MATH 201 - INTRODUCTION TO
PROBABILITY AND STATISTICS -
Combinatorial Analysis
Question Bank - Set 1
Liberty University
Question 1
Question
In a group of 10 students, how many ways can we form a committee of 4 stu-
dents?
Solution
Step 1: To solve this problem, we will use the combination formula. The number
of ways to choose robjects from a group of nobjects is given by the formula:
(n
r)=n!
r!(n−r)!
Step 2: Here, we want to choose 4 students from a group of 10 students.
Using the combination formula, we have:
(10
4)=10!
4!(10 −4)!
Step 3: Calculating the factorials in the formula, we get:
(10
4)=10 ×9×8×7
4×3×2×1
Step 4: Simplifying further:
(10
4)= 210
Step 5: Therefore, there are 210 ways to form a committee of 4 students
from a group of 10 students.
Question 2
Question
A committee of 5 people is to be formed from a group of 10 students (5 males
and 5 females). Calculate the number of ways the committee can be formed if
it must consist of 3 males and 2 females.
Solution
Step 1: Calculate the number of ways to choose 3 males from the 5 available
males. There are (5
3)=5!
3!(5−3)! = 10 ways to choose 3 males from 5.
Step 2: Calculate the number of ways to choose 2 females from the 5 available
females. There are (5
2)=5!
2!(5−2)! = 10 ways to choose 2 females from 5.
Step 3: Multiply the results from Step 1 and Step 2 to find the total number
of ways to form the committee. Total number of ways = 10 ×10 = 100.
Therefore, there are 100 ways to form a committee of 5 people consisting of
3 males and 2 females from a group of 10 students.
Question 3
Question
In a deck of playing cards, how many ways can you choose 5 cards such that 3
are hearts and 2 are spades?
Solution
Step 1: Determine the number of ways to choose 3 hearts out of the 13 hearts
in the deck. Since there are 13 hearts in a deck of playing cards, the number of
ways to choose 3 hearts is given by the combination formula:
(13
3)=13!
3!(13 −3)! =13 ×12 ×11
3×2×1= 286
Step 2: Determine the number of ways to choose 2 spades out of the 13
spades in the deck. Similarly, the number of ways to choose 2 spades is given
by the combination formula:
(13
2)=13!
2!(13 −2)! =13 ×12
2×1= 78
Step 3: Determine the total number of ways to choose 5 cards (3 hearts
and 2 spades). To determine the total number of ways to choose 5 cards with
3 hearts and 2 spades, you need to multiply the number of ways to choose 3
hearts and 2 spades.
Total ways =(13
3)×(13
2)= 286 ×78 = 22236
2
Therefore, there are 22,236 ways to choose 5 cards from a deck of playing
cards such that 3 are hearts and 2 are spades.
Question 4
Question
In a group of 12 students, how many ways are there to form a committee of
4 students if the committee must include exactly 1 male student and 1 female
student?
Solution
To solve this problem, we can break it down into two steps: first, selecting one
male student and one female student for the committee, and then selecting the
remaining two students from the remaining group.
Step 1: Selecting one male student and one female student Since
there are 12 students in total, the number of ways to select 1 male student out
of 6 male students is (6
1)= 6 ways. Similarly, the number of ways to select 1
female student out of 6 female students is (6
1)= 6 ways.
So, the total number of ways to select one male student and one female
student is 6×6 = 36 ways.
Step 2: Selecting the remaining two students After selecting one male
student and one female student, there are 10 students remaining. We need to
select 2 students from this group.
The number of ways to select 2 students out of 10 students is (10
2)=
10!
2!(10−2)! = 45 ways.
Combining both steps To find the total number of ways to form the
committee of 4 students with exactly 1 male and 1 female, we multiply the
number of ways in Step 1 and Step 2:
Total number of ways = 36 ×45 = 1620 ways.
Therefore, there are 1620 ways to form a committee of 4 students with
exactly 1 male student and 1 female student from a group of 12 students.
Question 5
Question
In how many ways can you arrange the letters of the word ”UNIVERSITY”
such that no two vowels are adjacent?
Solution
Step 1: Calculate the total number of ways to arrange the letters of the word
”UNIVERSITY”. There are 10 letters in the word ”UNIVERSITY”, but the
3
letter ”I” appears twice. Therefore, the total number of ways to arrange the
letters is 10!
2! .
Step 2: Calculate the number of ways to arrange the vowels (U, I, E) such
that no two vowels are adjacent. Consider the vowels (U, I, E) as one group.
This group can be arranged in 3! ways. Within this group, the vowels I can be
arranged in 2! ways.
Step 3: Calculate the number of ways to arrange the consonants (N, V, R,
S, T, Y). There are 6 consonants left to be arranged after grouping the vowels.
These can be arranged in 6! ways.
Step 4: Calculate the total number of ways to arrange the letters such that
no two vowels are adjacent. The number of ways to arrange the vowels without
adjacency is the product of the number of ways to arrange the vowels (Step 2)
and the number of ways to arrange the consonants (Step 3).
Therefore, the total number of ways to arrange the letters of the word ”UNI-
VERSITY” such that no two vowels are adjacent is
3! ×2! ×6! = 6 ×2×720 = 8640.
Question 6
Question
In a group of 10 friends, how many ways can we select a committee of 3 members
and a subcommittee of 2 members from the same group?
Solution
Step 1: Calculate the number of ways to select a committee of 3 members from
10 friends. This can be calculated using the combination formula (n
k)=n!
k!(n−k)! ,
where nis the total number of friends and kis the number of members in the
committee.
Number of ways to select committee of 3 members =(10
3)=10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
Step 2: Calculate the number of ways to select a subcommittee of 2 members
from the same group. This can be calculated in the same way as in Step 1.
Number of ways to select subcommittee of 2 members =(10
2)=10!
2!(10 −2)! =10 ×9
2×1= 45
Step 3: Multiply the results from Step 1 and Step 2 to find the total number
of ways to select a committee of 3 members and a subcommittee of 2 members
from the group of 10 friends.
Total number of ways to select both committees = 120 ×45 = 5400
Therefore, there are 5400 ways to select a committee of 3 members and a
subcommittee of 2 members from a group of 10 friends.
4
Question 7
Question
A committee of 5 people is to be chosen from a group of 10 students. How
many different committees can be formed if 2 particular students refuse to serve
together on the same committee?
Solution
Step 1: First, we find the total number of ways to choose a committee of 5
people from 10 students.
(10
5)=10!
5!(10 −5)! =10 ×9×8×7×6
5×4×3×2×1= 252
Step 2: Next, we find the number of ways in which the 2 particular students
can serve together on the same committee. Since the 2 students must be on
the same committee, we treat them as a single unit. So, there are 8 remaining
students from which to choose 3 more to accompany the pair.
(8
3)=8!
3!(8 −3)! =8×7×6
3×2×1= 56
Step 3: Finally, we subtract the number of committees where the 2 particular
students are together from the total number of committees to find the number
of committees where they are not together.
252 −56 = 196
Therefore, there are 196 different committees that can be formed if the 2
particular students refuse to serve together on the same committee.
Question 8
Question
In a group of 12 people, how many ways can we choose a committee of 3 people
to serve as president, vice president, and secretary?
Solution
Step 1: To find the number of ways to select a president, we have 12 options.
Step 2: After selecting the president, there are 11 remaining people for the vice
president position. Step 3: Finally, for the secretary position, there are 10
remaining people. Step 4: To find the total number of ways to choose the
committee, we multiply the number of choices for each position:
12 ×11 ×10 = 1320
5
So, there are 1320 ways to choose a committee of 3 people to serve as president,
vice president, and secretary from a group of 12 people.
Question 9
Question
In a group of 10 students, how many ways can we form a committee of 3 students
and a subcommittee of 2 students from that committee?
Solution
To determine the number of ways we can form a committee of 3 students and
a subcommittee of 2 students from that committee, we’ll use combinatorial
analysis:
Step 1: Calculate the number of ways to choose a committee of 3 students
out of 10. We can do this by using the combination formula: (n
r)=n!
r!(n−r)! .
So, the number of ways to choose a committee of 3 students from a group of 10
is: (10
3)=10!
3!(10−3)! =10∗9∗8
3∗2∗1= 120 ways.
Step 2: Calculate the number of ways to choose a subcommittee of 2 students
from the committee of 3. Similarly, using the combination formula, the number
of ways to select 2 students from a committee of 3 is: (3
2)=3!
2!(3−2)! =3
2= 3
ways.
Step 3: Multiply the results from Step 1 and Step 2 to find the total number
of ways. The total number of ways to form a committee of 3 students and a
subcommittee of 2 students from the initial group of 10 students is: 120×3 = 360
ways.
Therefore, there are 360 different ways to form a committee of 3 students
and a subcommittee of 2 students from a group of 10 students.
Question 10
Question
In a mathematics competition, there are 6 multiple-choice questions with 4
answer choices each. If a student guesses randomly on every question, what is
the probability that the student gets at least 5 questions correct?
Solution
Step 1: Calculate the total number of possible outcomes when guessing ran-
domly on every question.
Since there are 4 answer choices for each of the 6 questions, the total number
of possible outcomes is 46.
Step 2: Calculate the number of ways to get exactly 5 questions correct.
6
To get exactly 5 questions correct, the student must guess the correct answer
for 5 questions and the incorrect answer for the remaining 1 question. There
are (6
5)ways to choose the 5 questions to answer correctly, and for each of these
combinations, there is only 1 way to answer the remaining question incorrectly.
Therefore, the number of ways to get exactly 5 questions correct is (6
5)×1
= 6 ways.
Step 3: Calculate the number of ways to get all 6 questions correct.
