MATH 201 - INTRODUCTION TO
PROBABILITY AND STATISTICS -
Combinatorial Analysis
Question Bank - Set 7
Liberty University
Question 1
Question
In a group of 10 students, how many ways can we choose a committee of 4
students if 2 of the students, Alice and Bob, refuse to be on the committee
together?
Solution
Step 1: First, let’s calculate the total number of ways to choose a committee of
4 students from 10 students. This is a combination problem, so we can use the
formula for combinations:
Total ways to choose committee =(10
4)=10!
4!(10 −4)!
Step 2: Plug in the values into the formula:
(10
4)=10!
4!(10 −4)! =10 ×9×8×7
4×3×2×1= 210
Step 3: Now, we need to subtract the number of ways in which Alice and Bob
are on the committee together from the total number of ways. Let’s calculate
the number of ways in which Alice and Bob are on the committee together.
Step 4: If Alice and Bob are on the committee together, we treat them as a
single entity. So, we have 9 entities instead of 10 to choose from. We choose 3
more students from the remaining 8:
Ways to choose committee with Alice and Bob together =(8
3)=8!
3!(8 −3)!
Step 5: Calculate the number of ways:
(8
3)=8!
3!(8 −3)! =8×7×6
3×2×1= 56
Step 6: Subtract the number of ways with Alice and Bob together from the
total number of ways:
Number of ways to choose committee without Alice and Bob together = 210−56 = 154
Therefore, there are 154 ways to choose a committee of 4 students from a
group of 10 students where Alice and Bob are not on the committee together.
Question 2
Question
A group of five friends (Alice, Bob, Charlie, David, and Eve) are planning to
take a road trip in two cars. Each car can fit a maximum of three people.
However, Alice refuses to travel in the same car as Bob, and Eve refuses to
travel in the same car as David. In how many ways can the friends split into
two groups to go on the road trip?
Solution
Step 1: Consider the possible combinations of friends that can go in the first car.
- Alice and Bob cannot be in the same car. So we have the following options:
1. Alice, Charlie, David 2. Alice, Charlie, Eve 3. Alice, David, Eve 4. Bob,
Charlie, David 5. Bob, Charlie, Eve 6. Bob, David, Eve
Step 2: Calculate the number of ways to split the friends. - For each com-
bination identified in Step 1, there are 3 ways to arrange the friends in the first
car (since there are 3 seats). - The remaining friends will go in the second car.
Step 3: Calculate the total number of ways. - Count the number of possible
combinations for the second car (from the remaining friends). - Multiply the
number of combinations for the first car and the second car.
Step 4: Calculate the final answer. - Count the total number of ways to split
the friends into two cars for the road trip.
Therefore, the total number of ways the friends can split into two groups to
go on the road trip is 6 * (3 ways to arrange friends in the first car) = 18 ways.
Question 3
Question
In a group of 10 people, 4 are women and 6 are men. A committee of 3 people is
selected at random from the group. What is the probability that the committee
consists of 2 women and 1 man?
2
Solution
Step 1: Calculate the total number of ways to select a committee of 3 people
from a group of 10. Step 2: Calculate the number of ways to select 2 women
from the 4 available. Step 3: Calculate the number of ways to select 1 man
from the 6 available. Step 4: Determine the total number of ways to form a
committee with 2 women and 1 man. Step 5: Find the probability of selecting
a committee with 2 women and 1 man.
Step 1: The total number of ways to select a committee of 3 people from a
group of 10 is given by the combination formula:
(10
3)=10!
3!(10 −3)! = 120
Step 2: The number of ways to select 2 women from the 4 available is given
by:
(4
2)=4!
2!(4 −2)! = 6
Step 3: The number of ways to select 1 man from the 6 available is given
by:
(6
1)=6!
1!(6 −1)! = 6
Step 4: The total number of ways to form a committee with 2 women and
1 man is the product of the number of ways to select 2 women and 1 man:
6×6 = 36
Step 5: The probability of selecting a committee with 2 women and 1 man
is:
Number of ways to form committee with 2 women and 1 man
Total number of ways to select a committee of 3 people =36
120 =3
10 = 0.3
Therefore, the probability that the committee consists of 2 women and 1
man is 3
10 or 0.3.
Question 4
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 3 men and 2 women, how many different
committees can be formed?
3
Solution
Step 1: Calculate the number of ways to choose 3 men out of 8. Step 2: Calculate
the number of ways to choose 2 women out of 6. Step 3: Multiply the results
from steps 1 and 2 to find the total number of different committees that can be
formed.
Step 1: The number of ways to choose 3 men out of 8 can be calculated
using the combination formula:
8C3=8!
3!(8 −3)! =8×7×6
3×2×1= 56
So, there are 56 ways to choose 3 men out of 8.
Step 2: The number of ways to choose 2 women out of 6 can be calculated
using the combination formula:
6C2=6!
2!(6 −2)! =6×5
2×1= 15
So, there are 15 ways to choose 2 women out of 6.
Step 3: To find the total number of different committees that can be formed,
we multiply the results from steps 1 and 2:
56 ×15 = 840
Therefore, there are 840 different committees that can be formed consisting
of 3 men and 2 women.
Question 5
Question
A committee of 5 people is to be selected from a group of 10 women and 8 men.
1. How many different committees can be formed if the committee must
consist of 3 women and 2 men?
2. How many ways can the committee be formed if there are no restrictions
on the gender composition?
Solution
1. Step 1: Calculate the number of ways to choose 3 women out of 10:
(10
3)=10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
Step 2: Calculate the number of ways to choose 2 men out of 8:
(8
2)=8!
2!(8 −2)! =8×7
2×1= 28
4
Step 3: Multiply the number of ways to choose women and men:
Total ways = 120 ×28 = 3360
Therefore, there are 3360 different committees that can be formed con-
sisting of 3 women and 2 men.
2. Step 1: Calculate the number of ways to choose 5 people out of 18:
(18
5)=18!
5!(18 −5)! =18 ×17 ×16 ×15 ×14
5×4×3×2×1= 8568
Therefore, there are 8568 different committees that can be formed with
no restrictions on the gender composition.
Question 6
Question
A committee of 5 people is to be formed from a group of 10 men and 8 women.
Find the probability that the committee has at least 3 women.
Solution
Let’s calculate the total number of ways to form a committee of 5 people from
18 individuals first.
Total number of ways =(18
5)
=18!
5!13!
= 8568
Now, let’s calculate the number of ways to form a committee with at least
3 women.
Number of ways to form a committee with 3 women =(8
3)·(10
2)
=8!
3!5! ·10!
2!8!
= 1120
Number of ways to form a committee with 4 women =(8
4)·(10
1)
=8!
4!4! ·10!
1!9!
= 1680
5
Number of ways to form a committee with 5 women =(8
5)·(10
0)
=8!
5!3! ·10!
0!10!
= 56
Therefore, the total number of ways to form a committee with at least 3
women is 1120 + 1680 + 56 = 2856.
The probability of forming a committee with at least 3 women is:
2856
8568 =119
358
Therefore, the probability that the committee has at least 3 women is 119
358 .
Question 7
Question
A committee of 5 students is to be formed from a group of 8 graduate students
and 6 undergraduate students. Determine the number of ways this committee
can be formed if it must consist of 3 graduate students and 2 undergraduate
students.
Solution
Step 1: Calculate the number of ways to choose 3 graduate students from the
8 available. Step 2: Calculate the number of ways to choose 2 undergraduate
students from the 6 available. Step 3: Multiply the results from Step 1 and Step
2 to find the total number of ways the committee can be formed.
Step 1: Choosing 3 graduate students from 8 can be done using combina-
tions, denoted as (n
r), where nis the total number of items and ris the number
of items to choose:
(8
3)=8!
3!(8 −3)! =8×7×6
3×2×1= 56
So, there are 56 ways to choose 3 graduate students from 8.
Step 2: Choosing 2 undergraduate students from 6 can be done in a similar
manner:
(6
2)=6!
2!(6 −2)! =6×5
2×1= 15
So, there are 15 ways to choose 2 undergraduate students from 6.
6
Step 3: To find the total number of ways to form the committee, multiply
the results from Step 1 and Step 2:
56 ×15 = 840
Therefore, there are 840 ways to form a committee of 5 students consisting of
3 graduate students and 2 undergraduate students from the group of 8 graduate
students and 6 undergraduate students.
Question 8
Question
A committee of 4 people is to be formed from a group of 8 men and 6 women.
Find the probability that the committee consists of 2 men and 2 women.
Solution
Step 1: Find the total number of ways to form a committee of 4 people from
14. To find this, we will use the combination formula:
Total number of ways =(14
4)=14!
4!(14 −4)! =14 ×13 ×12 ×11
4×3×2×1= 1001
Step 2: Find the number of ways to choose 2 men out of 8. Using the
combination formula, we have:
(8
2)=8!
2!(8 −2)! =8×7
2×1= 28
Step 3: Find the number of ways to choose 2 women out of 6. Using the
combination formula, we have:
(6
2)=6!
2!(6 −2)! =6×5
2×1= 15
Step 4: Find the total number of ways to form a committee with 2 men
and 2 women. Since we’re looking for the intersection of choosing 2 men and 2
women, we multiply the number of ways to choose men and women:
Number of ways = 28 ×15 = 420
Step 5: Find the probability of forming a committee with 2 men and 2
women. The probability is given by:
Probability =Number of ways with 2 men and 2 women
Total number of ways =420
1001 ≈0.4196
7
Question 9
Question
A committee of 5 people is to be formed from a group of 10 men and 8 women.
How many different committees can be formed if there must be at least 2 men
and at least 2 women on the committee?
