, there are 2700 different committees that can be formed.

Question 34

Question

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

5 students where 2 of them must be assigned specific roles (president and vice

president)?

Solution

Step 1: Calculate the number of ways to choose the president and vice president.

Since there are 10 students and 2 specific roles, the number of ways to choose

the president and vice president is 10 ×9 = 90.

Step 2: Calculate the number of ways to choose the remaining 3 committee

members. After the president and vice president are chosen, there are 8 students

remaining to choose from for the 3 remaining committee positions. The number

of ways to choose these 3 members is 8 ×7×6 = 336.

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

of ways to form the committee. The total number of ways to form the committee

is 90 ×336 = 30240.

Therefore, there are 30,240 ways to form a committee of 5 students where 2

of them are assigned specific roles (president and vice president) from a group

of 10 students.

Question 35

Question

A committee of 5 people is to be formed from a group of 12 students. If 3

of the students are girls and the remaining are boys, how many ways can the

committee be formed if it must include exactly 2 girls?

Solution

Step 1: Calculate the number of ways to choose 2 girls from the 3 available.

Step 2: Calculate the number of ways to choose 3 boys from the 9 available.

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

the committee can be formed.

Step 1: To choose 2 girls from 3, we use a combination:

3

2=3!

2!(3 −2)! = 3.

So, there are 3 ways to choose 2 girls.

Question 2

Question

In a group of 10 students, how many ways can a committee of 4 students be

formed if a particular student, Alice, must be included?

Solution

Step 1: Select the particular student (Alice) who must be included in the com-

mittee. There is only 1 way to select Alice.

Step 2: Select the remaining 3 students to complete the committee. Since

we have 9 students left after Alice is selected, we need to choose 3 students from

the remaining 9. This can be done in 9

3ways.

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

number of ways. Total number of ways = (Number of ways in Step 1) ×(Number

of ways in Step 2)

Total number of ways = 1 ×9

3

Total number of ways = 1 ×9!

3!6!

Total number of ways = 9×8×7

3×2×1

Total number of ways = 84

Therefore, there are 84 ways to form a committee of 4 students with Alice

in it.

Question 3

Question

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

many ways can the committee be formed if the committee must include at least

one male student and at least one female student?

Solution

To find the number of ways the committee can be formed, we need to consider

the different cases separately.

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

5 students Since the order in which the students are selected does not matter,

this is a combination problem. We can use the formula for combinations:

Number of ways = n

r=n!

r!(n−r)!

where nis the total number of students (10) and ris the number of students

in the committee (5).

10

5=10!

5!(10 −5)! =10!

5!5! =10 ×9×8×7×6

5×4×3×2×1= 252

So, there are 252 ways to form a committee of 5 students from a group of

10 students.

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

at least one male and at least one female student

Let’s consider the different scenarios:

1. Committee with 1 male and 4 female students 2. Committee with 2

male and 3 female students 3. Committee with 3 male and 2 female students 4.

Committee with 4 male and 1 female student

For each scenario, we will calculate the number of ways to select the students

and then sum those up to find the total number of ways.



Scenario 1: Committee with 1 male and 4 female students

5

1×5

4= 5 ×5 = 25



Scenario 2: Committee with 2 male and 3 female students

5

2×5

3= 10 ×10 = 100



Scenario 3: Committee with 3 male and 2 female students

5

3×5

2= 10 ×10 = 100



Scenario 4: Committee with 4 male and 1 female student

5

4×5

1= 5 ×5 = 25

Adding up the possibilities from each scenario gives us the total number

of ways to form a committee with at least one male and at least one female

student:

25 + 100 + 100 + 25 = 250

Therefore, there are 250 ways to form a committee of 5 students from a

group of 10 students, ensuring that the committee includes at least one male

and at least one female student.

Question 4

Question

You are organizing a bookshelf with 10 history books, 5 science books, and 3

math books. In how many ways can you arrange the books on the shelf if books

of the same subject must be kept together?

Solution

Step 1: First, let’s calculate the number of ways the history books can be

arranged among themselves. There are 10 history books, which can be arranged

among themselves in 10! ways.

Step 2: Next, let’s calculate the number of ways the science books can be

arranged among themselves. There are 5 science books, which can be arranged

among themselves in 5! ways.

Step 3: Now, let’s calculate the number of ways the math books can be

arranged among themselves. There are 3 math books, which can be arranged

among themselves in 3! ways.