To get all 6 questions correct, the student must guess the correct answer for
all 6 questions. There is only 1 way to do this.
Step 4: Calculate the probability of getting at least 5 questions correct.
The probability of getting at least 5 questions correct is the sum of the
probabilities of getting exactly 5 questions correct and getting all 6 questions
correct, divided by the total number of possible outcomes.
The probability is 6+1
46=7
4096.
Question 11
Question
A committee of 5 people is to be formed from a group of 6 men and 4 women.
In how many ways can the committee be formed if it must contain at least 2
men and 2 women?
Solution
Step 1: Calculate the number of ways to select exactly 2 men and 2 women for
the committee. There are (6
2)ways to choose 2 out of 6 men, and (4
2)ways to
choose 2 out of 4 women. Therefore, the number of ways to select exactly 2 men
and 2 women is (6
2)×(4
2).
Step 2: Calculate the number of ways to select 3 men and 2 women for the
committee. There are (6
3)ways to choose 3 out of 6 men, and (4
2)ways to choose
2 out of 4 women. Therefore, the number of ways to select 3 men and 2 women
is (6
3)×(4
2).
Step 3: Calculate the number of ways to select 4 men and 1 woman for the
committee. There are (6
4)ways to choose 4 out of 6 men, and (4
1)ways to choose
1 out of 4 women. Therefore, the number of ways to select 4 men and 1 woman
is (6
4)×(4
1).
Step 4: Add the results from Step 1, Step 2, and Step 3 to find the total
number of ways to form the committee. Total number of ways = (6
2)×(4
2)+
(6
3)×(4
2)+(6
4)×(4
1).
Finally, calculate the total number of ways to form the committee by eval-
uating the expression.
7
Question 12
Question
In how many ways can a committee of 5 people be selected from a group of 10
people if 2 of them refuse to serve together?
Solution
Step 1: First, we find the total number of ways to select a committee of 5 people
from a group of 10 people. This can be done using the combination formula:
(n
r)=n!
r!(n−r)!
Total ways to select a committee of 5 people =(10
5)
=10!
5!(10 −5)!
=10 ×9×8×7×6
5×4×3×2×1
= 252
So, there are 252 total ways to select a committee of 5 people from a group
of 10 people.
Step 2: Next, we find the number of ways to select a committee of 5 people in
which the 2 people who refuse to serve together are selected. This can be done
by considering those two people as one unit, and then selecting the remaining
3 people from the group of 8 people left.
Ways to select the 2 people who refuse to serve together = 1 (considered as one unit)
Ways to select the remaining 3 people from the group of 8 people left =(8
3)
=8!
3!(8 −3)!
=8×7×6
3×2×1
= 56
Therefore, there are 56 ways to select a committee of 5 people where the 2
people who refuse to serve together are selected.
Step 3: Finally, we subtract the number of ways to select a committee where
the 2 people who refuse to serve together are selected from the total number of
ways to select a committee to find the number of ways to select a committee
where the 2 people do not serve together.
8
Ways to select a committee where the 2 people do not serve together =Total ways −Ways with the 2 people selected together
= 252 −56
= 196
Therefore, there are 196 ways to select a committee of 5 people from a group
of 10 people such that the 2 people who refuse to serve together are not both
selected.
Question 13
Question
In how many ways can you arrange the letters in the word ”STATISTICS” if no
two vowels can be next to each other?
Solution
Step 1: Count the total number of ways to arrange the letters in ”STATISTICS”.
There are 10 letters in the word ”STATISTICS”, but the letter ’S’ appears 3
times and the letter ’T’ appears 3 times. So, the total number of ways to arrange
the letters is 10!
3!3! .
Step 2: Count the number of ways when the vowels are grouped together.
Let’s consider the group of vowels ’AII’, which is treated as a single unit. So,
we have 6 units to consider: - AII - S - T - T - S - C - T - S - T - C Now, the
number of ways to arrange these 6 units is 6!.
Step 3: Count the number of ways when the vowels are treated as distinct
letters. Now we will consider the vowels ’A’, ’I’, ’I’ as distinct and not grouped
together. So, we have 7 units to consider: - A - I - I - S - T - S - T - S - T - C
Now, the number of ways to arrange these 7 units is 7!.
Step 4: Calculating the total number of ways when no two vowels can be
next to each other. The total number of ways can be calculated as total ways -
ways when the vowels are grouped together + ways when the vowels are treated
as distinct. So, the total number of ways =10!
3!3! −6! + 7!.
Question 14
Question
A committee of 5 people is to be formed from a group of 8 men and 6 women. If
the committee must include at least 2 women, how many different committees
can be formed?
9
Solution
Step 1: Calculate the number of ways to form a committee with 2, 3, 4, or 5
women.
Let’s consider the following cases: - Case 1: Committee with 2 women and
3 men There are (6
2)ways to choose 2 women from 6, and (8
3)ways to choose 3
men from 8. The total number of committees in this case is (6
2)×(8
3).
- Case 2: Committee with 3 women and 2 men There are (6
3)ways to choose
3 women from 6, and (8
2)ways to choose 2 men from 8. The total number of
committees in this case is (6
3)×(8
2).
- Case 3: Committee with 4 women and 1 man There are (6
4)ways to choose
4 women from 6, and (8
1)ways to choose 1 man from 8. The total number of
committees in this case is (6
4)×(8
1).
- Case 4: Committee with 5 women and 0 men There is only (6
5)way to
choose all 5 women from 6. The total number of committees in this case is (6
5).
Step 2: Calculate the total number of committees by summing up the results
from the different cases.
The total number of committees is:
(6
2)×(8
3)+(6
3)×(8
2)+(6
4)×(8
1)+(6
5)
= 15 ×56 + 20 ×28 + 15 ×8+6
= 840 + 560 + 120 + 6
= 1526
Therefore, there are 1526 different committees that can be formed with at
least 2 women.
Question 15
Question
In a deck of 52 playing cards, how many different ways can you choose 5 cards
if you want exactly 2 of them to be aces?
Solution
Step 1: Calculate the number of ways to choose 2 aces out of the 4 available.
- Since there are 4 aces in a deck of 52 cards, the number of ways to choose 2
aces is given by (4
2)=4!
2!(4−2)! = 6.
Step 2: Calculate the number of ways to choose the remaining 3 non-ace
cards from the 48 remaining cards (52 total cards minus the 4 aces). - The
number of ways to choose 3 non-ace cards from the remaining 48 cards is (48
3)=
48!
3!(48−3)! = 17,296.
10
Step 3: Multiply the results from Step 1 and Step 2 to determine the total
number of ways to choose 5 cards with exactly 2 aces. - The total number of
ways to choose 5 cards with exactly 2 aces is 6×17,296 = 103,776.
Therefore, there are 103,776 different ways to choose 5 cards from a deck of
52 cards if exactly 2 of them are aces.
Question 16
Question
A committee of 5 people is to be formed from a group of 7 men and 4 women.
In how many ways can the committee be formed if it must contain at least 2
men and 2 women?
Solution
Step 1: Calculate the number of ways to choose 2 men and 3 women. Step 2:
Calculate the number of ways to choose 3 men and 2 women. Step 3: Add the
results from Step 1 and Step 2 to find the total number of ways to form the
committee.
Step 1: To choose 2 men from 7, we use the combination formula (n
k)=
n!
k!(n−k)! . Therefore, the number of ways to choose 2 men from 7 is:
(7
2)=7!
2!(7 −2)! =7×6
2×1= 21
To choose 3 women from 4, we use the combination formula. So, the number
of ways to choose 3 women from 4 is:
(4
3)=4!
3!(4 −3)! =4
1= 4
Therefore, the total number of ways to choose 2 men and 3 women is 21×4 =
84 ways.
Step 2: To choose 3 men from 7, the number of ways is:
(7
3)=7!
3!(7 −3)! =7×6×5
3×2×1= 35
To choose 2 women from 4, the number of ways is:
(4
2)=4!
2!(4 −2)! =4×3
2×1= 6
Therefore, the total number of ways to choose 3 men and 2 women is 35×6 =
210 ways.
Step 3: Finally, the total number of ways to form the committee with at
least 2 men and 2 women is the sum of the results from Step 1 and Step 2:
84 + 210 = 294 ways.
11
Question 17
Question
A committee of 5 people is to be formed from a group of 10 students (5 boys
and 5 girls). If the committee must consist of exactly 3 girls and 2 boys, how
many different ways can the committee be formed?
Solution
Step 1: Calculate the number of ways to select 3 girls from a group of 5 girls.
There are (5
3)ways to select 3 girls from 5 girls. (5
3)=5!
3!(5−3)! =5×4×3
3×2×1= 10
ways.
Step 2: Calculate the number of ways to select 2 boys from a group of 5 boys.
There are (5
2)ways to select 2 boys from 5 boys. (5
2)=5!
2!(5−2)! =5×4
2×1= 10
ways.
Step 3: Multiply the number of ways in Step 1 and Step 2 to find the total
number of ways to form the committee. Total number of ways = (5
3)×(5
2)=
10 ×10 = 100.
Therefore, there are 100 different ways to form a committee with exactly 3
girls and 2 boys from a group of 10 students.
Question 18
Question
In a group of 10 students, how many ways can we choose a committee of 4
students if 2 of them refuse to serve together?
Solution
Step 1: First, we find the total number of ways to choose a committee of
4 students from a group of 10 students. Since the order of selection does
not matter and repetitions are not allowed, we use the combination formula:
C(n, r) = n!
r!(n−r)! . Here, n= 10 and r= 4. The total number of ways is:
C(10,4) = 10!
4!(10 −4)! =10!