Solution
Step 1: Calculate the number of ways to form a committee with exactly 2 men
and 3 women.
• There are (10
2)ways to choose 2 men from 10.
• There are (8
3)ways to choose 3 women from 8.
So, the number of ways to form a committee with exactly 2 men and 3 women
is (10
2)×(8
3).
Step 2: Calculate the number of ways to form a committee with 3 men and
2 women.
• There are (10
3)ways to choose 3 men from 10.
• There are (8
2)ways to choose 2 women from 8.
So, the number of ways to form a committee with 3 men and 2 women is
(10
3)×(8
2).
Step 3: Calculate the number of ways to form a committee with 4 men and
1 woman.
• There are (10
4)ways to choose 4 men from 10.
• There are (8
1)ways to choose 1 woman from 8.
So, the number of ways to form a committee with 4 men and 1 woman is
(10
4)×(8
1).
Step 4: Calculate the total number of ways to form a committee with at
least 2 men and at least 2 women.
• Add the results from Step 1, Step 2, and Step 3.
So, the total number of ways to form a committee with at least 2 men and at
least 2 women is
(10
2)×(8
3)+(10
3)×(8
2)+(10
4)×(8
1).
8
Question 10
Question
A committee of 5 people is to be selected from a group of 10 men and 8 women.
How many ways can the committee be formed if it must consist of 3 men and 2
women?
Solution
To form the committee consisting of 3 men and 2 women, we need to calculate
the number of ways we can choose 3 men from 10 men and 2 women from 8
women.
Step 1: Calculate the number of ways to choose 3 men from 10
men (10
3)=10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
There are 120 ways to choose 3 men from 10 men.
Step 2: Calculate the number of ways to choose 2 women from 8
women (8
2)=8!
2!(8 −2)! =8×7
2×1= 28
There are 28 ways to choose 2 women from 8 women.
Step 3: Calculate the total number of ways to form the committee
To get the total number of ways to form the committee consisting of 3 men and
2 women, we multiply the number of ways to choose men and the number of
ways to choose women.
Total ways = 120 ×28 = 3360
Therefore, the committee can be formed in 3360 ways.
Question 11
Question
In how many ways can 5 identical red balls, 4 identical blue balls, and 3 identical
green balls be arranged in a row if balls of the same color cannot be adjacent
to each other?
Solution
Step 1: Calculate the total number of ways to arrange the balls without any
restrictions.
There are a total of 12 balls, consisting of 5 red balls, 4 blue balls, and 3 green
balls. The number of ways to arrange these 12 balls without any restrictions
9
is given by the formula for permutations of objects with repetitions, which is
12!
5!4!3! .
Step 2: Calculate the number of ways to arrange the balls with at least two
adjacent balls of the same color.
Firstly, calculate the number of ways to arrange the red balls with at least
two adjacent red balls. Treat the 5 red balls as a single entity, which can be
arranged in 5! ways. Within this arrangement, the red balls themselves can be
arranged in 5! ways. Therefore, the total number of ways to arrange the red
balls with at least two adjacent red balls is 5! ×5!.
Similarly, calculate the number of ways to arrange the blue balls with at
least two adjacent blue balls (4! ×4!) and the number of ways to arrange the
green balls with at least two adjacent green balls (3! ×3!).
Step 3: Subtract the number of ways with at least two adjacent balls of the
same color from the total number of ways to arrange the balls.
The total number of ways to arrange the balls without any restrictions is
12!
5!4!3! , and the number of ways with at least two adjacent balls of the same color
is 5! ×5! + 4! ×4! + 3! ×3!.
Subtracting the number of ways with at least two adjacent balls of the same
color from the total number of ways gives us the final answer.
Therefore, the number of ways the balls can be arranged in a row if balls of
the same color cannot be adjacent to each other is 12!
5!4!3! −(5!×5!+4!×4!+3!×3!).
Question 12
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 women, how many different committees
can be formed?
Solution
Step 1: Find the number of committees with exactly 2, 3, 4, or 5 women.
• With exactly 2 women: Choose 2 women out of 6 and 3 men out of 8.
(6
2)×(8
3)= 15 ×56 = 840
• With exactly 3 women: Choose 3 women out of 6 and 2 men out of 8.
(6
3)×(8
2)= 20 ×28 = 560
• With exactly 4 women: Choose 4 women out of 6 and 1 man out of 8.
(6
4)×(8
1)= 15 ×8 = 120
10
• With all 5 women: Choose 5 women out of 6 and 0 men out of 8.
(6
5)×(8
0)= 6 ×1 = 6
Step 2: Add up the number of committees from each case.
840 + 560 + 120 + 6 = 1526
Therefore, there are 1526 different committees that can be formed with at
least 2 women.
Question 13
Question
A committee of 5 people is to be formed from a group of 9 students and 7
professors. If the committee must consist of at least 2 professors, how many
different committees can be formed?
Solution
Step 1: Calculate the number of ways to choose the committee with exactly 2
professors. There are 7 professors to choose from and since there must be at least
2 professors, we choose 2 professors from 7, and the remaining 3 members must
be students chosen from 9. So, the number of ways to choose the committee
with exactly 2 professors is given by:
(7
2)×(9
3)=7!
2!(7 −2)! ×9!
3!(9 −3)!
Step 2: Calculate the number of ways to choose the committee with exactly
3 professors. Similar to the previous step, the number of ways to choose the
committee with exactly 3 professors is given by:
(7
3)×(9
2)=7!
3!(7 −3)! ×9!
2!(9 −2)!
Step 3: Calculate the total number of committees with at least 2 professors.
Adding the results from Step 1 and Step 2 will give us the total number of
committees with at least 2 professors:
7!
2!(7 −2)! ×9!
3!(9 −3)! +7!
3!(7 −3)! ×9!
2!(9 −2)!
Step 4: Calculate the total number of ways to form a committee with at
least 2 professors. Evaluate the expression obtained in Step 3 to find the total
number of different committees that can be formed. This will give you the final
answer.
11
Question 14
Question
In a group of 10 students, how many ways can we split them into two groups of
5 students each for a team competition?
Solution
To determine the number of ways we can split the 10 students into two groups
of 5 for a team competition, we will use combinatorial analysis principles.
Step 1: Calculate the total number of ways to choose 5 students out of 10.
This can be done using the combination formula: (n
k)=n!
k!(n−k)! , where nis the
total number of students and kis the number of students we want to choose. In
this case, n= 10 and k= 5. So, the total number of ways to choose 5 students
out of 10 is: (10
5)=10!
5!(10 −5)!
Calculating this gives us:
(10
5)=10!
5!5! =10 ×9×8×7×6
5×4×3×2×1= 252
So, there are 252 ways to choose 5 students out of 10.
Step 2: Divide by 2 to account for the fact that the order of the groups
doesn’t matter. Since we are splitting the students into two equal groups (each
with 5 students), the order of the groups doesn’t matter. Therefore, we need to
divide the total number of ways by 2 to avoid overcounting. Thus, the number
of ways to split 10 students into two groups of 5 for a team competition is:
252
2= 126
Therefore, there are 126 ways to split the 10 students into two groups of 5
students each for the team competition.
Question 15
Question
A committee of 6 people is to be formed from a group of 10 individuals. If 4
of the individuals are women and 6 are men, what is the probability that the
committee consists of exactly 3 women and 3 men?
12
Solution
Step 1: Calculate the total number of ways to form a committee of 6 people
from a group of 10 individuals. This is given by the combination formula (n
k)=
n!
k!(n−k)! .
Total number of ways =(10
6)
=10!
6!(10 −6)!
=10 ×9×8×7×6×5
6×5×4×3×2×1
= 210
Step 2: Calculate the number of ways to choose 3 women from 4 women and
3 men from 6 men. This is given by the product of the combination of women
and men.
Number of ways to choose 3 women =(4
3)= 4
Number of ways to choose 3 men =(6
3)= 20
Total number of ways to choose 3 women and 3 men = 4 ×20 = 80
Step 3: Calculate the probability of forming a committee with 3 women and
3 men by dividing the number of favorable outcomes by the total number of
outcomes.
Probability =Number of ways to choose 3 women and 3 men
Total number of ways
=80
210
=8
21 ≈0.381
Therefore, the probability that the committee consists of exactly 3 women
and 3 men is 8
21 or approximately 0.381.
Question 16
Question
In a committee of 6 people, there are 3 men and 3 women. If the committee
must select a president, a vice-president, and a treasurer, how many ways can
these positions be filled if the president cannot be a woman and the treasurer
cannot be a man?
13
Solution
Step 1: Determine the number of ways to select the president (a man) from the
3 men. Since the president must be a man, there are 3 choices for this position.
Step 2: Determine the number of ways to select the treasurer (a woman)
from the 3 women. Since the treasurer must be a woman, there are 3 choices
for this position.
Step 3: Determine the number of ways to select the vice-president. After
the president and treasurer have been selected, there are 4 remaining people (2
men and 2 women) for the vice-president position. Thus, there are 4 choices for
this position.
Step 4: Calculate the total number of ways to fill the positions. To find the
total number of ways to fill the positions, multiply the number of choices for
each position: 3×3×4 = 36.
Therefore, there are 36 ways to fill the positions of president, vice-president,
and treasurer in the committee.
Question 17
Question
A committee of 5 people is to be formed from a group of 7 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: Calculate the number of ways to choose 2 men from 7 men. There are
(7
2)= 21 ways to choose 2 men from 7 men.
Step 2: Calculate the number of ways to choose 2 women from 6 women.
There are (6
2)= 15 ways to choose 2 women from 6 women.
Step 3: Calculate the number of ways to choose 1 person (either a man or a
woman) from the remaining people. There are 5 ways to choose 1 person from
the remaining group of 5 people.