Step 4: Since the books of each subject must be kept together, we treat each

set of subject books as a single entity. So, we have 3 entities: history books,

science books, and math books. These entities can be arranged in 3! ways.

Step 5: Combining all the permutations calculated in the previous steps, the

total number of ways the books can be arranged on the shelf is:

10! ×5! ×3! ×3!

Step 6: Simplifying, we get:

10!×5!×3!×3! = 10×9×8×7×6×5×4×3×2×1×5×4×3×2×1×3×2×1×3×2×1

Step 7: Finally, calculating the result:

10×9×8×7×6×5×4×3×2×1×5×4×3×2×1×3×2×1×3×2×1 = 3628800

Therefore, the books can be arranged in 3,628,800 ways on the shelf.

Question 5

Question

Consider a group of 10 people, consisting of 4 men and 6 women. If a committee

of 3 people is to be formed, 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 select a committee of 3 people from a

group of 10. Given that there are 10 people in total, to select a committee of 3

people, we calculate the number of ways that can be done using combinations.

Total number of ways to select a committee of 3 people from 10 = 10

3=10!

3!7! = 120

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

woman. There are 4 men and 6 women in the group. To select 2 men from

4, there are 4

2= 6 ways. To select 1 woman from 6, there are 6

1= 6 ways.

Therefore, the number of ways to select a committee with 2 men and 1 woman

is 6 ∗6 = 36.

Step 3: Calculate the probability of selecting a committee with 2 men and 1

woman. The probability is given by the number of favorable outcomes divided

by the total number of outcomes.

Probability = Number of ways to select a committee with 2 men and 1 woman

Total number of ways to select a committee of 3 people from 10

Probability = 36

120 =3

10 = 0.3

Therefore, the probability of forming a committee with 2 men and 1 woman is

0.3.

Question 6

Question

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

the committee must consist of at least 3 women, how many different committees

can be formed?

Solution

Step 1: Calculate the number of committees with exactly 3 women. Choose 3

women from 10: 10

3Choose 2 men from 8: 8

2Total number of committees

with exactly 3 women: 10

3×8

2

Step 2: Calculate the number of committees with exactly 4 women. Choose

4 women from 10: 10

4Choose 1 man from 8: 8

1Total number of committees

with exactly 4 women: 10

4×8

1

Step 3: Calculate the number of committees with exactly 5 women. Choose

5 women from 10: 10

5Total number of committees with exactly 5 women: 10

5

Step 4: Sum up the results from Steps 1, 2, and 3 to find the total number

of committees. Total number of committees = 10

3×8

2+10

4×8

1+10

5

Perform the calculations to find the final answer.

Question 7

Question

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

If at least 2 women must be on the committee, how many different committees

can be formed?

Solution

Step 1: Find the total number of ways to form a committee with no restrictions.

Since we are choosing a committee of 5 people from 8 men and 6 women, we

can calculate the total number of ways without any restrictions using combina-

tions.

Total number of ways = 14

5

Step 2: Find the number of ways to form a committee with less than 2

women.

If less than 2 women are on the committee, that means either 0 women or

1 woman is on the committee. We will calculate the number of ways for each

case separately.

For the case of 0 women:

Number of ways to choose 0 women = 6

0×8

5

For the case of 1 woman:

Number of ways to choose 1 woman = 6

1×8

4

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

Subtracting the total number of ways with less than 2 women from the total

number of ways without restrictions will give us the number of ways to form a

committee with at least 2 women.

Number of ways to form a committee with at least 2 women = Total number of ways−(Number of ways with 0 women + Number of ways with 1 woman)

=14

5−6

0×8

5+6

1×8

4

Now, compute the final answer to find the number of different committees

that can be formed.

Question 8

Question

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

if exactly 2 of the members refuse to work together on the same committee?

Solution

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

from 12 people. The total number of ways to choose a committee of 5 members

from 12 people is given by 12

5=12!

5!(12−5)! = 792.

Step 2: Calculate the number of ways to choose a committee when the 2

members refuse to work together. Since the 2 members refuse to work together,

we have two cases to consider: Case 1: Choose 1 member from the 2 who

refuse to work together and 4 members from the remaining 10. The number

of ways to choose this committee is 2

1×10

4= 2 ×10!

4!(10−4)! = 420. Case 2:

Exclude both members who refuse to work together and choose all 5 members

from the remaining 10. The number of ways to choose this committee is 10

5=

10!