4!6! = 210
Step 2: Next, we find the number of ways to choose a committee of 4 stu-
dents such that the 2 students who refuse to serve together are included in the
committee. There are 8 students left to choose from, and we need to select 2
more students to complete the committee. Using the combination formula, the
number of ways to choose 2 students from 8 is:
C(8,2) = 8!
2!(8 −2)! =8!
2!6! = 28
12
Step 3: Finally, we calculate the number of ways to choose a committee of
4 students where the 2 students who refuse to serve together are not included.
This can be done by subtracting the number of ways to choose a committee
with the 2 students from the total number of ways to choose a committee:
210 −28 = 182
Therefore, there are 182 ways to choose a committee of 4 students from a
group of 10 students if 2 of them refuse to serve together.
Question 19
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 3 men and exactly 2 women, how many
different committees can be formed?
Solution
Step 1: Calculate the number of ways to choose 3 men out of 7. There are (7
3)
ways to choose 3 men from a group of 7.
Step 2: Calculate the number of ways to choose 2 women out of 6. There
are (6
2)ways to choose 2 women from a group of 6.
Step 3: Multiply the results of Step 1 and Step 2 to determine the total
number of committees that can be formed with at least 3 men and exactly 2
women. (7
3)×(6
2)=7!
3!4! ×6!
2!4!
Step 4: Simplify the expression. 7!
3!4! ×6!
2!4! =7×6×5
3×2×1×6×5
2×1
Step 5: Compute the final answer. 7×6×5
3×2×1×6×5
2×1= 35 ×15 = 525
Therefore, there are 525 different committees that can be formed with at
least 3 men and exactly 2 women.
Question 20
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 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 10. Step 2:
Calculate the number of ways to choose 2 men out of 8. Step 3: Calculate
the number of ways to choose the remaining person (can be either a man or a
13
woman). Step 4: Multiply the results from Steps 1, 2, and 3 to find the total
number of committees.
Step 1: There are (10
2)ways to choose 2 women out of 10. (10
2)=10!
2!(10−2)! =
10×9
2×1= 45
Step 2: There are (8
2)ways to choose 2 men out of 8. (8
2)=8!
2!(8−2)! =8×7
2×1=
28
Step 3: Since the committee must consist of at least 2 women and 2 men, the
remaining person can be either a man or a woman. So, there are 10 women and
8 men left to choose from. There are 10 + 8 = 18 ways to choose the remaining
person.
Step 4: Multiply the results from Steps 1, 2, and 3 to find the total number
of committees: Total number of committees = 45 ×28 ×18 = 22680
Therefore, there are 22,680 different committees that can be formed with at
least 2 women and 2 men.
Question 21
Question
A committee of 5 people is to be formed from a group of 10 men and 8 women.
How many ways can the committee be formed if at least 2 men and 2 women
must be included?
Solution
Step 1: Calculate the number of ways to choose 2 men and 2 women out of 10
men and 8 women.
There are (10
2)ways to choose 2 men from the 10 men, and (8
2)ways to
choose 2 women from the 8 women.
So, the number of ways to choose 2 men and 2 women is (10
2)×(8
2).
Step 2: Calculate the number of ways to choose the remaining person on the
committee.
After selecting 2 men and 2 women, there are 6 people left to choose from -
8 men and 6 women.
We need to choose 1 more person to complete the committee, so there are
(14
1)ways to choose 1 more person.
Step 3: Calculate the total number of ways the committee can be formed.
To find the total number of ways the committee can be formed, multiply the
number of ways from Step 1 and Step 2.
Therefore, the total number of ways the committee can be formed if at least
2 men and 2 women must be included is:
(10
2)×(8
2)×(14
1)
14
Question 22
Question
In a class of 30 students, how many ways can a group of 5 students be selected
to form a project team if two particular students, Alice and Bob, must both be
on the team?
Solution
To solve this problem, we can use the concept of combinations, denoted by (n
k),
which represents the number of ways to choose kobjects from a set of nobjects
without regard to order.
Step 1: First, we select Alice and Bob to be on the team. This leaves us
with the task of selecting 3 more students from the remaining 28 students.
Step 2: The number of ways to select 3 students from 28 is given by (28
3).
Step 3: To find the total number of ways to form the project team with
Alice and Bob on it, we multiply the results from Step 2 and 3.
Step 4: Therefore, the total number of ways to select a group of 5 students
with Alice and Bob on the team is (28
3)=28!
3!(28−3)! =28×27×26
3×2×1= 3276.
Hence, there are 3276 ways to select a project team of 5 students where both
Alice and Bob must be included.
Question 23
Question
A committee of 5 members is to be formed from a group of 8 men and 7 women.
How many different committees can be formed if the committee must consist of
at least 2 men and 2 women?
Solution
Step 1: Calculate the number of ways to choose 2 men from 8 men. Step
2: Calculate the number of ways to choose 2 women from 7 women. Step 3:
Calculate the number of ways to choose the remaining member (either man or
woman) to complete the committee. Step 4: Use the multiplication principle to
find the total number of possible committees.
Step 1: There are (8
2)ways to choose 2 men from 8 men.
(8
2)=8!
2!(8 −2)! =8×7
2×1= 28
Step 2: There are (7
2)ways to choose 2 women from 7 women.
(7
2)=7!
2!(7 −2)! =7×6
2×1= 21
15
Step 3: We need at least 2 men and 2 women in the committee, so the
remaining member must be either a man or a woman. Therefore, the remaining
member can be chosen from the remaining 6 men and 5 women in (6
1)ways if
the remaining member is a man, or in (5
1)ways if the remaining member is a
woman. (6
1)= 6 ways (if the remaining member is a man)
(5
1)= 5 ways (if the remaining member is a woman)
Step 4: To find the total number of possible committees, multiply the number
of ways for each step together: Total number of committees = 28 ×21 ×(6 + 5)
= 588 ×11
= 6468
Therefore, there are 6468 different committees that can be formed if the
committee must consist of at least 2 men and 2 women.
Question 24
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 at least 2 women, how many different committees
can be formed?
Solution
Step 1: Calculate the number of ways to select a committee with exactly 2
women and 2 men. There are (8
2)ways to select 2 women from the 8 available,
and (6
2)ways to select 2 men from the 6 available. Therefore, the number of
ways to select a committee with exactly 2 women and 2 men is given by:
(8
2)×(6
2)= 28 ×15 = 420
Step 2: Calculate the number of ways to select a committee with 3 women
and 1 man. There are (8
3)ways to select 3 women from the 8 available, and (6
1)
ways to select 1 man from the 6 available. Therefore, the number of ways to
select a committee with 3 women and 1 man is given by:
(8
3)×(6
1)= 56 ×6 = 336
Step 3: Calculate the number of ways to select a committee with all 4
members being women. There are (8
4)ways to select 4 women from the 8
16
available. Therefore, the number of ways to select a committee with all women
is given by:
(8
4)= 70
Step 4: Add the results from Steps 1, 2, and 3 to find the total number of
different committees that can be formed.
420 + 336 + 70 = 826
Therefore, there are 826 different committees that can be formed from the
group of 8 women and 6 men, where the committee must consist of at least 2
women.
Question 25
Question
A committee of 5 people is to be formed from a group of 8 men and 6 women.
How many different committees can be formed if there must be at least 2 women
on the committee?
Solution
Step 1: Calculate the number of ways to choose 2 women out of 6. There are
(6
2)= 15 ways to choose 2 women from 6.
Step 2: Calculate the number of ways to choose the remaining 3 people (men
and women). Since there must be at least 2 women on the committee, we can
have either 2 women and 3 men or all 3 women. 2 women and 3 men: (8
3)= 56
ways to choose 3 men from 8. All 3 women: 1 way.
Step 3: Calculate the total number of committees that can be formed. The
total number of committees that can be formed is the product of the choices
made in steps 1 and 2. For committees with 2 women and 3 men: 15 ×56 = 840
committees. For committees with all 3 women: 1×1 = 1 committee.
Therefore, there are a total of 840 + 1 = 841 different committees that can
be formed with at least 2 women.
17
Question 2
Question
A committee of 5 people is to be formed from a group of 10 students (5 males
and 5 females). Calculate the number of ways the committee can be formed if
it must consist of 3 males and 2 females.
Solution
Step 1: Calculate the number of ways to choose 3 males from the 5 available
males. There are (5
3)=5!
3!(5−3)! = 10 ways to choose 3 males from 5.
Step 2: Calculate the number of ways to choose 2 females from the 5 available
females. There are (5
2)=5!
2!(5−2)! = 10 ways to choose 2 females from 5.
Step 3: Multiply the results from Step 1 and Step 2 to find the total number
of ways to form the committee. Total number of ways = 10 ×10 = 100.
Therefore, there are 100 ways to form a committee of 5 people consisting of
3 males and 2 females from a group of 10 students.
Question 3
Question
In a deck of playing cards, how many ways can you choose 5 cards such that 3
are hearts and 2 are spades?
Solution
Step 1: Determine the number of ways to choose 3 hearts out of the 13 hearts
in the deck. Since there are 13 hearts in a deck of playing cards, the number of
ways to choose 3 hearts is given by the combination formula:
(13
3)=13!
3!(13 −3)! =13 ×12 ×11
3×2×1= 286
Step 2: Determine the number of ways to choose 2 spades out of the 13
spades in the deck. Similarly, the number of ways to choose 2 spades is given
by the combination formula:
(13
2)=13!
2!(13 −2)! =13 ×12
2×1= 78
Step 3: Determine the total number of ways to choose 5 cards (3 hearts
and 2 spades). To determine the total number of ways to choose 5 cards with
3 hearts and 2 spades, you need to multiply the number of ways to choose 3
hearts and 2 spades.