Step 4: Multiply the results from Step 1, Step 2, and Step 3 to find the total
number of ways to form the committee. Therefore, the total number of ways to
form the committee is 21 ×15 ×5 = 1575.
Thus, there are 1575 different committees that can be formed with at least
2 men and 2 women from the group of 7 men and 6 women.
Question 18
Question
In a group of 10 people, how many ways are there to select a committee of
3 people, where one person will be the president, one person will be the vice
14
president, and one person will be the secretary?
Solution
Step 1: First, we choose the president from the 10 people. There are 10 ways
to do this.
Step 2: After selecting the president, we choose the vice president from the
remaining 9 people. There are 9 ways to do this.
Step 3: Finally, after selecting the president and vice president, we choose
the secretary from the remaining 8 people. There are 8 ways to do this.
Step 4: To find the total number of ways to select the committee of 3 people
with specific roles, we multiply the number of ways at each step. Therefore, the
total number of ways to select the committee is 10 ×9×8 = 720 .
Question 19
Question
In a math class at Liberty University, there are 12 students: 6 males and 6
females. The professor randomly selects 3 students to compete in a math com-
petition. What is the probability that all 3 students selected are females?
Solution
Step 1: Find the total number of ways to select 3 students out of 12. There are
(12
3)=12!
3!(12−3)! =12×11×10
3×2×1= 220 ways to select 3 students out of 12.
Step 2: Find the number of ways to select 3 female students out of 6. Since
there are 6 females in the class, there are (6
3)=6!
3!(6−3)! =6×5×4
3×2×1= 20 ways to
select 3 female students out of 6.
Step 3: Find the probability of selecting 3 female students. The probability
of selecting 3 female students is given by Number of ways to select 3 females
Total number of ways to select 3 students =
20
220 =1
11 .
Question 20
Question
In a group of 10 friends, how many ways can we select a committee of 4 members?
Solution
Step 1: To solve this problem, we will use the combination formula:
C(n, k) = n!
k!(n−k)!
15
where C(n, k)represents the number of ways to choose kitems from a set of n
distinct items.
Step 2: Substituting n= 10 and k= 4 into the formula, we get:
C(10,4) = 10!
4!(10 −4)!
Step 3: Simplifying the expression in the numerator:
10! = 10 ×9×8×7×6×5×4!
Step 4: Simplifying the expression in the denominator:
4! = 4 ×3×2×1
Step 5: Substituting the simplified forms back into the formula:
C(10,4) = 10 ×9×8×7×6×5×4!
4×3×2×1×6!
Step 6: Canceling out the common factors:
C(10,4) = 10 ×9×8×7×5
4×3×2×1
Step 7: Calculating the final result:
C(10,4) = 30240
24 = 1260
Therefore, there are 1260 ways to select a committee of 4 members from a
group of 10 friends.
Question 21
Question
A committee of 5 people is to be formed from a group of 10 men and 8 women.
In how many ways can the committee be formed if it must consist of at least 2
women?
Solution
Step 1: Calculate the number of ways to form a committee with only 1 woman,
then subtract that from the total number of ways to form a committee with at
least 2 women.
Step 2: Calculate the number of ways to form a committee with only 1
woman. This can be done by selecting 1 woman from the 8 available women
and selecting 4 men from the 10 available men.
(8
1)×(10
4)= 8 ×210 = 1680 ways
16
Step 3: Calculate the total number of ways to form a committee with at
least 2 women. This can be done by selecting 2 women from the 8 available
women and selecting 3 men from the 10 available men, then adding the cases
where we select 3 women and 2 men from the available options.
(8
2)×(10
3)+(8
3)×(10
2)= 28 ×120 + 56 ×45 = 3360 + 2520 = 5880 ways
Step 4: Finally, subtract the number of ways to form a committee with only
1 woman from the total number of ways to form a committee with at least 2
women.
5880 −1680 = 4200
Therefore, there are 4200 ways to form a committee of 5 people from the
group of 10 men and 8 women if it must consist of at least 2 women.
Question 22
Question
In a group of 10 friends, how many ways can you choose a committee of 4 people
to represent the group?
Solution
Step 1: To find the number of ways to choose a committee of 4 people out of 10
friends, we will use the combination formula. The combination formula is given
by:
C(n, k) = n!
k!(n−k)!
where nis the total number of friends and kis the number of people in the
committee.
Step 2: Substituting n= 10 and k= 4 into the formula, we get:
C(10,4) = 10!
4!(10 −4)!
Step 3: Calculating the factorials:
10! = 10 ×9×8×7×6×5×4×3×2×1
4! = 4 ×3×2×1
6! = 6 ×5×4×3×2×1
Step 4: Substituting the factorials back into the formula:
C(10,4) = 10 ×9×8×7×6×5×4×3×2×1
(4 ×3×2×1)(6 ×5×4×3×2×1)
17
Step 5: Simplifying the expression, we get:
C(10,4) = 10 ×9×8×7
4×3×2×1=10 ×9×7×2
2×1= 210
Therefore, there are 210 ways to choose a committee of 4 people from a
group of 10 friends.
Question 23
Question
A committee of 5 people is to be formed from a group of 10 men and 8 women. If
the committee must consist of at least 3 men and 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: Multiply the results
from Step 1 and Step 2 to find the total number of committees that can be
formed.
Step 1: The number of ways to choose 3 men from 10 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 women from 8 is given by the
combination formula:
(8
2)=8!
2!(8 −2)! =8×7
2×1= 28
Step 3: Multiplying the results from Step 1 and Step 2: 120 ×28 = 3360
Therefore, there are 3360 different committees that can be formed with at
least 3 men and 2 women.
Question 24
Question
A committee of 5 people is to be formed from a group of 10 men and 8 women.
If the committee is to consist of 3 men and 2 women, how many different com-
mittees can be formed?
18
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: 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 men from 10 is given by the
combination formula:
(10
3)=10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
So, there are 120 ways to choose 3 men from 10.
Step 2: The number of ways to choose 2 women from 8 is given by the
combination formula:
(8
2)=8!
2!(8 −2)! =8×7
2×1= 28
So, there are 28 ways to choose 2 women from 8.
Step 3: To find the total number of different committees that can be formed,
we multiply the results from Step 1 and Step 2: 120 ×28 = 3360
Therefore, there are 3360 different committees that can be formed consisting
of 3 men and 2 women from the given group.
Question 25
Question
A committee of 5 people is to be formed from a group of 10 men and 8 women.
In how many ways can the committee be formed if it must consist of at least 3
men?
Solution
Step 1: Find the number of ways to choose exactly 3 men out of 10. There are
(10
3)ways to select 3 men from a group of 10.
Step 2: Find the number of ways to choose 2 more members (men or women)
to complete the committee. Since the committee must consist of at least 3 men,
we need to select 2 more members to join the 3 men already chosen. We can
select either 2 women from the 8 available, or 1 woman and 1 man. Therefore,
the total number of ways to choose 2 more members is (8
2)+(8
1)×(10
1).
Step 3: Calculate the total number of ways to form the committee. The
total number of ways to form the committee is the product of the choices made
in Step 1 and Step 2: Total = (10
3)×((8
2)+(8
1)×(10
1)).
Calculating this expression gives us the final answer.
19
Step 5: Calculate the number of ways:
(8
3)=8!
3!(8 −3)! =8×7×6
3×2×1= 56
Step 6: Subtract the number of ways with Alice and Bob together from the
total number of ways:
Number of ways to choose committee without Alice and Bob together = 210−56 = 154
Therefore, there are 154 ways to choose a committee of 4 students from a
group of 10 students where Alice and Bob are not on the committee together.
Question 2
Question
A group of five friends (Alice, Bob, Charlie, David, and Eve) are planning to
take a road trip in two cars. Each car can fit a maximum of three people.
However, Alice refuses to travel in the same car as Bob, and Eve refuses to
travel in the same car as David. In how many ways can the friends split into
two groups to go on the road trip?
Solution
Step 1: Consider the possible combinations of friends that can go in the first car.
- Alice and Bob cannot be in the same car. So we have the following options:
1. Alice, Charlie, David 2. Alice, Charlie, Eve 3. Alice, David, Eve 4. Bob,
Charlie, David 5. Bob, Charlie, Eve 6. Bob, David, Eve
Step 2: Calculate the number of ways to split the friends. - For each com-
bination identified in Step 1, there are 3 ways to arrange the friends in the first
car (since there are 3 seats). - The remaining friends will go in the second car.
Step 3: Calculate the total number of ways. - Count the number of possible
combinations for the second car (from the remaining friends). - Multiply the
number of combinations for the first car and the second car.
Step 4: Calculate the final answer. - Count the total number of ways to split
the friends into two cars for the road trip.
Therefore, the total number of ways the friends can split into two groups to
go on the road trip is 6 * (3 ways to arrange friends in the first car) = 18 ways.
Question 3
Question
In a group of 10 people, 4 are women and 6 are men. A committee of 3 people is
selected at random from the group. What is the probability that the committee
consists of 2 women and 1 man?
2
Solution
Step 1: Calculate the total number of ways to select a committee of 3 people
from a group of 10. Step 2: Calculate the number of ways to select 2 women
from the 4 available. Step 3: Calculate the number of ways to select 1 man
from the 6 available. Step 4: Determine the total number of ways to form a
committee with 2 women and 1 man. Step 5: Find the probability of selecting
a committee with 2 women and 1 man.
Step 1: The total number of ways to select a committee of 3 people from a
group of 10 is given by the combination formula:
(10
3)=10!
3!(10 −3)! = 120
Step 2: The number of ways to select 2 women from the 4 available is given
by:
(4
2)=4!