5!(10−5)! = 252.

Step 3: Calculate the total number of ways to choose a committee of 5

members if exactly 2 refuse to work together. The total number of ways is the

sum of the two cases calculated in Step 2: 420 + 252 = 672.

Therefore, there are 672 ways to choose a committee of 5 members from a

group of 12 people if exactly 2 of the members refuse to work together.

Question 9

Question

In a group of 10 students, how many ways are there to choose a committee of 5

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

together?

Solution

Step 1: Firstly, we find the total number of ways to choose a committee of 5

students from 10 without any restrictions. This is given by the combination

formula n

k=n!

k!(n−k)! .

Total ways = 10

5=10!

5!5! = 252

Step 2: Next, we find the number of ways that Alice and Bob are both on

the committee. This means that we are choosing 3 students from the remaining

8 students. We use the combination formula again.

Ways Alice and Bob are on committee together = 8

3=8!

3!5! = 56

Step 3: Then, we find the number of ways that Alice and Bob are both

not on the committee. This involves choosing 5 students from the remaining 8

students (excluding Alice and Bob).

Ways Alice and Bob are not on committee together = 8

5=8!

5!3! = 56

Step 4: Finally, we find the total number of ways to choose a committee of

5 students with the given restriction.

Number of ways = Total ways−Ways Alice and Bob are on committee together−Ways Alice and Bob are not on committee together

Number of ways = 252 −56 −56 = 140

Therefore, there are 140 ways to choose a committee of 5 students from a

group of 10 students if Alice and Bob refuse to serve on the committee together.

Question 10

Question

A committee of 5 people is to be formed from a group of 7 men and 4 women.

If at least 2 women must be in the committee, how many different committees

can be formed?

Solution

Step 1: Calculate the total number of committees that can be formed with no

restrictions. Since we need to select 5 people from a group of 7 men and 4

women, we have a total of 11 people to choose from. This can be done in 11

5

ways.

Step 2: Calculate the number of committees with only 1 woman. If there is

only 1 woman in the committee, we have 4 choices for the woman and 3 choices

for the remaining 4 members from the 7 men. So the number of committees

with only 1 woman is 4 ×7

4.

Step 3: Calculate the number of committees with no women. If there are

no women in the committee, we need to select all 5 members from the 7 men,

which can be done in 7

5ways.

Step 4: Calculate the total number of committees with at least 2 women.

The total number of committees with at least 2 women is the total number

of committees minus the number of committees with only 1 woman and the

number of committees with no women: 11

5−4×7

4−7

5

Calculating this expression gives us the total number of different committees

that can be formed with at least 2 women.

Question 11

Question

In a group of 10 people, how many ways can we select a committee of 3 people

if 2 of the people are a couple and must serve on the committee together?

Solution

Step 1: First we select the couple to be on the committee together. There is

only 1 way to select the couple.

Step 2: Next, we select the third person to join the couple on the committee.

Since we have already selected 2 people, there are 8 remaining people to choose

from. So, there are 8 ways to select the third person.

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

number of ways. Therefore, the total number of ways to select the committee

is 1 ×8 = 8 .

Question 12

Question

A committee of 5 people is to be formed from a group of 7 men and 6 women.

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

committees can be formed?

Solution

Step 1: Find the total number of ways to form a committee of 5 people. Since

we must choose either 2, 3, 4, or 5 men and the remaining members will be

women, we’ll break this step down into sub-cases.

Case 1: Choosing 2 men and 3 women There are 7

2ways to choose 2 men

from 7 and 6

3ways to choose 3 women from 6. This results in 7

2×6

3ways

to form a committee with 2 men and 3 women.

Case 2: Choosing 3 men and 2 women Similarly, there are 7

3ways to

choose 3 men from 7 and 6

2ways to choose 2 women from 6. This results in

7

3×6

2ways to form a committee with 3 men and 2 women.

Case 3: Choosing 4 men and 1 woman There are 7

4ways to choose 4 men

from 7 and 6

1ways to choose 1 woman from 6. This results in 7

4×6

1ways

to form a committee with 4 men and 1 woman.

Case 4: Choosing 5 men There is only 7

5way to choose all 5 men from 7,

as there are no women left. This results in 7

5ways to form a committee with

5 men.