Total ways =(13
3)×(13
2)= 286 ×78 = 22236
2
Therefore, there are 22,236 ways to choose 5 cards from a deck of playing
cards such that 3 are hearts and 2 are spades.
Question 4
Question
In a group of 12 students, how many ways are there to form a committee of
4 students if the committee must include exactly 1 male student and 1 female
student?
Solution
To solve this problem, we can break it down into two steps: first, selecting one
male student and one female student for the committee, and then selecting the
remaining two students from the remaining group.
Step 1: Selecting one male student and one female student Since
there are 12 students in total, the number of ways to select 1 male student out
of 6 male students is (6
1)= 6 ways. Similarly, the number of ways to select 1
female student out of 6 female students is (6
1)= 6 ways.
So, the total number of ways to select one male student and one female
student is 6×6 = 36 ways.
Step 2: Selecting the remaining two students After selecting one male
student and one female student, there are 10 students remaining. We need to
select 2 students from this group.
The number of ways to select 2 students out of 10 students is (10
2)=
10!
2!(10−2)! = 45 ways.
Combining both steps To find the total number of ways to form the
committee of 4 students with exactly 1 male and 1 female, we multiply the
number of ways in Step 1 and Step 2:
Total number of ways = 36 ×45 = 1620 ways.
Therefore, there are 1620 ways to form a committee of 4 students with
exactly 1 male student and 1 female student from a group of 12 students.
Question 5
Question
In how many ways can you arrange the letters of the word ”UNIVERSITY”
such that no two vowels are adjacent?
Solution
Step 1: Calculate the total number of ways to arrange the letters of the word
”UNIVERSITY”. There are 10 letters in the word ”UNIVERSITY”, but the
3
letter ”I” appears twice. Therefore, the total number of ways to arrange the
letters is 10!
2! .
Step 2: Calculate the number of ways to arrange the vowels (U, I, E) such
that no two vowels are adjacent. Consider the vowels (U, I, E) as one group.
This group can be arranged in 3! ways. Within this group, the vowels I can be
arranged in 2! ways.
Step 3: Calculate the number of ways to arrange the consonants (N, V, R,
S, T, Y). There are 6 consonants left to be arranged after grouping the vowels.
These can be arranged in 6! ways.
Step 4: Calculate the total number of ways to arrange the letters such that
no two vowels are adjacent. The number of ways to arrange the vowels without
adjacency is the product of the number of ways to arrange the vowels (Step 2)
and the number of ways to arrange the consonants (Step 3).
Therefore, the total number of ways to arrange the letters of the word ”UNI-
VERSITY” such that no two vowels are adjacent is
3! ×2! ×6! = 6 ×2×720 = 8640.
Question 6
Question
In a group of 10 friends, how many ways can we select a committee of 3 members
and a subcommittee of 2 members from the same group?
Solution
Step 1: Calculate the number of ways to select a committee of 3 members from
10 friends. This can be calculated using the combination formula (n
k)=n!
k!(n−k)! ,
where nis the total number of friends and kis the number of members in the
committee.
Number of ways to select committee of 3 members =(10
3)=10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
Step 2: Calculate the number of ways to select a subcommittee of 2 members
from the same group. This can be calculated in the same way as in Step 1.
Number of ways to select subcommittee of 2 members =(10
2)=10!
2!(10 −2)! =10 ×9
2×1= 45
Step 3: Multiply the results from Step 1 and Step 2 to find the total number
of ways to select a committee of 3 members and a subcommittee of 2 members
from the group of 10 friends.
Total number of ways to select both committees = 120 ×45 = 5400
Therefore, there are 5400 ways to select a committee of 3 members and a
subcommittee of 2 members from a group of 10 friends.
4
Question 7
Question
A committee of 5 people is to be chosen from a group of 10 students. How
many different committees can be formed if 2 particular students refuse to serve
together on the same committee?
Solution
Step 1: First, we find the total number of ways to choose a committee of 5
people from 10 students.
(10
5)=10!
5!(10 −5)! =10 ×9×8×7×6
5×4×3×2×1= 252
Step 2: Next, we find the number of ways in which the 2 particular students
can serve together on the same committee. Since the 2 students must be on
the same committee, we treat them as a single unit. So, there are 8 remaining
students from which to choose 3 more to accompany the pair.
(8
3)=8!
3!(8 −3)! =8×7×6
3×2×1= 56
Step 3: Finally, we subtract the number of committees where the 2 particular
students are together from the total number of committees to find the number
of committees where they are not together.
252 −56 = 196
Therefore, there are 196 different committees that can be formed if the 2
particular students refuse to serve together on the same committee.
Question 8
Question
In a group of 12 people, how many ways can we choose a committee of 3 people
to serve as president, vice president, and secretary?
Solution
Step 1: To find the number of ways to select a president, we have 12 options.
Step 2: After selecting the president, there are 11 remaining people for the vice
president position. Step 3: Finally, for the secretary position, there are 10
remaining people. Step 4: To find the total number of ways to choose the
committee, we multiply the number of choices for each position:
12 ×11 ×10 = 1320
5
So, there are 1320 ways to choose a committee of 3 people to serve as president,
vice president, and secretary from a group of 12 people.
Question 9
Question
In a group of 10 students, how many ways can we form a committee of 3 students
and a subcommittee of 2 students from that committee?
Solution
To determine the number of ways we can form a committee of 3 students and
a subcommittee of 2 students from that committee, we’ll use combinatorial
analysis:
Step 1: Calculate the number of ways to choose a committee of 3 students
out of 10. We can do this by using the combination formula: (n
r)=n!
r!(n−r)! .
So, the number of ways to choose a committee of 3 students from a group of 10
is: (10
3)=10!
3!(10−3)! =10∗9∗8
3∗2∗1= 120 ways.
Step 2: Calculate the number of ways to choose a subcommittee of 2 students
from the committee of 3. Similarly, using the combination formula, the number
of ways to select 2 students from a committee of 3 is: (3
2)=3!
2!(3−2)! =3
2= 3
ways.
Step 3: Multiply the results from Step 1 and Step 2 to find the total number
of ways. The total number of ways to form a committee of 3 students and a
subcommittee of 2 students from the initial group of 10 students is: 120×3 = 360
ways.
Therefore, there are 360 different ways to form a committee of 3 students
and a subcommittee of 2 students from a group of 10 students.
Question 10
Question
In a mathematics competition, there are 6 multiple-choice questions with 4
answer choices each. If a student guesses randomly on every question, what is
the probability that the student gets at least 5 questions correct?
Solution
Step 1: Calculate the total number of possible outcomes when guessing ran-
domly on every question.
Since there are 4 answer choices for each of the 6 questions, the total number
of possible outcomes is 46.
Step 2: Calculate the number of ways to get exactly 5 questions correct.
6
To get exactly 5 questions correct, the student must guess the correct answer
for 5 questions and the incorrect answer for the remaining 1 question. There
are (6
5)ways to choose the 5 questions to answer correctly, and for each of these
combinations, there is only 1 way to answer the remaining question incorrectly.
Therefore, the number of ways to get exactly 5 questions correct is (6
5)×1
= 6 ways.
Step 3: Calculate the number of ways to get all 6 questions correct.
To get all 6 questions correct, the student must guess the correct answer for
all 6 questions. There is only 1 way to do this.
Step 4: Calculate the probability of getting at least 5 questions correct.
The probability of getting at least 5 questions correct is the sum of the
probabilities of getting exactly 5 questions correct and getting all 6 questions
correct, divided by the total number of possible outcomes.
The probability is 6+1
46=7
4096.
Question 11
Question
A committee of 5 people is to be formed from a group of 6 men and 4 women.
In how many ways can the committee be formed if it must contain at least 2
men and 2 women?
Solution
Step 1: Calculate the number of ways to select exactly 2 men and 2 women for
the committee. There are (6
2)ways to choose 2 out of 6 men, and (4
2)ways to
choose 2 out of 4 women. Therefore, the number of ways to select exactly 2 men
and 2 women is (6
2)×(4
2).
Step 2: Calculate the number of ways to select 3 men and 2 women for the
committee. There are (6
3)ways to choose 3 out of 6 men, and (4
2)ways to choose
2 out of 4 women. Therefore, the number of ways to select 3 men and 2 women
is (6
3)×(4
2).
Step 3: Calculate the number of ways to select 4 men and 1 woman for the
committee. There are (6
4)ways to choose 4 out of 6 men, and (4
1)ways to choose
1 out of 4 women. Therefore, the number of ways to select 4 men and 1 woman
is (6
4)×(4
1).
Step 4: Add the results from Step 1, Step 2, and Step 3 to find the total
number of ways to form the committee. Total number of ways = (6
2)×(4
2)+
(6
3)×(4
2)+(6
4)×(4
1).
Finally, calculate the total number of ways to form the committee by eval-
uating the expression.
7
Question 12
Question
In how many ways can a committee of 5 people be selected from a group of 10
people if 2 of them refuse to serve together?
Solution
Step 1: First, we find the total number of ways to select a committee of 5 people
from a group of 10 people. This can be done using the combination formula:
(n
r)=n!
r!(n−r)!
Total ways to select a committee of 5 people =(10
5)
=10!
5!(10 −5)!
=10 ×9×8×7×6
5×4×3×2×1
= 252
So, there are 252 total ways to select a committee of 5 people from a group
of 10 people.
Step 2: Next, we find the number of ways to select a committee of 5 people in
which the 2 people who refuse to serve together are selected. This can be done
by considering those two people as one unit, and then selecting the remaining
3 people from the group of 8 people left.
Ways to select the 2 people who refuse to serve together = 1 (considered as one unit)
Ways to select the remaining 3 people from the group of 8 people left =(8
3)
=8!
3!(8 −3)!