2!(4 −2)! = 6
Step 3: The number of ways to select 1 man from the 6 available is given
by:
(6
1)=6!
1!(6 −1)! = 6
Step 4: The total number of ways to form a committee with 2 women and
1 man is the product of the number of ways to select 2 women and 1 man:
6×6 = 36
Step 5: The probability of selecting a committee with 2 women and 1 man
is:
Number of ways to form committee with 2 women and 1 man
Total number of ways to select a committee of 3 people =36
120 =3
10 = 0.3
Therefore, the probability that the committee consists of 2 women and 1
man is 3
10 or 0.3.
Question 4
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 3 men and 2 women, how many different
committees can be formed?
3
Solution
Step 1: Calculate the number of ways to choose 3 men out of 8. Step 2: Calculate
the number of ways to choose 2 women out of 6. Step 3: Multiply the results
from steps 1 and 2 to find the total number of different committees that can be
formed.
Step 1: The number of ways to choose 3 men out of 8 can be calculated
using the combination formula:
8C3=8!
3!(8 −3)! =8×7×6
3×2×1= 56
So, there are 56 ways to choose 3 men out of 8.
Step 2: The number of ways to choose 2 women out of 6 can be calculated
using the combination formula:
6C2=6!
2!(6 −2)! =6×5
2×1= 15
So, there are 15 ways to choose 2 women out of 6.
Step 3: To find the total number of different committees that can be formed,
we multiply the results from steps 1 and 2:
56 ×15 = 840
Therefore, there are 840 different committees that can be formed consisting
of 3 men and 2 women.
Question 5
Question
A committee of 5 people is to be selected from a group of 10 women and 8 men.
1. How many different committees can be formed if the committee must
consist of 3 women and 2 men?
2. How many ways can the committee be formed if there are no restrictions
on the gender composition?
Solution
1. Step 1: Calculate the number of ways to choose 3 women out of 10:
(10
3)=10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
Step 2: Calculate the number of ways to choose 2 men out of 8:
(8
2)=8!
2!(8 −2)! =8×7
2×1= 28
4
Step 3: Multiply the number of ways to choose women and men:
Total ways = 120 ×28 = 3360
Therefore, there are 3360 different committees that can be formed con-
sisting of 3 women and 2 men.
2. Step 1: Calculate the number of ways to choose 5 people out of 18:
(18
5)=18!
5!(18 −5)! =18 ×17 ×16 ×15 ×14
5×4×3×2×1= 8568
Therefore, there are 8568 different committees that can be formed with
no restrictions on the gender composition.
Question 6
Question
A committee of 5 people is to be formed from a group of 10 men and 8 women.
Find the probability that the committee has at least 3 women.
Solution
Let’s calculate the total number of ways to form a committee of 5 people from
18 individuals first.
Total number of ways =(18
5)
=18!
5!13!
= 8568
Now, let’s calculate the number of ways to form a committee with at least
3 women.
Number of ways to form a committee with 3 women =(8
3)·(10
2)
=8!
3!5! ·10!
2!8!
= 1120
Number of ways to form a committee with 4 women =(8
4)·(10
1)
=8!
4!4! ·10!
1!9!
= 1680
5
Number of ways to form a committee with 5 women =(8
5)·(10
0)
=8!
5!3! ·10!
0!10!
= 56
Therefore, the total number of ways to form a committee with at least 3
women is 1120 + 1680 + 56 = 2856.
The probability of forming a committee with at least 3 women is:
2856
8568 =119
358
Therefore, the probability that the committee has at least 3 women is 119
358 .
Question 7
Question
A committee of 5 students is to be formed from a group of 8 graduate students
and 6 undergraduate students. Determine the number of ways this committee
can be formed if it must consist of 3 graduate students and 2 undergraduate
students.
Solution
Step 1: Calculate the number of ways to choose 3 graduate students from the
8 available. Step 2: Calculate the number of ways to choose 2 undergraduate
students from the 6 available. Step 3: Multiply the results from Step 1 and Step
2 to find the total number of ways the committee can be formed.
Step 1: Choosing 3 graduate students from 8 can be done using combina-
tions, denoted as (n
r), where nis the total number of items and ris the number
of items to choose:
(8
3)=8!
3!(8 −3)! =8×7×6
3×2×1= 56
So, there are 56 ways to choose 3 graduate students from 8.
Step 2: Choosing 2 undergraduate students from 6 can be done in a similar
manner:
(6
2)=6!
2!(6 −2)! =6×5
2×1= 15
So, there are 15 ways to choose 2 undergraduate students from 6.
6
Step 3: To find the total number of ways to form the committee, multiply
the results from Step 1 and Step 2:
56 ×15 = 840
Therefore, there are 840 ways to form a committee of 5 students consisting of
3 graduate students and 2 undergraduate students from the group of 8 graduate
students and 6 undergraduate students.
Question 8
Question
A committee of 4 people is to be formed from a group of 8 men and 6 women.
Find the probability that the committee consists of 2 men and 2 women.
Solution
Step 1: Find the total number of ways to form a committee of 4 people from
14. To find this, we will use the combination formula:
Total number of ways =(14
4)=14!
4!(14 −4)! =14 ×13 ×12 ×11
4×3×2×1= 1001
Step 2: Find the number of ways to choose 2 men out of 8. Using the
combination formula, we have:
(8
2)=8!
2!(8 −2)! =8×7
2×1= 28
Step 3: Find the number of ways to choose 2 women out of 6. Using the
combination formula, we have:
(6
2)=6!
2!(6 −2)! =6×5
2×1= 15
Step 4: Find the total number of ways to form a committee with 2 men
and 2 women. Since we’re looking for the intersection of choosing 2 men and 2
women, we multiply the number of ways to choose men and women:
Number of ways = 28 ×15 = 420
Step 5: Find the probability of forming a committee with 2 men and 2
women. The probability is given by:
Probability =Number of ways with 2 men and 2 women
Total number of ways =420
1001 ≈0.4196
7
Question 9
Question
A committee of 5 people is to be formed from a group of 10 men and 8 women.
How many different committees can be formed if there must be at least 2 men
and at least 2 women on the committee?
Solution
Step 1: Calculate the number of ways to form a committee with exactly 2 men
and 3 women.
• There are (10
2)ways to choose 2 men from 10.
• There are (8
3)ways to choose 3 women from 8.
So, the number of ways to form a committee with exactly 2 men and 3 women
is (10
2)×(8
3).
Step 2: Calculate the number of ways to form a committee with 3 men and
2 women.
• There are (10
3)ways to choose 3 men from 10.
• There are (8
2)ways to choose 2 women from 8.
So, the number of ways to form a committee with 3 men and 2 women is
(10
3)×(8
2).
Step 3: Calculate the number of ways to form a committee with 4 men and
1 woman.
• There are (10
4)ways to choose 4 men from 10.
• There are (8
1)ways to choose 1 woman from 8.
So, the number of ways to form a committee with 4 men and 1 woman is
(10
4)×(8
1).
Step 4: Calculate the total number of ways to form a committee with at
least 2 men and at least 2 women.
• Add the results from Step 1, Step 2, and Step 3.
So, the total number of ways to form a committee with at least 2 men and at
least 2 women is
(10
2)×(8
3)+(10
3)×(8
2)+(10
4)×(8
1).
8
Question 10
Question
A committee of 5 people is to be selected from a group of 10 men and 8 women.
How many ways can the committee be formed if it must consist of 3 men and 2
women?
Solution
To form the committee consisting of 3 men and 2 women, we need to calculate
the number of ways we can choose 3 men from 10 men and 2 women from 8
women.
Step 1: Calculate the number of ways to choose 3 men from 10
men (10
3)=10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
There are 120 ways to choose 3 men from 10 men.
Step 2: Calculate the number of ways to choose 2 women from 8
women (8
2)=8!
2!(8 −2)! =8×7
2×1= 28
There are 28 ways to choose 2 women from 8 women.
Step 3: Calculate the total number of ways to form the committee
To get the total number of ways to form the committee consisting of 3 men and
2 women, we multiply the number of ways to choose men and the number of
ways to choose women.
Total ways = 120 ×28 = 3360
Therefore, the committee can be formed in 3360 ways.
Question 11
Question
In how many ways can 5 identical red balls, 4 identical blue balls, and 3 identical
green balls be arranged in a row if balls of the same color cannot be adjacent
to each other?
Solution
Step 1: Calculate the total number of ways to arrange the balls without any
restrictions.
There are a total of 12 balls, consisting of 5 red balls, 4 blue balls, and 3 green
balls. The number of ways to arrange these 12 balls without any restrictions
9
is given by the formula for permutations of objects with repetitions, which is
12!
5!4!3! .
Step 2: Calculate the number of ways to arrange the balls with at least two
adjacent balls of the same color.
Firstly, calculate the number of ways to arrange the red balls with at least
two adjacent red balls. Treat the 5 red balls as a single entity, which can be
arranged in 5! ways. Within this arrangement, the red balls themselves can be
arranged in 5! ways. Therefore, the total number of ways to arrange the red
balls with at least two adjacent red balls is 5! ×5!.
Similarly, calculate the number of ways to arrange the blue balls with at
least two adjacent blue balls (4! ×4!) and the number of ways to arrange the
green balls with at least two adjacent green balls (3! ×3!).
Step 3: Subtract the number of ways with at least two adjacent balls of the
same color from the total number of ways to arrange the balls.
The total number of ways to arrange the balls without any restrictions is
12!
5!4!3! , and the number of ways with at least two adjacent balls of the same color
is 5! ×5! + 4! ×4! + 3! ×3!.
Subtracting the number of ways with at least two adjacent balls of the same
color from the total number of ways gives us the final answer.