Step 2: Add up the results from all the cases to find the total number

of different committees that can be formed. Total number of committees =

7

2×6

3+7

3×6

2+7

4×6

1+7

5

Calculating the values:

7

2= 21,6

3= 20,7

3= 35,6

2= 15,7

4= 35,6

1= 6,7

5= 21

Total number of committees = 21 ×20 + 35 ×15 + 35 ×6 + 21 = 420 + 525 +

210 + 21 = 1176

Therefore, there are 1176 different committees that can be formed.

Question 13

Question

A committee of 5 people is to be formed from a group of 8 students and 4

professors. If at least one student must be on the committee, how many different

committees can be formed?

Solution

Step 1: Find the total number of ways to form a committee with no restrictions.

The total number of ways to form a committee with 5 people from 8 students

and 4 professors is given by the combination formula:

8+4

5=12

5=12!

5!7! = 792

Step 2: Find the number of ways to form a committee with only professors.

The number of ways to select 5 professors from 4 professors is given by the

combination formula: 4

5= 0

Step 3: Find the number of ways to form a committee with only students.

The number of ways to select 5 students from 8 students is given by the combi-

nation formula: 8

5= 56

Step 4: Find the number of ways to form a committee with at least one

student. The total number of ways to form a committee with 5 people with

at least one student is given by the total number of ways minus the number of

ways with only professors:

792 −0 = 792

Therefore, there are 792 different committees that can be formed with at

least one student.

Question 14

Question

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

of 4 students if exactly 2 of the students, Alice and Bob, refuse to be on the

committee together?

Solution

Step 1: First, we find the total number of ways to select a committee of 4 stu-

dents from a group of 10. The total number of ways is given by the combination

formula: 10

4=10!

4!(10 −4)! =10 ×9×8×7

4×3×2×1= 210

Step 2: Next, we find the number of ways where Alice and Bob are both on

the committee. The number of ways to select the remaining 2 students from

the remaining 8 students is:

8

2=8!

2!(8 −2)! =8×7

2×1= 28

Step 3: Finally, we subtract the number of ways both Alice and Bob are on

the committee from the total number of ways.

210 −28 = 182

Therefore, there are 182 ways to select a committee of 4 students from a

group of 10 where Alice and Bob do not want to be on the committee together.

Question 15

Question

In how many ways can 5 red, 4 blue, and 3 green balls be arranged in a row

such that no two green balls are adjacent?

Solution

Step 1: Find the total number of ways to arrange all the balls without restric-

tions. In this case, we have a total of 5 + 4 + 3 = 12 balls to arrange. The total

number of ways to arrange them is 12!.

Step 2: Find the number of ways the green balls can be arranged where they

are adjacent. Since no two green balls can be adjacent, let’s treat the 3 green

balls as a single entity. This means we have 12 −3 + 1 = 10 entities to arrange:

the 3 green balls as a group and the other 9 non-green balls. The total number

of ways to arrange these 10 entities is 10!. However, within the green group,

the balls can be arranged in 3! ways. So the total number of ways to arrange

all balls with the greens adjacent is 10! ×3!.

Step 3: Find the number of ways the green balls can be arranged with the

constraint that they are not adjacent. Subtract the number of arrangements

with the green balls adjacent from the total number of arrangements:

12! −10! ×3!

Therefore, the number of ways the 5 red, 4 blue, and 3 green balls can be

arranged in a row such that no two green balls are adjacent is 12! −10! ×3!.

Question 16

Question

In how many ways can a committee of 4 people be selected from a group of

8 women and 6 men if the committee must consist of at least 2 women and 2

men?

Solution

Step 1: Calculate the number of ways to select 2 women from 8. There are 8

2

ways to select 2 women from 8.

Step 2: Calculate the number of ways to select 2 men from 6. Similarly,

there are 6

2ways to select 2 men from 6.

Step 3: Calculate the number of ways to select the remaining 1 woman and

1 man. There are 6 remaining women and 4 remaining men, so there are 6

1·4

1

ways to select the remaining 1 woman and 1 man.

Step 4: Calculate the total number of ways to select 2 women and 2 men.

Multiplying the results from Step 1, Step 2, and Step 3, we have: 8

2·6

1·

4

1= 28 ·15 ·6·4 = 16800.

Therefore, there are 16800 ways to select a committee of 4 people consisting

of at least 2 women and 2 men from a group of 8 women and 6 men.