=8×7×6
3×2×1
= 56
Therefore, there are 56 ways to select a committee of 5 people where the 2
people who refuse to serve together are selected.
Step 3: Finally, we subtract the number of ways to select a committee where
the 2 people who refuse to serve together are selected from the total number of
ways to select a committee to find the number of ways to select a committee
where the 2 people do not serve together.
8
Ways to select a committee where the 2 people do not serve together =Total ways −Ways with the 2 people selected together
= 252 −56
= 196
Therefore, there are 196 ways to select a committee of 5 people from a group
of 10 people such that the 2 people who refuse to serve together are not both
selected.
Question 13
Question
In how many ways can you arrange the letters in the word ”STATISTICS” if no
two vowels can be next to each other?
Solution
Step 1: Count the total number of ways to arrange the letters in ”STATISTICS”.
There are 10 letters in the word ”STATISTICS”, but the letter ’S’ appears 3
times and the letter ’T’ appears 3 times. So, the total number of ways to arrange
the letters is 10!
3!3! .
Step 2: Count the number of ways when the vowels are grouped together.
Let’s consider the group of vowels ’AII’, which is treated as a single unit. So,
we have 6 units to consider: - AII - S - T - T - S - C - T - S - T - C Now, the
number of ways to arrange these 6 units is 6!.
Step 3: Count the number of ways when the vowels are treated as distinct
letters. Now we will consider the vowels ’A’, ’I’, ’I’ as distinct and not grouped
together. So, we have 7 units to consider: - A - I - I - S - T - S - T - S - T - C
Now, the number of ways to arrange these 7 units is 7!.
Step 4: Calculating the total number of ways when no two vowels can be
next to each other. The total number of ways can be calculated as total ways -
ways when the vowels are grouped together + ways when the vowels are treated
as distinct. So, the total number of ways =10!
3!3! −6! + 7!.
Question 14
Question
A committee of 5 people is to be formed from a group of 8 men and 6 women. If
the committee must include at least 2 women, how many different committees
can be formed?
9
Solution
Step 1: Calculate the number of ways to form a committee with 2, 3, 4, or 5
women.
Let’s consider the following cases: - Case 1: Committee with 2 women and
3 men There are (6
2)ways to choose 2 women from 6, and (8
3)ways to choose 3
men from 8. The total number of committees in this case is (6
2)×(8
3).
- Case 2: Committee with 3 women and 2 men There are (6
3)ways to choose
3 women from 6, and (8
2)ways to choose 2 men from 8. The total number of
committees in this case is (6
3)×(8
2).
- Case 3: Committee with 4 women and 1 man There are (6
4)ways to choose
4 women from 6, and (8
1)ways to choose 1 man from 8. The total number of
committees in this case is (6
4)×(8
1).
- Case 4: Committee with 5 women and 0 men There is only (6
5)way to
choose all 5 women from 6. The total number of committees in this case is (6
5).
Step 2: Calculate the total number of committees by summing up the results
from the different cases.
The total number of committees is:
(6
2)×(8
3)+(6
3)×(8
2)+(6
4)×(8
1)+(6
5)
= 15 ×56 + 20 ×28 + 15 ×8+6
= 840 + 560 + 120 + 6
= 1526
Therefore, there are 1526 different committees that can be formed with at
least 2 women.
Question 15
Question
In a deck of 52 playing cards, how many different ways can you choose 5 cards
if you want exactly 2 of them to be aces?
Solution
Step 1: Calculate the number of ways to choose 2 aces out of the 4 available.
- Since there are 4 aces in a deck of 52 cards, the number of ways to choose 2
aces is given by (4
2)=4!
2!(4−2)! = 6.
Step 2: Calculate the number of ways to choose the remaining 3 non-ace
cards from the 48 remaining cards (52 total cards minus the 4 aces). - The
number of ways to choose 3 non-ace cards from the remaining 48 cards is (48
3)=
48!
3!(48−3)! = 17,296.
10
Step 3: Multiply the results from Step 1 and Step 2 to determine the total
number of ways to choose 5 cards with exactly 2 aces. - The total number of
ways to choose 5 cards with exactly 2 aces is 6×17,296 = 103,776.
Therefore, there are 103,776 different ways to choose 5 cards from a deck of
52 cards if exactly 2 of them are aces.
Question 16
Question
A committee of 5 people is to be formed from a group of 7 men and 4 women.
In how many ways can the committee be formed if it must contain at least 2
men and 2 women?
Solution
Step 1: Calculate the number of ways to choose 2 men and 3 women. Step 2:
Calculate the number of ways to choose 3 men and 2 women. Step 3: Add the
results from Step 1 and Step 2 to find the total number of ways to form the
committee.
Step 1: To choose 2 men from 7, we use the combination formula (n
k)=
n!
k!(n−k)! . Therefore, the number of ways to choose 2 men from 7 is:
(7
2)=7!
2!(7 −2)! =7×6
2×1= 21
To choose 3 women from 4, we use the combination formula. So, the number
of ways to choose 3 women from 4 is:
(4
3)=4!
3!(4 −3)! =4
1= 4
Therefore, the total number of ways to choose 2 men and 3 women is 21×4 =
84 ways.
Step 2: To choose 3 men from 7, the number of ways is:
(7
3)=7!
3!(7 −3)! =7×6×5
3×2×1= 35
To choose 2 women from 4, the number of ways is:
(4
2)=4!
2!(4 −2)! =4×3
2×1= 6
Therefore, the total number of ways to choose 3 men and 2 women is 35×6 =
210 ways.
Step 3: Finally, the total number of ways to form the committee with at
least 2 men and 2 women is the sum of the results from Step 1 and Step 2:
84 + 210 = 294 ways.
11
Question 17
Question
A committee of 5 people is to be formed from a group of 10 students (5 boys
and 5 girls). If the committee must consist of exactly 3 girls and 2 boys, how
many different ways can the committee be formed?
Solution
Step 1: Calculate the number of ways to select 3 girls from a group of 5 girls.
There are (5
3)ways to select 3 girls from 5 girls. (5
3)=5!
3!(5−3)! =5×4×3
3×2×1= 10
ways.
Step 2: Calculate the number of ways to select 2 boys from a group of 5 boys.
There are (5
2)ways to select 2 boys from 5 boys. (5
2)=5!
2!(5−2)! =5×4
2×1= 10
ways.
Step 3: Multiply the number of ways in Step 1 and Step 2 to find the total
number of ways to form the committee. Total number of ways = (5
3)×(5
2)=
10 ×10 = 100.
Therefore, there are 100 different ways to form a committee with exactly 3
girls and 2 boys from a group of 10 students.
Question 18
Question
In a group of 10 students, how many ways can we choose a committee of 4
students if 2 of them refuse to serve together?
Solution
Step 1: First, we find the total number of ways to choose a committee of
4 students from a group of 10 students. Since the order of selection does
not matter and repetitions are not allowed, we use the combination formula:
C(n, r) = n!
r!(n−r)! . Here, n= 10 and r= 4. The total number of ways is:
C(10,4) = 10!
4!(10 −4)! =10!
4!6! = 210
Step 2: Next, we find the number of ways to choose a committee of 4 stu-
dents such that the 2 students who refuse to serve together are included in the
committee. There are 8 students left to choose from, and we need to select 2
more students to complete the committee. Using the combination formula, the
number of ways to choose 2 students from 8 is:
C(8,2) = 8!
2!(8 −2)! =8!
2!6! = 28
12
Step 3: Finally, we calculate the number of ways to choose a committee of
4 students where the 2 students who refuse to serve together are not included.
This can be done by subtracting the number of ways to choose a committee
with the 2 students from the total number of ways to choose a committee:
210 −28 = 182
Therefore, there are 182 ways to choose a committee of 4 students from a
group of 10 students if 2 of them refuse to serve together.
Question 19
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 3 men and exactly 2 women, how many
different committees can be formed?
Solution
Step 1: Calculate the number of ways to choose 3 men out of 7. There are (7
3)
ways to choose 3 men from a group of 7.
Step 2: Calculate the number of ways to choose 2 women out of 6. There
are (6
2)ways to choose 2 women from a group of 6.
Step 3: Multiply the results of Step 1 and Step 2 to determine the total
number of committees that can be formed with at least 3 men and exactly 2
women. (7
3)×(6
2)=7!
3!4! ×6!
2!4!
Step 4: Simplify the expression. 7!
3!4! ×6!
2!4! =7×6×5
3×2×1×6×5
2×1
Step 5: Compute the final answer. 7×6×5
3×2×1×6×5
2×1= 35 ×15 = 525
Therefore, there are 525 different committees that can be formed with at
least 3 men and exactly 2 women.
Question 20
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 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 10. Step 2:
Calculate the number of ways to choose 2 men out of 8. Step 3: Calculate
the number of ways to choose the remaining person (can be either a man or a
13
woman). Step 4: Multiply the results from Steps 1, 2, and 3 to find the total
number of committees.
Step 1: There are (10
2)ways to choose 2 women out of 10. (10
2)=10!
2!(10−2)! =
10×9
2×1= 45
Step 2: There are (8
2)ways to choose 2 men out of 8. (8
2)=8!
2!(8−2)! =8×7
2×1=
28
Step 3: Since the committee must consist of at least 2 women and 2 men, the
remaining person can be either a man or a woman. So, there are 10 women and
8 men left to choose from. There are 10 + 8 = 18 ways to choose the remaining
person.
Step 4: Multiply the results from Steps 1, 2, and 3 to find the total number
of committees: Total number of committees = 45 ×28 ×18 = 22680
Therefore, there are 22,680 different committees that can be formed with at
least 2 women and 2 men.