Therefore, the number of ways the balls can be arranged in a row if balls of
the same color cannot be adjacent to each other is 12!
5!4!3! −(5!×5!+4!×4!+3!×3!).
Question 12
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 women, how many different committees
can be formed?
Solution
Step 1: Find the number of committees with exactly 2, 3, 4, or 5 women.
• With exactly 2 women: Choose 2 women out of 6 and 3 men out of 8.
(6
2)×(8
3)= 15 ×56 = 840
• With exactly 3 women: Choose 3 women out of 6 and 2 men out of 8.
(6
3)×(8
2)= 20 ×28 = 560
• With exactly 4 women: Choose 4 women out of 6 and 1 man out of 8.
(6
4)×(8
1)= 15 ×8 = 120
10
• With all 5 women: Choose 5 women out of 6 and 0 men out of 8.
(6
5)×(8
0)= 6 ×1 = 6
Step 2: Add up the number of committees from each case.
840 + 560 + 120 + 6 = 1526
Therefore, there are 1526 different committees that can be formed with at
least 2 women.
Question 13
Question
A committee of 5 people is to be formed from a group of 9 students and 7
professors. If the committee must consist of at least 2 professors, how many
different committees can be formed?
Solution
Step 1: Calculate the number of ways to choose the committee with exactly 2
professors. There are 7 professors to choose from and since there must be at least
2 professors, we choose 2 professors from 7, and the remaining 3 members must
be students chosen from 9. So, the number of ways to choose the committee
with exactly 2 professors is given by:
(7
2)×(9
3)=7!
2!(7 −2)! ×9!
3!(9 −3)!
Step 2: Calculate the number of ways to choose the committee with exactly
3 professors. Similar to the previous step, the number of ways to choose the
committee with exactly 3 professors is given by:
(7
3)×(9
2)=7!
3!(7 −3)! ×9!
2!(9 −2)!
Step 3: Calculate the total number of committees with at least 2 professors.
Adding the results from Step 1 and Step 2 will give us the total number of
committees with at least 2 professors:
7!
2!(7 −2)! ×9!
3!(9 −3)! +7!
3!(7 −3)! ×9!
2!(9 −2)!
Step 4: Calculate the total number of ways to form a committee with at
least 2 professors. Evaluate the expression obtained in Step 3 to find the total
number of different committees that can be formed. This will give you the final
answer.
11
Question 14
Question
In a group of 10 students, how many ways can we split them into two groups of
5 students each for a team competition?
Solution
To determine the number of ways we can split the 10 students into two groups
of 5 for a team competition, we will use combinatorial analysis principles.
Step 1: Calculate the total number of ways to choose 5 students out of 10.
This can be done using the combination formula: (n
k)=n!
k!(n−k)! , where nis the
total number of students and kis the number of students we want to choose. In
this case, n= 10 and k= 5. So, the total number of ways to choose 5 students
out of 10 is: (10
5)=10!
5!(10 −5)!
Calculating this gives us:
(10
5)=10!
5!5! =10 ×9×8×7×6
5×4×3×2×1= 252
So, there are 252 ways to choose 5 students out of 10.
Step 2: Divide by 2 to account for the fact that the order of the groups
doesn’t matter. Since we are splitting the students into two equal groups (each
with 5 students), the order of the groups doesn’t matter. Therefore, we need to
divide the total number of ways by 2 to avoid overcounting. Thus, the number
of ways to split 10 students into two groups of 5 for a team competition is:
252
2= 126
Therefore, there are 126 ways to split the 10 students into two groups of 5
students each for the team competition.
Question 15
Question
A committee of 6 people is to be formed from a group of 10 individuals. If 4
of the individuals are women and 6 are men, what is the probability that the
committee consists of exactly 3 women and 3 men?
12
Solution
Step 1: Calculate the total number of ways to form a committee of 6 people
from a group of 10 individuals. This is given by the combination formula (n
k)=
n!
k!(n−k)! .
Total number of ways =(10
6)
=10!
6!(10 −6)!
=10 ×9×8×7×6×5
6×5×4×3×2×1
= 210
Step 2: Calculate the number of ways to choose 3 women from 4 women and
3 men from 6 men. This is given by the product of the combination of women
and men.
Number of ways to choose 3 women =(4
3)= 4
Number of ways to choose 3 men =(6
3)= 20
Total number of ways to choose 3 women and 3 men = 4 ×20 = 80
Step 3: Calculate the probability of forming a committee with 3 women and
3 men by dividing the number of favorable outcomes by the total number of
outcomes.
Probability =Number of ways to choose 3 women and 3 men
Total number of ways
=80
210
=8
21 ≈0.381
Therefore, the probability that the committee consists of exactly 3 women
and 3 men is 8
21 or approximately 0.381.
Question 16
Question
In a committee of 6 people, there are 3 men and 3 women. If the committee
must select a president, a vice-president, and a treasurer, how many ways can
these positions be filled if the president cannot be a woman and the treasurer
cannot be a man?
13
Solution
Step 1: Determine the number of ways to select the president (a man) from the
3 men. Since the president must be a man, there are 3 choices for this position.
Step 2: Determine the number of ways to select the treasurer (a woman)
from the 3 women. Since the treasurer must be a woman, there are 3 choices
for this position.
Step 3: Determine the number of ways to select the vice-president. After
the president and treasurer have been selected, there are 4 remaining people (2
men and 2 women) for the vice-president position. Thus, there are 4 choices for
this position.
Step 4: Calculate the total number of ways to fill the positions. To find the
total number of ways to fill the positions, multiply the number of choices for
each position: 3×3×4 = 36.
Therefore, there are 36 ways to fill the positions of president, vice-president,
and treasurer in the committee.
Question 17
Question
A committee of 5 people is to be formed from a group of 7 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: Calculate the number of ways to choose 2 men from 7 men. There are
(7
2)= 21 ways to choose 2 men from 7 men.
Step 2: Calculate the number of ways to choose 2 women from 6 women.
There are (6
2)= 15 ways to choose 2 women from 6 women.
Step 3: Calculate the number of ways to choose 1 person (either a man or a
woman) from the remaining people. There are 5 ways to choose 1 person from
the remaining group of 5 people.
Step 4: Multiply the results from Step 1, Step 2, and Step 3 to find the total
number of ways to form the committee. Therefore, the total number of ways to
form the committee is 21 ×15 ×5 = 1575.
Thus, there are 1575 different committees that can be formed with at least
2 men and 2 women from the group of 7 men and 6 women.
Question 18
Question
In a group of 10 people, how many ways are there to select a committee of
3 people, where one person will be the president, one person will be the vice
14
president, and one person will be the secretary?
Solution
Step 1: First, we choose the president from the 10 people. There are 10 ways
to do this.
Step 2: After selecting the president, we choose the vice president from the
remaining 9 people. There are 9 ways to do this.
Step 3: Finally, after selecting the president and vice president, we choose
the secretary from the remaining 8 people. There are 8 ways to do this.
Step 4: To find the total number of ways to select the committee of 3 people
with specific roles, we multiply the number of ways at each step. Therefore, the
total number of ways to select the committee is 10 ×9×8 = 720 .
Question 19
Question
In a math class at Liberty University, there are 12 students: 6 males and 6
females. The professor randomly selects 3 students to compete in a math com-
petition. What is the probability that all 3 students selected are females?
Solution
Step 1: Find the total number of ways to select 3 students out of 12. There are
(12
3)=12!
3!(12−3)! =12×11×10
3×2×1= 220 ways to select 3 students out of 12.
Step 2: Find the number of ways to select 3 female students out of 6. Since
there are 6 females in the class, there are (6
3)=6!
3!(6−3)! =6×5×4
3×2×1= 20 ways to
select 3 female students out of 6.
Step 3: Find the probability of selecting 3 female students. The probability
of selecting 3 female students is given by Number of ways to select 3 females
Total number of ways to select 3 students =
20
220 =1
11 .
Question 20
Question
In a group of 10 friends, how many ways can we select a committee of 4 members?
Solution
Step 1: To solve this problem, we will use the combination formula:
C(n, k) = n!
k!(n−k)!
15
where C(n, k)represents the number of ways to choose kitems from a set of n
distinct items.
Step 2: Substituting n= 10 and k= 4 into the formula, we get:
C(10,4) = 10!
4!(10 −4)!
Step 3: Simplifying the expression in the numerator:
10! = 10 ×9×8×7×6×5×4!
Step 4: Simplifying the expression in the denominator:
4! = 4 ×3×2×1
Step 5: Substituting the simplified forms back into the formula:
C(10,4) = 10 ×9×8×7×6×5×4!
4×3×2×1×6!
Step 6: Canceling out the common factors:
C(10,4) = 10 ×9×8×7×5
4×3×2×1
Step 7: Calculating the final result:
C(10,4) = 30240
24 = 1260
Therefore, there are 1260 ways to select a committee of 4 members from a
group of 10 friends.
Question 21
Question
A committee of 5 people is to be formed from a group of 10 men and 8 women.
In how many ways can the committee be formed if it must consist of at least 2
women?
Solution
Step 1: Calculate the number of ways to form a committee with only 1 woman,
then subtract that from the total number of ways to form a committee with at
least 2 women.
Step 2: Calculate the number of ways to form a committee with only 1
woman. This can be done by selecting 1 woman from the 8 available women
and selecting 4 men from the 10 available men.
(8
1)×(10
4)= 8 ×210 = 1680 ways
16
Step 3: Calculate the total number of ways to form a committee with at
least 2 women. This can be done by selecting 2 women from the 8 available
women and selecting 3 men from the 10 available men, then adding the cases
where we select 3 women and 2 men from the available options.