Question 17

Question

How many ways are there to arrange the letters in the word ”MISSISSIPPI”

such that no two consecutive letters are the same?

Solution

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

SIPPI”. Let’s denote the word ”MISSISSIPPI” as M1, I4, S4, P2. The total

number of ways to arrange these letters is given by:

11!

1!4!4!2! = 34,650

Step 2: Calculate the total number of ways with consecutive letters the

same. To calculate the number of ways with consecutive letters the same, we

treat each set of consecutive identical letters as a single entity. Consider the

sets M, IIII, SSSS, PP. We treat each set as a single entity to calculate the total

number of ways:

7! = 5,040

Step 3: Calculate the total number of ways with no consecutive letters the

same. The total number of ways with no consecutive letters the same is the

difference between the total number of arrangements and the number of ar-

rangements with consecutive letters the same:

34,650 −5,040 = 29,610

Therefore, there are 29,610 ways to arrange the letters in ”MISSISSIPPI”

such that no two consecutive letters are the same.

Question 18

Question

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

committee must consist of 4 members, with at least one woman and one man.

How many different committees can be formed?

Solution

Step 1: Calculate the total number of committees that can be formed. Since

the committee must consist of 4 members, we need to find the number of ways

to choose 4 people from the total group of 18 (10 women and 8 men). This can

be calculated using combinations:

Total number of committees = 18

4

Step 2: Calculate the number of committees without any women. If the

committee has no women, then all 4 members must be men. We need to find

the number of ways to choose 4 men from the group of 8 men:

Number of committees without any women = 8

4

Step 3: Calculate the number of committees without any men. If the com-

mittee has no men, then all 4 members must be women. We need to find the

number of ways to choose 4 women from the group of 10 women:

Number of committees without any men = 10

4

Step 4: Calculate the number of committees with at least one woman and

one man. To find the number of committees with at least one woman and one

man, we can subtract the number of committees without any women or without

any men from the total number of committees:

Number of committees with at least one woman and one man = Total number of committees−Number of committees without any women−Number of committees without any men

=18

4−8

4−10

4

Question 19

Question

In how many ways can 5 mathematics books, 4 physics books, and 3 chemistry

books be arranged on a bookshelf if books on the same subject must be kept

together, but the subjects can be in any order?

Solution

Step 1: Calculate the number of ways to arrange the mathematics books. There

are 5 mathematics books, and they must be kept together. They can be arranged

among themselves in 5! ways.

Step 2: Calculate the number of ways to arrange the physics books. Simi-

larly, there are 4 physics books, and they must be kept together. They can be

arranged among themselves in 4! ways.

Step 3: Calculate the number of ways to arrange the chemistry books. There

are 3 chemistry books, and they must be kept together. They can be arranged

among themselves in 3! ways.

Step 4: Calculate the number of ways the subjects can be arranged on the

bookshelf. Now, we have 3 groups of books (math, physics, and chemistry)

which can be arranged on the bookshelf. Since the subjects can be in any order,

these 3 groups can be arranged among themselves in 3! ways.

Step 5: Multiply the number of ways from each step to find the total number

of arrangements. The total number of arrangements is given by: 5!×4!×3!×3! =

120 ×24 ×6×6 = 51840

Therefore, there are 51,840 ways to arrange the books on the bookshelf.

Question 20

Question

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

of the students are math majors and the other 7 are computer science majors,

determine the number of ways to form the committee if at least one math major

must be included.

Solution

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

out of 10. Step 2: Determine the number of ways to form a committee with no

math majors. Step 3: Subtract the result from Step 2 from the total to find the

number of ways to form a committee with at least one math major.

Step 1: The total number of ways to form a committee of 5 students out of

10 is given by the combination formula:

10

5=10!

5!(10 −5)! = 252.

Step 2: To find the number of ways to form a committee with no math

majors, we need to choose 5 students from the 7 computer science majors:

7

5=7!

5!(7 −5)! = 21.

Step 3: To find the number of ways to form a committee with at least one

math major, we subtract the result from Step 2 from the total:

252 −21 = 231.

Therefore, there are 231 ways to form a committee of 5 students from a

group of 10 students with at least one math major.

Question 21

Question

In how many ways can 5 different math books, 4 different history books, and

3 different science books be arranged on a bookshelf if the books within each

subject must remain together?

Solution

Step 1: Calculate the number of ways to arrange the math books. Since there

are 5 different math books, the number of ways to arrange them is 5!.