Question 21
Question
A committee of 5 people is to be formed from a group of 10 men and 8 women.
How many ways can the committee be formed if at least 2 men and 2 women
must be included?
Solution
Step 1: Calculate the number of ways to choose 2 men and 2 women out of 10
men and 8 women.
There are (10
2)ways to choose 2 men from the 10 men, and (8
2)ways to
choose 2 women from the 8 women.
So, the number of ways to choose 2 men and 2 women is (10
2)×(8
2).
Step 2: Calculate the number of ways to choose the remaining person on the
committee.
After selecting 2 men and 2 women, there are 6 people left to choose from -
8 men and 6 women.
We need to choose 1 more person to complete the committee, so there are
(14
1)ways to choose 1 more person.
Step 3: Calculate the total number of ways the committee can be formed.
To find the total number of ways the committee can be formed, multiply the
number of ways from Step 1 and Step 2.
Therefore, the total number of ways the committee can be formed if at least
2 men and 2 women must be included is:
(10
2)×(8
2)×(14
1)
14
Question 22
Question
In a class of 30 students, how many ways can a group of 5 students be selected
to form a project team if two particular students, Alice and Bob, must both be
on the team?
Solution
To solve this problem, we can use the concept of combinations, denoted by (n
k),
which represents the number of ways to choose kobjects from a set of nobjects
without regard to order.
Step 1: First, we select Alice and Bob to be on the team. This leaves us
with the task of selecting 3 more students from the remaining 28 students.
Step 2: The number of ways to select 3 students from 28 is given by (28
3).
Step 3: To find the total number of ways to form the project team with
Alice and Bob on it, we multiply the results from Step 2 and 3.
Step 4: Therefore, the total number of ways to select a group of 5 students
with Alice and Bob on the team is (28
3)=28!
3!(28−3)! =28×27×26
3×2×1= 3276.
Hence, there are 3276 ways to select a project team of 5 students where both
Alice and Bob must be included.
Question 23
Question
A committee of 5 members is to be formed from a group of 8 men and 7 women.
How many different committees can be formed if the committee must consist of
at least 2 men and 2 women?
Solution
Step 1: Calculate the number of ways to choose 2 men from 8 men. Step
2: Calculate the number of ways to choose 2 women from 7 women. Step 3:
Calculate the number of ways to choose the remaining member (either man or
woman) to complete the committee. Step 4: Use the multiplication principle to
find the total number of possible committees.
Step 1: There are (8
2)ways to choose 2 men from 8 men.
(8
2)=8!
2!(8 −2)! =8×7
2×1= 28
Step 2: There are (7
2)ways to choose 2 women from 7 women.
(7
2)=7!
2!(7 −2)! =7×6
2×1= 21
15
Step 3: We need at least 2 men and 2 women in the committee, so the
remaining member must be either a man or a woman. Therefore, the remaining
member can be chosen from the remaining 6 men and 5 women in (6
1)ways if
the remaining member is a man, or in (5
1)ways if the remaining member is a
woman. (6
1)= 6 ways (if the remaining member is a man)
(5
1)= 5 ways (if the remaining member is a woman)
Step 4: To find the total number of possible committees, multiply the number
of ways for each step together: Total number of committees = 28 ×21 ×(6 + 5)
= 588 ×11
= 6468
Therefore, there are 6468 different committees that can be formed if the
committee must consist of at least 2 men and 2 women.
Question 24
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 at least 2 women, how many different committees
can be formed?
Solution
Step 1: Calculate the number of ways to select a committee with exactly 2
women and 2 men. There are (8
2)ways to select 2 women from the 8 available,
and (6
2)ways to select 2 men from the 6 available. Therefore, the number of
ways to select a committee with exactly 2 women and 2 men is given by:
(8
2)×(6
2)= 28 ×15 = 420
Step 2: Calculate the number of ways to select a committee with 3 women
and 1 man. There are (8
3)ways to select 3 women from the 8 available, and (6
1)
ways to select 1 man from the 6 available. Therefore, the number of ways to
select a committee with 3 women and 1 man is given by:
(8
3)×(6
1)= 56 ×6 = 336
Step 3: Calculate the number of ways to select a committee with all 4
members being women. There are (8
4)ways to select 4 women from the 8
16
available. Therefore, the number of ways to select a committee with all women
is given by:
(8
4)= 70
Step 4: Add the results from Steps 1, 2, and 3 to find the total number of
different committees that can be formed.
420 + 336 + 70 = 826
Therefore, there are 826 different committees that can be formed from the
group of 8 women and 6 men, where the committee must consist of at least 2
women.
Question 25
Question
A committee of 5 people is to be formed from a group of 8 men and 6 women.
How many different committees can be formed if there must be at least 2 women
on the committee?
Solution
Step 1: Calculate the number of ways to choose 2 women out of 6. There are
(6
2)= 15 ways to choose 2 women from 6.
Step 2: Calculate the number of ways to choose the remaining 3 people (men
and women). Since there must be at least 2 women on the committee, we can
have either 2 women and 3 men or all 3 women. 2 women and 3 men: (8
3)= 56
ways to choose 3 men from 8. All 3 women: 1 way.
Step 3: Calculate the total number of committees that can be formed. The
total number of committees that can be formed is the product of the choices
made in steps 1 and 2. For committees with 2 women and 3 men: 15 ×56 = 840
committees. For committees with all 3 women: 1×1 = 1 committee.
Therefore, there are a total of 840 + 1 = 841 different committees that can
be formed with at least 2 women.
17
Question 2
Question
A committee of 5 people is to be formed from a group of 10 students (5 males
and 5 females). Calculate the number of ways the committee can be formed if
it must consist of 3 males and 2 females.
Solution
Step 1: Calculate the number of ways to choose 3 males from the 5 available
males. There are (5
3)=5!
3!(5−3)! = 10 ways to choose 3 males from 5.
Step 2: Calculate the number of ways to choose 2 females from the 5 available
females. There are (5
2)=5!
2!(5−2)! = 10 ways to choose 2 females from 5.
Step 3: Multiply the results from Step 1 and Step 2 to find the total number
of ways to form the committee. Total number of ways = 10 ×10 = 100.
Therefore, there are 100 ways to form a committee of 5 people consisting of
3 males and 2 females from a group of 10 students.
Question 3
Question
In a deck of playing cards, how many ways can you choose 5 cards such that 3
are hearts and 2 are spades?
Solution
Step 1: Determine the number of ways to choose 3 hearts out of the 13 hearts
in the deck. Since there are 13 hearts in a deck of playing cards, the number of
ways to choose 3 hearts is given by the combination formula:
(13
3)=13!
3!(13 −3)! =13 ×12 ×11
3×2×1= 286
Step 2: Determine the number of ways to choose 2 spades out of the 13
spades in the deck. Similarly, the number of ways to choose 2 spades is given
by the combination formula:
(13
2)=13!
2!(13 −2)! =13 ×12
2×1= 78
Step 3: Determine the total number of ways to choose 5 cards (3 hearts
and 2 spades). To determine the total number of ways to choose 5 cards with
3 hearts and 2 spades, you need to multiply the number of ways to choose 3
hearts and 2 spades.
Total ways =(13
3)×(13
2)= 286 ×78 = 22236
2
Therefore, there are 22,236 ways to choose 5 cards from a deck of playing
cards such that 3 are hearts and 2 are spades.
Question 4
Question
In a group of 12 students, how many ways are there to form a committee of
4 students if the committee must include exactly 1 male student and 1 female
student?
Solution
To solve this problem, we can break it down into two steps: first, selecting one
male student and one female student for the committee, and then selecting the
remaining two students from the remaining group.
Step 1: Selecting one male student and one female student Since
there are 12 students in total, the number of ways to select 1 male student out
of 6 male students is (6
1)= 6 ways. Similarly, the number of ways to select 1
female student out of 6 female students is (6
1)= 6 ways.
So, the total number of ways to select one male student and one female
student is 6×6 = 36 ways.
Step 2: Selecting the remaining two students After selecting one male
student and one female student, there are 10 students remaining. We need to
select 2 students from this group.
The number of ways to select 2 students out of 10 students is (10
2)=
10!
2!(10−2)! = 45 ways.
Combining both steps To find the total number of ways to form the
committee of 4 students with exactly 1 male and 1 female, we multiply the
number of ways in Step 1 and Step 2:
Total number of ways = 36 ×45 = 1620 ways.
Therefore, there are 1620 ways to form a committee of 4 students with
exactly 1 male student and 1 female student from a group of 12 students.
Question 5
Question
In how many ways can you arrange the letters of the word ”UNIVERSITY”
such that no two vowels are adjacent?
Solution
Step 1: Calculate the total number of ways to arrange the letters of the word
”UNIVERSITY”. There are 10 letters in the word ”UNIVERSITY”, but the
3
letter ”I” appears twice. Therefore, the total number of ways to arrange the
letters is 10!
2! .
Step 2: Calculate the number of ways to arrange the vowels (U, I, E) such
that no two vowels are adjacent. Consider the vowels (U, I, E) as one group.
This group can be arranged in 3! ways. Within this group, the vowels I can be
arranged in 2! ways.
Step 3: Calculate the number of ways to arrange the consonants (N, V, R,
S, T, Y). There are 6 consonants left to be arranged after grouping the vowels.
These can be arranged in 6! ways.
Step 4: Calculate the total number of ways to arrange the letters such that
no two vowels are adjacent. The number of ways to arrange the vowels without
adjacency is the product of the number of ways to arrange the vowels (Step 2)
and the number of ways to arrange the consonants (Step 3).
Therefore, the total number of ways to arrange the letters of the word ”UNI-
VERSITY” such that no two vowels are adjacent is
3! ×2! ×6! = 6 ×2×720 = 8640.