(8
2)×(10
3)+(8
3)×(10
2)= 28 ×120 + 56 ×45 = 3360 + 2520 = 5880 ways
Step 4: Finally, subtract the number of ways to form a committee with only
1 woman from the total number of ways to form a committee with at least 2
women.
5880 −1680 = 4200
Therefore, there are 4200 ways to form a committee of 5 people from the
group of 10 men and 8 women if it must consist of at least 2 women.
Question 22
Question
In a group of 10 friends, how many ways can you choose a committee of 4 people
to represent the group?
Solution
Step 1: To find the number of ways to choose a committee of 4 people out of 10
friends, we will use the combination formula. The combination formula is given
by:
C(n, k) = n!
k!(n−k)!
where nis the total number of friends and kis the number of people in the
committee.
Step 2: Substituting n= 10 and k= 4 into the formula, we get:
C(10,4) = 10!
4!(10 −4)!
Step 3: Calculating the factorials:
10! = 10 ×9×8×7×6×5×4×3×2×1
4! = 4 ×3×2×1
6! = 6 ×5×4×3×2×1
Step 4: Substituting the factorials back into the formula:
C(10,4) = 10 ×9×8×7×6×5×4×3×2×1
(4 ×3×2×1)(6 ×5×4×3×2×1)
17
Step 5: Simplifying the expression, we get:
C(10,4) = 10 ×9×8×7
4×3×2×1=10 ×9×7×2
2×1= 210
Therefore, there are 210 ways to choose a committee of 4 people from a
group of 10 friends.
Question 23
Question
A committee of 5 people is to be formed from a group of 10 men and 8 women. If
the committee must consist of at least 3 men and 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: Multiply the results
from Step 1 and Step 2 to find the total number of committees that can be
formed.
Step 1: The number of ways to choose 3 men from 10 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 women from 8 is given by the
combination formula:
(8
2)=8!
2!(8 −2)! =8×7
2×1= 28
Step 3: Multiplying the results from Step 1 and Step 2: 120 ×28 = 3360
Therefore, there are 3360 different committees that can be formed with at
least 3 men and 2 women.
Question 24
Question
A committee of 5 people is to be formed from a group of 10 men and 8 women.
If the committee is to consist of 3 men and 2 women, how many different com-
mittees can be formed?
18
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: 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 men from 10 is given by the
combination formula:
(10
3)=10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
So, there are 120 ways to choose 3 men from 10.
Step 2: The number of ways to choose 2 women from 8 is given by the
combination formula:
(8
2)=8!
2!(8 −2)! =8×7
2×1= 28
So, there are 28 ways to choose 2 women from 8.
Step 3: To find the total number of different committees that can be formed,
we multiply the results from Step 1 and Step 2: 120 ×28 = 3360
Therefore, there are 3360 different committees that can be formed consisting
of 3 men and 2 women from the given group.
Question 25
Question
A committee of 5 people is to be formed from a group of 10 men and 8 women.
In how many ways can the committee be formed if it must consist of at least 3
men?
Solution
Step 1: Find the number of ways to choose exactly 3 men out of 10. There are
(10
3)ways to select 3 men from a group of 10.
Step 2: Find the number of ways to choose 2 more members (men or women)
to complete the committee. Since the committee must consist of at least 3 men,
we need to select 2 more members to join the 3 men already chosen. We can
select either 2 women from the 8 available, or 1 woman and 1 man. Therefore,
the total number of ways to choose 2 more members is (8
2)+(8
1)×(10
1).
Step 3: Calculate the total number of ways to form the committee. The
total number of ways to form the committee is the product of the choices made
in Step 1 and Step 2: Total = (10
3)×((8
2)+(8
1)×(10
1)).
Calculating this expression gives us the final answer.
19
Step 5: Calculate the number of ways:
(8
3)=8!
3!(8 −3)! =8×7×6
3×2×1= 56
Step 6: Subtract the number of ways with Alice and Bob together from the
total number of ways:
Number of ways to choose committee without Alice and Bob together = 210−56 = 154
Therefore, there are 154 ways to choose a committee of 4 students from a
group of 10 students where Alice and Bob are not on the committee together.
Question 2
Question
A group of five friends (Alice, Bob, Charlie, David, and Eve) are planning to
take a road trip in two cars. Each car can fit a maximum of three people.
However, Alice refuses to travel in the same car as Bob, and Eve refuses to
travel in the same car as David. In how many ways can the friends split into
two groups to go on the road trip?
Solution
Step 1: Consider the possible combinations of friends that can go in the first car.
- Alice and Bob cannot be in the same car. So we have the following options:
1. Alice, Charlie, David 2. Alice, Charlie, Eve 3. Alice, David, Eve 4. Bob,
Charlie, David 5. Bob, Charlie, Eve 6. Bob, David, Eve
Step 2: Calculate the number of ways to split the friends. - For each com-
bination identified in Step 1, there are 3 ways to arrange the friends in the first
car (since there are 3 seats). - The remaining friends will go in the second car.
Step 3: Calculate the total number of ways. - Count the number of possible
combinations for the second car (from the remaining friends). - Multiply the
number of combinations for the first car and the second car.
Step 4: Calculate the final answer. - Count the total number of ways to split
the friends into two cars for the road trip.
Therefore, the total number of ways the friends can split into two groups to
go on the road trip is 6 * (3 ways to arrange friends in the first car) = 18 ways.
Question 3
Question
In a group of 10 people, 4 are women and 6 are men. A committee of 3 people is
selected at random from the group. What is the probability that the committee
consists of 2 women and 1 man?
2
Solution
Step 1: Calculate the total number of ways to select a committee of 3 people
from a group of 10. Step 2: Calculate the number of ways to select 2 women
from the 4 available. Step 3: Calculate the number of ways to select 1 man
from the 6 available. Step 4: Determine the total number of ways to form a
committee with 2 women and 1 man. Step 5: Find the probability of selecting
a committee with 2 women and 1 man.
Step 1: The total number of ways to select a committee of 3 people from a
group of 10 is given by the combination formula:
(10
3)=10!
3!(10 −3)! = 120
Step 2: The number of ways to select 2 women from the 4 available is given
by:
(4
2)=4!
2!(4 −2)! = 6
Step 3: The number of ways to select 1 man from the 6 available is given
by:
(6
1)=6!
1!(6 −1)! = 6
Step 4: The total number of ways to form a committee with 2 women and
1 man is the product of the number of ways to select 2 women and 1 man:
6×6 = 36
Step 5: The probability of selecting a committee with 2 women and 1 man
is:
Number of ways to form committee with 2 women and 1 man
Total number of ways to select a committee of 3 people =36
120 =3
10 = 0.3
Therefore, the probability that the committee consists of 2 women and 1
man is 3
10 or 0.3.
Question 4
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 3 men and 2 women, how many different
committees can be formed?
3
Solution
Step 1: Calculate the number of ways to choose 3 men out of 8. Step 2: Calculate
the number of ways to choose 2 women out of 6. Step 3: Multiply the results
from steps 1 and 2 to find the total number of different committees that can be
formed.
Step 1: The number of ways to choose 3 men out of 8 can be calculated
using the combination formula:
8C3=8!
3!(8 −3)! =8×7×6
3×2×1= 56
So, there are 56 ways to choose 3 men out of 8.
Step 2: The number of ways to choose 2 women out of 6 can be calculated
using the combination formula:
6C2=6!
2!(6 −2)! =6×5
2×1= 15
So, there are 15 ways to choose 2 women out of 6.
Step 3: To find the total number of different committees that can be formed,
we multiply the results from steps 1 and 2:
56 ×15 = 840
Therefore, there are 840 different committees that can be formed consisting
of 3 men and 2 women.
Question 5
Question
A committee of 5 people is to be selected from a group of 10 women and 8 men.
1. How many different committees can be formed if the committee must
consist of 3 women and 2 men?
2. How many ways can the committee be formed if there are no restrictions
on the gender composition?
Solution
1. Step 1: Calculate the number of ways to choose 3 women out of 10:
(10
3)=10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
Step 2: Calculate the number of ways to choose 2 men out of 8:
(8
2)=8!
2!(8 −2)! =8×7
2×1= 28
4
Step 3: Multiply the number of ways to choose women and men:
Total ways = 120 ×28 = 3360
Therefore, there are 3360 different committees that can be formed con-
sisting of 3 women and 2 men.
2. Step 1: Calculate the number of ways to choose 5 people out of 18:
(18
5)=18!
5!(18 −5)! =18 ×17 ×16 ×15 ×14
5×4×3×2×1= 8568
Therefore, there are 8568 different committees that can be formed with
no restrictions on the gender composition.
Question 6
Question
A committee of 5 people is to be formed from a group of 10 men and 8 women.
Find the probability that the committee has at least 3 women.
Solution
Let’s calculate the total number of ways to form a committee of 5 people from
18 individuals first.
Total number of ways =(18
5)
=18!
5!13!
= 8568
Now, let’s calculate the number of ways to form a committee with at least
3 women.
Number of ways to form a committee with 3 women =(8
3)·(10
2)
=8!
3!5! ·10!
2!8!
= 1120
Number of ways to form a committee with 4 women =(8
4)·(10
1)
=8!
4!4! ·10!
1!9!
= 1680
5
Number of ways to form a committee with 5 women =(8
5)·(10
0)
=8!
5!3! ·10!
0!10!
= 56
Therefore, the total number of ways to form a committee with at least 3
women is 1120 + 1680 + 56 = 2856.
The probability of forming a committee with at least 3 women is:
2856
8568 =119
358
Therefore, the probability that the committee has at least 3 women is 119
358 .
Question 7
Question
A committee of 5 students is to be formed from a group of 8 graduate students
and 6 undergraduate students. Determine the number of ways this committee
can be formed if it must consist of 3 graduate students and 2 undergraduate
students.