Question 6
Question
In a group of 10 friends, how many ways can we select a committee of 3 members
and a subcommittee of 2 members from the same group?
Solution
Step 1: Calculate the number of ways to select a committee of 3 members from
10 friends. This can be calculated using the combination formula (n
k)=n!
k!(n−k)! ,
where nis the total number of friends and kis the number of members in the
committee.
Number of ways to select committee of 3 members =(10
3)=10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
Step 2: Calculate the number of ways to select a subcommittee of 2 members
from the same group. This can be calculated in the same way as in Step 1.
Number of ways to select subcommittee of 2 members =(10
2)=10!
2!(10 −2)! =10 ×9
2×1= 45
Step 3: Multiply the results from Step 1 and Step 2 to find the total number
of ways to select a committee of 3 members and a subcommittee of 2 members
from the group of 10 friends.
Total number of ways to select both committees = 120 ×45 = 5400
Therefore, there are 5400 ways to select a committee of 3 members and a
subcommittee of 2 members from a group of 10 friends.
4
Question 7
Question
A committee of 5 people is to be chosen from a group of 10 students. How
many different committees can be formed if 2 particular students refuse to serve
together on the same committee?
Solution
Step 1: First, we find the total number of ways to choose a committee of 5
people from 10 students.
(10
5)=10!
5!(10 −5)! =10 ×9×8×7×6
5×4×3×2×1= 252
Step 2: Next, we find the number of ways in which the 2 particular students
can serve together on the same committee. Since the 2 students must be on
the same committee, we treat them as a single unit. So, there are 8 remaining
students from which to choose 3 more to accompany the pair.
(8
3)=8!
3!(8 −3)! =8×7×6
3×2×1= 56
Step 3: Finally, we subtract the number of committees where the 2 particular
students are together from the total number of committees to find the number
of committees where they are not together.
252 −56 = 196
Therefore, there are 196 different committees that can be formed if the 2
particular students refuse to serve together on the same committee.
Question 8
Question
In a group of 12 people, how many ways can we choose a committee of 3 people
to serve as president, vice president, and secretary?
Solution
Step 1: To find the number of ways to select a president, we have 12 options.
Step 2: After selecting the president, there are 11 remaining people for the vice
president position. Step 3: Finally, for the secretary position, there are 10
remaining people. Step 4: To find the total number of ways to choose the
committee, we multiply the number of choices for each position:
12 ×11 ×10 = 1320
5
So, there are 1320 ways to choose a committee of 3 people to serve as president,
vice president, and secretary from a group of 12 people.
Question 9
Question
In a group of 10 students, how many ways can we form a committee of 3 students
and a subcommittee of 2 students from that committee?
Solution
To determine the number of ways we can form a committee of 3 students and
a subcommittee of 2 students from that committee, we’ll use combinatorial
analysis:
Step 1: Calculate the number of ways to choose a committee of 3 students
out of 10. We can do this by using the combination formula: (n
r)=n!
r!(n−r)! .
So, the number of ways to choose a committee of 3 students from a group of 10
is: (10
3)=10!
3!(10−3)! =10∗9∗8
3∗2∗1= 120 ways.
Step 2: Calculate the number of ways to choose a subcommittee of 2 students
from the committee of 3. Similarly, using the combination formula, the number
of ways to select 2 students from a committee of 3 is: (3
2)=3!
2!(3−2)! =3
2= 3
ways.
Step 3: Multiply the results from Step 1 and Step 2 to find the total number
of ways. The total number of ways to form a committee of 3 students and a
subcommittee of 2 students from the initial group of 10 students is: 120×3 = 360
ways.
Therefore, there are 360 different ways to form a committee of 3 students
and a subcommittee of 2 students from a group of 10 students.
Question 10
Question
In a mathematics competition, there are 6 multiple-choice questions with 4
answer choices each. If a student guesses randomly on every question, what is
the probability that the student gets at least 5 questions correct?
Solution
Step 1: Calculate the total number of possible outcomes when guessing ran-
domly on every question.
Since there are 4 answer choices for each of the 6 questions, the total number
of possible outcomes is 46.
Step 2: Calculate the number of ways to get exactly 5 questions correct.
6
To get exactly 5 questions correct, the student must guess the correct answer
for 5 questions and the incorrect answer for the remaining 1 question. There
are (6
5)ways to choose the 5 questions to answer correctly, and for each of these
combinations, there is only 1 way to answer the remaining question incorrectly.
Therefore, the number of ways to get exactly 5 questions correct is (6
5)×1
= 6 ways.
Step 3: Calculate the number of ways to get all 6 questions correct.
To get all 6 questions correct, the student must guess the correct answer for
all 6 questions. There is only 1 way to do this.
Step 4: Calculate the probability of getting at least 5 questions correct.
The probability of getting at least 5 questions correct is the sum of the
probabilities of getting exactly 5 questions correct and getting all 6 questions
correct, divided by the total number of possible outcomes.
The probability is 6+1
46=7
4096.
Question 11
Question
A committee of 5 people is to be formed from a group of 6 men and 4 women.
In how many ways can the committee be formed if it must contain at least 2
men and 2 women?
Solution
Step 1: Calculate the number of ways to select exactly 2 men and 2 women for
the committee. There are (6
2)ways to choose 2 out of 6 men, and (4
2)ways to
choose 2 out of 4 women. Therefore, the number of ways to select exactly 2 men
and 2 women is (6
2)×(4
2).
Step 2: Calculate the number of ways to select 3 men and 2 women for the
committee. There are (6
3)ways to choose 3 out of 6 men, and (4
2)ways to choose
2 out of 4 women. Therefore, the number of ways to select 3 men and 2 women
is (6
3)×(4
2).
Step 3: Calculate the number of ways to select 4 men and 1 woman for the
committee. There are (6
4)ways to choose 4 out of 6 men, and (4
1)ways to choose
1 out of 4 women. Therefore, the number of ways to select 4 men and 1 woman
is (6
4)×(4
1).
Step 4: Add the results from Step 1, Step 2, and Step 3 to find the total
number of ways to form the committee. Total number of ways = (6
2)×(4
2)+
(6
3)×(4
2)+(6
4)×(4
1).
Finally, calculate the total number of ways to form the committee by eval-
uating the expression.
7
Question 12
Question
In how many ways can a committee of 5 people be selected from a group of 10
people if 2 of them refuse to serve together?
Solution
Step 1: First, we find the total number of ways to select a committee of 5 people
from a group of 10 people. This can be done using the combination formula:
(n
r)=n!
r!(n−r)!
Total ways to select a committee of 5 people =(10
5)
=10!
5!(10 −5)!
=10 ×9×8×7×6
5×4×3×2×1
= 252
So, there are 252 total ways to select a committee of 5 people from a group
of 10 people.
Step 2: Next, we find the number of ways to select a committee of 5 people in
which the 2 people who refuse to serve together are selected. This can be done
by considering those two people as one unit, and then selecting the remaining
3 people from the group of 8 people left.
Ways to select the 2 people who refuse to serve together = 1 (considered as one unit)
Ways to select the remaining 3 people from the group of 8 people left =(8
3)
=8!
3!(8 −3)!
=8×7×6
3×2×1
= 56
Therefore, there are 56 ways to select a committee of 5 people where the 2
people who refuse to serve together are selected.
Step 3: Finally, we subtract the number of ways to select a committee where
the 2 people who refuse to serve together are selected from the total number of
ways to select a committee to find the number of ways to select a committee
where the 2 people do not serve together.
8
Ways to select a committee where the 2 people do not serve together =Total ways −Ways with the 2 people selected together
= 252 −56
= 196
Therefore, there are 196 ways to select a committee of 5 people from a group
of 10 people such that the 2 people who refuse to serve together are not both
selected.
Question 13
Question
In how many ways can you arrange the letters in the word ”STATISTICS” if no
two vowels can be next to each other?
Solution
Step 1: Count the total number of ways to arrange the letters in ”STATISTICS”.
There are 10 letters in the word ”STATISTICS”, but the letter ’S’ appears 3
times and the letter ’T’ appears 3 times. So, the total number of ways to arrange
the letters is 10!
3!3! .
Step 2: Count the number of ways when the vowels are grouped together.
Let’s consider the group of vowels ’AII’, which is treated as a single unit. So,
we have 6 units to consider: - AII - S - T - T - S - C - T - S - T - C Now, the
number of ways to arrange these 6 units is 6!.
Step 3: Count the number of ways when the vowels are treated as distinct
letters. Now we will consider the vowels ’A’, ’I’, ’I’ as distinct and not grouped
together. So, we have 7 units to consider: - A - I - I - S - T - S - T - S - T - C
Now, the number of ways to arrange these 7 units is 7!.
Step 4: Calculating the total number of ways when no two vowels can be
next to each other. The total number of ways can be calculated as total ways -
ways when the vowels are grouped together + ways when the vowels are treated
as distinct. So, the total number of ways =10!
3!3! −6! + 7!.
Question 14
Question
A committee of 5 people is to be formed from a group of 8 men and 6 women. If
the committee must include at least 2 women, how many different committees
can be formed?
9
Solution
Step 1: Calculate the number of ways to form a committee with 2, 3, 4, or 5
women.
Let’s consider the following cases: - Case 1: Committee with 2 women and
3 men There are (6
2)ways to choose 2 women from 6, and (8
3)ways to choose 3
men from 8. The total number of committees in this case is (6
2)×(8
3).
- Case 2: Committee with 3 women and 2 men There are (6
3)ways to choose
3 women from 6, and (8
2)ways to choose 2 men from 8. The total number of
committees in this case is (6
3)×(8
2).