Solution
Step 1: Calculate the number of ways to choose 3 graduate students from the
8 available. Step 2: Calculate the number of ways to choose 2 undergraduate
students from the 6 available. Step 3: Multiply the results from Step 1 and Step
2 to find the total number of ways the committee can be formed.
Step 1: Choosing 3 graduate students from 8 can be done using combina-
tions, denoted as (n
r), where nis the total number of items and ris the number
of items to choose:
(8
3)=8!
3!(8 −3)! =8×7×6
3×2×1= 56
So, there are 56 ways to choose 3 graduate students from 8.
Step 2: Choosing 2 undergraduate students from 6 can be done in a similar
manner:
(6
2)=6!
2!(6 −2)! =6×5
2×1= 15
So, there are 15 ways to choose 2 undergraduate students from 6.
6
Step 3: To find the total number of ways to form the committee, multiply
the results from Step 1 and Step 2:
56 ×15 = 840
Therefore, there are 840 ways to form a committee of 5 students consisting of
3 graduate students and 2 undergraduate students from the group of 8 graduate
students and 6 undergraduate students.
Question 8
Question
A committee of 4 people is to be formed from a group of 8 men and 6 women.
Find the probability that the committee consists of 2 men and 2 women.
Solution
Step 1: Find the total number of ways to form a committee of 4 people from
14. To find this, we will use the combination formula:
Total number of ways =(14
4)=14!
4!(14 −4)! =14 ×13 ×12 ×11
4×3×2×1= 1001
Step 2: Find the number of ways to choose 2 men out of 8. Using the
combination formula, we have:
(8
2)=8!
2!(8 −2)! =8×7
2×1= 28
Step 3: Find the number of ways to choose 2 women out of 6. Using the
combination formula, we have:
(6
2)=6!
2!(6 −2)! =6×5
2×1= 15
Step 4: Find the total number of ways to form a committee with 2 men
and 2 women. Since we’re looking for the intersection of choosing 2 men and 2
women, we multiply the number of ways to choose men and women:
Number of ways = 28 ×15 = 420
Step 5: Find the probability of forming a committee with 2 men and 2
women. The probability is given by:
Probability =Number of ways with 2 men and 2 women
Total number of ways =420
1001 ≈0.4196
7
Question 9
Question
A committee of 5 people is to be formed from a group of 10 men and 8 women.
How many different committees can be formed if there must be at least 2 men
and at least 2 women on the committee?
Solution
Step 1: Calculate the number of ways to form a committee with exactly 2 men
and 3 women.
• There are (10
2)ways to choose 2 men from 10.
• There are (8
3)ways to choose 3 women from 8.
So, the number of ways to form a committee with exactly 2 men and 3 women
is (10
2)×(8
3).
Step 2: Calculate the number of ways to form a committee with 3 men and
2 women.
• There are (10
3)ways to choose 3 men from 10.
• There are (8
2)ways to choose 2 women from 8.
So, the number of ways to form a committee with 3 men and 2 women is
(10
3)×(8
2).
Step 3: Calculate the number of ways to form a committee with 4 men and
1 woman.
• There are (10
4)ways to choose 4 men from 10.
• There are (8
1)ways to choose 1 woman from 8.
So, the number of ways to form a committee with 4 men and 1 woman is
(10
4)×(8
1).
Step 4: Calculate the total number of ways to form a committee with at
least 2 men and at least 2 women.
• Add the results from Step 1, Step 2, and Step 3.
So, the total number of ways to form a committee with at least 2 men and at
least 2 women is
(10
2)×(8
3)+(10
3)×(8
2)+(10
4)×(8
1).
8
Question 10
Question
A committee of 5 people is to be selected from a group of 10 men and 8 women.
How many ways can the committee be formed if it must consist of 3 men and 2
women?
Solution
To form the committee consisting of 3 men and 2 women, we need to calculate
the number of ways we can choose 3 men from 10 men and 2 women from 8
women.
Step 1: Calculate the number of ways to choose 3 men from 10
men (10
3)=10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
There are 120 ways to choose 3 men from 10 men.
Step 2: Calculate the number of ways to choose 2 women from 8
women (8
2)=8!
2!(8 −2)! =8×7
2×1= 28
There are 28 ways to choose 2 women from 8 women.
Step 3: Calculate the total number of ways to form the committee
To get the total number of ways to form the committee consisting of 3 men and
2 women, we multiply the number of ways to choose men and the number of
ways to choose women.
Total ways = 120 ×28 = 3360
Therefore, the committee can be formed in 3360 ways.
Question 11
Question
In how many ways can 5 identical red balls, 4 identical blue balls, and 3 identical
green balls be arranged in a row if balls of the same color cannot be adjacent
to each other?
Solution
Step 1: Calculate the total number of ways to arrange the balls without any
restrictions.
There are a total of 12 balls, consisting of 5 red balls, 4 blue balls, and 3 green
balls. The number of ways to arrange these 12 balls without any restrictions
9
is given by the formula for permutations of objects with repetitions, which is
12!
5!4!3! .
Step 2: Calculate the number of ways to arrange the balls with at least two
adjacent balls of the same color.
Firstly, calculate the number of ways to arrange the red balls with at least
two adjacent red balls. Treat the 5 red balls as a single entity, which can be
arranged in 5! ways. Within this arrangement, the red balls themselves can be
arranged in 5! ways. Therefore, the total number of ways to arrange the red
balls with at least two adjacent red balls is 5! ×5!.
Similarly, calculate the number of ways to arrange the blue balls with at
least two adjacent blue balls (4! ×4!) and the number of ways to arrange the
green balls with at least two adjacent green balls (3! ×3!).
Step 3: Subtract the number of ways with at least two adjacent balls of the
same color from the total number of ways to arrange the balls.
The total number of ways to arrange the balls without any restrictions is
12!
5!4!3! , and the number of ways with at least two adjacent balls of the same color
is 5! ×5! + 4! ×4! + 3! ×3!.
Subtracting the number of ways with at least two adjacent balls of the same
color from the total number of ways gives us the final answer.
Therefore, the number of ways the balls can be arranged in a row if balls of
the same color cannot be adjacent to each other is 12!
5!4!3! −(5!×5!+4!×4!+3!×3!).
Question 12
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 women, how many different committees
can be formed?
Solution
Step 1: Find the number of committees with exactly 2, 3, 4, or 5 women.
• With exactly 2 women: Choose 2 women out of 6 and 3 men out of 8.
(6
2)×(8
3)= 15 ×56 = 840
• With exactly 3 women: Choose 3 women out of 6 and 2 men out of 8.
(6
3)×(8
2)= 20 ×28 = 560
• With exactly 4 women: Choose 4 women out of 6 and 1 man out of 8.
(6
4)×(8
1)= 15 ×8 = 120
10
• With all 5 women: Choose 5 women out of 6 and 0 men out of 8.
(6
5)×(8
0)= 6 ×1 = 6
Step 2: Add up the number of committees from each case.
840 + 560 + 120 + 6 = 1526
Therefore, there are 1526 different committees that can be formed with at
least 2 women.
Question 13
Question
A committee of 5 people is to be formed from a group of 9 students and 7
professors. If the committee must consist of at least 2 professors, how many
different committees can be formed?
Solution
Step 1: Calculate the number of ways to choose the committee with exactly 2
professors. There are 7 professors to choose from and since there must be at least
2 professors, we choose 2 professors from 7, and the remaining 3 members must
be students chosen from 9. So, the number of ways to choose the committee
with exactly 2 professors is given by:
(7
2)×(9
3)=7!
2!(7 −2)! ×9!
3!(9 −3)!
Step 2: Calculate the number of ways to choose the committee with exactly
3 professors. Similar to the previous step, the number of ways to choose the
committee with exactly 3 professors is given by:
(7
3)×(9
2)=7!
3!(7 −3)! ×9!
2!(9 −2)!
Step 3: Calculate the total number of committees with at least 2 professors.
Adding the results from Step 1 and Step 2 will give us the total number of
committees with at least 2 professors:
7!
2!(7 −2)! ×9!
3!(9 −3)! +7!
3!(7 −3)! ×9!
2!(9 −2)!
Step 4: Calculate the total number of ways to form a committee with at
least 2 professors. Evaluate the expression obtained in Step 3 to find the total
number of different committees that can be formed. This will give you the final
answer.
11
Question 14
Question
In a group of 10 students, how many ways can we split them into two groups of
5 students each for a team competition?
Solution
To determine the number of ways we can split the 10 students into two groups
of 5 for a team competition, we will use combinatorial analysis principles.
Step 1: Calculate the total number of ways to choose 5 students out of 10.
This can be done using the combination formula: (n
k)=n!
k!(n−k)! , where nis the
total number of students and kis the number of students we want to choose. In
this case, n= 10 and k= 5. So, the total number of ways to choose 5 students
out of 10 is: (10
5)=10!
5!(10 −5)!
Calculating this gives us:
(10
5)=10!
5!5! =10 ×9×8×7×6
5×4×3×2×1= 252
So, there are 252 ways to choose 5 students out of 10.
Step 2: Divide by 2 to account for the fact that the order of the groups
doesn’t matter. Since we are splitting the students into two equal groups (each
with 5 students), the order of the groups doesn’t matter. Therefore, we need to
divide the total number of ways by 2 to avoid overcounting. Thus, the number
of ways to split 10 students into two groups of 5 for a team competition is:
252
2= 126
Therefore, there are 126 ways to split the 10 students into two groups of 5
students each for the team competition.
Question 15
Question
A committee of 6 people is to be formed from a group of 10 individuals. If 4
of the individuals are women and 6 are men, what is the probability that the
committee consists of exactly 3 women and 3 men?
12
Solution
Step 1: Calculate the total number of ways to form a committee of 6 people
from a group of 10 individuals. This is given by the combination formula (n
k)=
n!
k!(n−k)! .
Total number of ways =(10
6)
=10!
6!(10 −6)!
=10 ×9×8×7×6×5
6×5×4×3×2×1
= 210
Step 2: Calculate the number of ways to choose 3 women from 4 women and
3 men from 6 men. This is given by the product of the combination of women
and men.
Number of ways to choose 3 women =(4
3)= 4
Number of ways to choose 3 men =(6
3)= 20
Total number of ways to choose 3 women and 3 men = 4 ×20 = 80
Step 3: Calculate the probability of forming a committee with 3 women and
3 men by dividing the number of favorable outcomes by the total number of
outcomes.
Probability =Number of ways to choose 3 women and 3 men
Total number of ways
=80
210
=8
21 ≈0.381
Therefore, the probability that the committee consists of exactly 3 women
and 3 men is 8
21 or approximately 0.381.
Question 16
Question
In a committee of 6 people, there are 3 men and 3 women. If the committee
must select a president, a vice-president, and a treasurer, how many ways can
these positions be filled if the president cannot be a woman and the treasurer
cannot be a man?
13
Solution
Step 1: Determine the number of ways to select the president (a man) from the
3 men. Since the president must be a man, there are 3 choices for this position.
Step 2: Determine the number of ways to select the treasurer (a woman)
from the 3 women. Since the treasurer must be a woman, there are 3 choices
for this position.
Step 3: Determine the number of ways to select the vice-president. After
the president and treasurer have been selected, there are 4 remaining people (2
men and 2 women) for the vice-president position. Thus, there are 4 choices for
this position.
Step 4: Calculate the total number of ways to fill the positions. To find the
total number of ways to fill the positions, multiply the number of choices for
each position: 3×3×4 = 36.
Therefore, there are 36 ways to fill the positions of president, vice-president,
and treasurer in the committee.
Question 17
Question
A committee of 5 people is to be formed from a group of 7 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: Calculate the number of ways to choose 2 men from 7 men. There are
(7
2)= 21 ways to choose 2 men from 7 men.
Step 2: Calculate the number of ways to choose 2 women from 6 women.
There are (6
2)= 15 ways to choose 2 women from 6 women.
Step 3: Calculate the number of ways to choose 1 person (either a man or a
woman) from the remaining people. There are 5 ways to choose 1 person from
the remaining group of 5 people.
Step 4: Multiply the results from Step 1, Step 2, and Step 3 to find the total
number of ways to form the committee. Therefore, the total number of ways to
form the committee is 21 ×15 ×5 = 1575.
Thus, there are 1575 different committees that can be formed with at least
2 men and 2 women from the group of 7 men and 6 women.
Question 18
Question
In a group of 10 people, how many ways are there to select a committee of
3 people, where one person will be the president, one person will be the vice
14
president, and one person will be the secretary?
Solution
Step 1: First, we choose the president from the 10 people. There are 10 ways
to do this.
Step 2: After selecting the president, we choose the vice president from the
remaining 9 people. There are 9 ways to do this.
Step 3: Finally, after selecting the president and vice president, we choose
the secretary from the remaining 8 people. There are 8 ways to do this.
Step 4: To find the total number of ways to select the committee of 3 people
with specific roles, we multiply the number of ways at each step. Therefore, the
total number of ways to select the committee is 10 ×9×8 = 720 .
Question 19
Question
In a math class at Liberty University, there are 12 students: 6 males and 6
females. The professor randomly selects 3 students to compete in a math com-
petition. What is the probability that all 3 students selected are females?
Solution
Step 1: Find the total number of ways to select 3 students out of 12. There are
(12
3)=12!
3!(12−3)! =12×11×10
3×2×1= 220 ways to select 3 students out of 12.
Step 2: Find the number of ways to select 3 female students out of 6. Since
there are 6 females in the class, there are (6
3)=6!
3!(6−3)! =6×5×4
3×2×1= 20 ways to
select 3 female students out of 6.
Step 3: Find the probability of selecting 3 female students. The probability
of selecting 3 female students is given by Number of ways to select 3 females
Total number of ways to select 3 students =
20
220 =1
11 .
Question 20
Question
In a group of 10 friends, how many ways can we select a committee of 4 members?
Solution
Step 1: To solve this problem, we will use the combination formula:
C(n, k) = n!
k!(n−k)!
15
where C(n, k)represents the number of ways to choose kitems from a set of n
distinct items.
Step 2: Substituting n= 10 and k= 4 into the formula, we get:
C(10,4) = 10!
4!(10 −4)!
Step 3: Simplifying the expression in the numerator:
10! = 10 ×9×8×7×6×5×4!
Step 4: Simplifying the expression in the denominator:
4! = 4 ×3×2×1
Step 5: Substituting the simplified forms back into the formula:
C(10,4) = 10 ×9×8×7×6×5×4!
4×3×2×1×6!
Step 6: Canceling out the common factors:
C(10,4) = 10 ×9×8×7×5
4×3×2×1
Step 7: Calculating the final result:
C(10,4) = 30240
24 = 1260
Therefore, there are 1260 ways to select a committee of 4 members from a
group of 10 friends.
Question 21
Question
A committee of 5 people is to be formed from a group of 10 men and 8 women.
In how many ways can the committee be formed if it must consist of at least 2
women?
Solution
Step 1: Calculate the number of ways to form a committee with only 1 woman,
then subtract that from the total number of ways to form a committee with at
least 2 women.
Step 2: Calculate the number of ways to form a committee with only 1
woman. This can be done by selecting 1 woman from the 8 available women
and selecting 4 men from the 10 available men.
(8
1)×(10
4)= 8 ×210 = 1680 ways
16
Step 3: Calculate the total number of ways to form a committee with at
least 2 women. This can be done by selecting 2 women from the 8 available
women and selecting 3 men from the 10 available men, then adding the cases
where we select 3 women and 2 men from the available options.
(8
2)×(10
3)+(8
3)×(10
2)= 28 ×120 + 56 ×45 = 3360 + 2520 = 5880 ways
Step 4: Finally, subtract the number of ways to form a committee with only
1 woman from the total number of ways to form a committee with at least 2
women.
5880 −1680 = 4200
Therefore, there are 4200 ways to form a committee of 5 people from the
group of 10 men and 8 women if it must consist of at least 2 women.
Question 22
Question
In a group of 10 friends, how many ways can you choose a committee of 4 people
to represent the group?
Solution
Step 1: To find the number of ways to choose a committee of 4 people out of 10
friends, we will use the combination formula. The combination formula is given
by:
C(n, k) = n!
k!(n−k)!
where nis the total number of friends and kis the number of people in the
committee.
Step 2: Substituting n= 10 and k= 4 into the formula, we get:
C(10,4) = 10!
4!(10 −4)!
Step 3: Calculating the factorials:
10! = 10 ×9×8×7×6×5×4×3×2×1
4! = 4 ×3×2×1
6! = 6 ×5×4×3×2×1
Step 4: Substituting the factorials back into the formula:
C(10,4) = 10 ×9×8×7×6×5×4×3×2×1
(4 ×3×2×1)(6 ×5×4×3×2×1)
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Step 5: Simplifying the expression, we get:
C(10,4) = 10 ×9×8×7
4×3×2×1=10 ×9×7×2
2×1= 210
Therefore, there are 210 ways to choose a committee of 4 people from a
group of 10 friends.
Question 23
Question
A committee of 5 people is to be formed from a group of 10 men and 8 women. If
the committee must consist of at least 3 men and 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: Multiply the results
from Step 1 and Step 2 to find the total number of committees that can be
formed.
Step 1: The number of ways to choose 3 men from 10 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 women from 8 is given by the
combination formula:
(8
2)=8!
2!(8 −2)! =8×7
2×1= 28
Step 3: Multiplying the results from Step 1 and Step 2: 120 ×28 = 3360
Therefore, there are 3360 different committees that can be formed with at
least 3 men and 2 women.
Question 24
Question
A committee of 5 people is to be formed from a group of 10 men and 8 women.
If the committee is to consist of 3 men and 2 women, how many different com-
mittees can be formed?
18
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: 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 men from 10 is given by the
combination formula:
(10
3)=10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
So, there are 120 ways to choose 3 men from 10.
Step 2: The number of ways to choose 2 women from 8 is given by the
combination formula:
(8
2)=8!
2!(8 −2)! =8×7
2×1= 28
So, there are 28 ways to choose 2 women from 8.
Step 3: To find the total number of different committees that can be formed,
we multiply the results from Step 1 and Step 2: 120 ×28 = 3360
Therefore, there are 3360 different committees that can be formed consisting
of 3 men and 2 women from the given group.
Question 25
Question
A committee of 5 people is to be formed from a group of 10 men and 8 women.
In how many ways can the committee be formed if it must consist of at least 3
men?
Solution
Step 1: Find the number of ways to choose exactly 3 men out of 10. There are
(10
3)ways to select 3 men from a group of 10.
Step 2: Find the number of ways to choose 2 more members (men or women)
to complete the committee. Since the committee must consist of at least 3 men,
we need to select 2 more members to join the 3 men already chosen. We can
select either 2 women from the 8 available, or 1 woman and 1 man. Therefore,
the total number of ways to choose 2 more members is (8
2)+(8
1)×(10
1).
Step 3: Calculate the total number of ways to form the committee. The
total number of ways to form the committee is the product of the choices made
in Step 1 and Step 2: Total = (10
3)×((8
2)+(8
1)×(10
1)).
Calculating this expression gives us the final answer.
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