- Case 3: Committee with 4 women and 1 man There are (6
4)ways to choose
4 women from 6, and (8
1)ways to choose 1 man from 8. The total number of
committees in this case is (6
4)×(8
1).
- Case 4: Committee with 5 women and 0 men There is only (6
5)way to
choose all 5 women from 6. The total number of committees in this case is (6
5).
Step 2: Calculate the total number of committees by summing up the results
from the different cases.
The total number of committees is:
(6
2)×(8
3)+(6
3)×(8
2)+(6
4)×(8
1)+(6
5)
= 15 ×56 + 20 ×28 + 15 ×8+6
= 840 + 560 + 120 + 6
= 1526
Therefore, there are 1526 different committees that can be formed with at
least 2 women.
Question 15
Question
In a deck of 52 playing cards, how many different ways can you choose 5 cards
if you want exactly 2 of them to be aces?
Solution
Step 1: Calculate the number of ways to choose 2 aces out of the 4 available.
- Since there are 4 aces in a deck of 52 cards, the number of ways to choose 2
aces is given by (4
2)=4!
2!(4−2)! = 6.
Step 2: Calculate the number of ways to choose the remaining 3 non-ace
cards from the 48 remaining cards (52 total cards minus the 4 aces). - The
number of ways to choose 3 non-ace cards from the remaining 48 cards is (48
3)=
48!
3!(48−3)! = 17,296.
10
Step 3: Multiply the results from Step 1 and Step 2 to determine the total
number of ways to choose 5 cards with exactly 2 aces. - The total number of
ways to choose 5 cards with exactly 2 aces is 6×17,296 = 103,776.
Therefore, there are 103,776 different ways to choose 5 cards from a deck of
52 cards if exactly 2 of them are aces.
Question 16
Question
A committee of 5 people is to be formed from a group of 7 men and 4 women.
In how many ways can the committee be formed if it must contain at least 2
men and 2 women?
Solution
Step 1: Calculate the number of ways to choose 2 men and 3 women. Step 2:
Calculate the number of ways to choose 3 men and 2 women. Step 3: Add the
results from Step 1 and Step 2 to find the total number of ways to form the
committee.
Step 1: To choose 2 men from 7, we use the combination formula (n
k)=
n!
k!(n−k)! . Therefore, the number of ways to choose 2 men from 7 is:
(7
2)=7!
2!(7 −2)! =7×6
2×1= 21
To choose 3 women from 4, we use the combination formula. So, the number
of ways to choose 3 women from 4 is:
(4
3)=4!
3!(4 −3)! =4
1= 4
Therefore, the total number of ways to choose 2 men and 3 women is 21×4 =
84 ways.
Step 2: To choose 3 men from 7, the number of ways is:
(7
3)=7!
3!(7 −3)! =7×6×5
3×2×1= 35
To choose 2 women from 4, the number of ways is:
(4
2)=4!
2!(4 −2)! =4×3
2×1= 6
Therefore, the total number of ways to choose 3 men and 2 women is 35×6 =
210 ways.
Step 3: Finally, the total number of ways to form the committee with at
least 2 men and 2 women is the sum of the results from Step 1 and Step 2:
84 + 210 = 294 ways.
11
Question 17
Question
A committee of 5 people is to be formed from a group of 10 students (5 boys
and 5 girls). If the committee must consist of exactly 3 girls and 2 boys, how
many different ways can the committee be formed?
Solution
Step 1: Calculate the number of ways to select 3 girls from a group of 5 girls.
There are (5
3)ways to select 3 girls from 5 girls. (5
3)=5!
3!(5−3)! =5×4×3
3×2×1= 10
ways.
Step 2: Calculate the number of ways to select 2 boys from a group of 5 boys.
There are (5
2)ways to select 2 boys from 5 boys. (5
2)=5!
2!(5−2)! =5×4
2×1= 10
ways.
Step 3: Multiply the number of ways in Step 1 and Step 2 to find the total
number of ways to form the committee. Total number of ways = (5
3)×(5
2)=
10 ×10 = 100.
Therefore, there are 100 different ways to form a committee with exactly 3
girls and 2 boys from a group of 10 students.
Question 18
Question
In a group of 10 students, how many ways can we choose a committee of 4
students if 2 of them refuse to serve together?
Solution
Step 1: First, we find the total number of ways to choose a committee of
4 students from a group of 10 students. Since the order of selection does
not matter and repetitions are not allowed, we use the combination formula:
C(n, r) = n!
r!(n−r)! . Here, n= 10 and r= 4. The total number of ways is:
C(10,4) = 10!
4!(10 −4)! =10!
4!6! = 210
Step 2: Next, we find the number of ways to choose a committee of 4 stu-
dents such that the 2 students who refuse to serve together are included in the
committee. There are 8 students left to choose from, and we need to select 2
more students to complete the committee. Using the combination formula, the
number of ways to choose 2 students from 8 is:
C(8,2) = 8!
2!(8 −2)! =8!
2!6! = 28
12
Step 3: Finally, we calculate the number of ways to choose a committee of
4 students where the 2 students who refuse to serve together are not included.
This can be done by subtracting the number of ways to choose a committee
with the 2 students from the total number of ways to choose a committee:
210 −28 = 182
Therefore, there are 182 ways to choose a committee of 4 students from a
group of 10 students if 2 of them refuse to serve together.
Question 19
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 3 men and exactly 2 women, how many
different committees can be formed?
Solution
Step 1: Calculate the number of ways to choose 3 men out of 7. There are (7
3)
ways to choose 3 men from a group of 7.
Step 2: Calculate the number of ways to choose 2 women out of 6. There
are (6
2)ways to choose 2 women from a group of 6.
Step 3: Multiply the results of Step 1 and Step 2 to determine the total
number of committees that can be formed with at least 3 men and exactly 2
women. (7
3)×(6
2)=7!
3!4! ×6!
2!4!
Step 4: Simplify the expression. 7!
3!4! ×6!
2!4! =7×6×5
3×2×1×6×5
2×1
Step 5: Compute the final answer. 7×6×5
3×2×1×6×5
2×1= 35 ×15 = 525
Therefore, there are 525 different committees that can be formed with at
least 3 men and exactly 2 women.
Question 20
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 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 10. Step 2:
Calculate the number of ways to choose 2 men out of 8. Step 3: Calculate
the number of ways to choose the remaining person (can be either a man or a
13
woman). Step 4: Multiply the results from Steps 1, 2, and 3 to find the total
number of committees.
Step 1: There are (10
2)ways to choose 2 women out of 10. (10
2)=10!
2!(10−2)! =
10×9
2×1= 45
Step 2: There are (8
2)ways to choose 2 men out of 8. (8
2)=8!
2!(8−2)! =8×7
2×1=
28
Step 3: Since the committee must consist of at least 2 women and 2 men, the
remaining person can be either a man or a woman. So, there are 10 women and
8 men left to choose from. There are 10 + 8 = 18 ways to choose the remaining
person.
Step 4: Multiply the results from Steps 1, 2, and 3 to find the total number
of committees: Total number of committees = 45 ×28 ×18 = 22680
Therefore, there are 22,680 different committees that can be formed with at
least 2 women and 2 men.
Question 21
Question
A committee of 5 people is to be formed from a group of 10 men and 8 women.
How many ways can the committee be formed if at least 2 men and 2 women
must be included?
Solution
Step 1: Calculate the number of ways to choose 2 men and 2 women out of 10
men and 8 women.
There are (10
2)ways to choose 2 men from the 10 men, and (8
2)ways to
choose 2 women from the 8 women.
So, the number of ways to choose 2 men and 2 women is (10
2)×(8
2).
Step 2: Calculate the number of ways to choose the remaining person on the
committee.
After selecting 2 men and 2 women, there are 6 people left to choose from -
8 men and 6 women.
We need to choose 1 more person to complete the committee, so there are
(14
1)ways to choose 1 more person.
Step 3: Calculate the total number of ways the committee can be formed.
To find the total number of ways the committee can be formed, multiply the
number of ways from Step 1 and Step 2.
Therefore, the total number of ways the committee can be formed if at least
2 men and 2 women must be included is:
(10
2)×(8
2)×(14
1)
14
Question 22
Question
In a class of 30 students, how many ways can a group of 5 students be selected
to form a project team if two particular students, Alice and Bob, must both be
on the team?
Solution
To solve this problem, we can use the concept of combinations, denoted by (n
k),
which represents the number of ways to choose kobjects from a set of nobjects
without regard to order.
Step 1: First, we select Alice and Bob to be on the team. This leaves us
with the task of selecting 3 more students from the remaining 28 students.
Step 2: The number of ways to select 3 students from 28 is given by (28
3).
Step 3: To find the total number of ways to form the project team with
Alice and Bob on it, we multiply the results from Step 2 and 3.
Step 4: Therefore, the total number of ways to select a group of 5 students
with Alice and Bob on the team is (28
3)=28!
3!(28−3)! =28×27×26
3×2×1= 3276.
Hence, there are 3276 ways to select a project team of 5 students where both
Alice and Bob must be included.
Question 23
Question
A committee of 5 members is to be formed from a group of 8 men and 7 women.
How many different committees can be formed if the committee must consist of
at least 2 men and 2 women?
Solution
Step 1: Calculate the number of ways to choose 2 men from 8 men. Step
2: Calculate the number of ways to choose 2 women from 7 women. Step 3:
Calculate the number of ways to choose the remaining member (either man or
woman) to complete the committee. Step 4: Use the multiplication principle to
find the total number of possible committees.
Step 1: There are (8
2)ways to choose 2 men from 8 men.
(8
2)=8!
2!(8 −2)! =8×7
2×1= 28
Step 2: There are (7
2)ways to choose 2 women from 7 women.
(7
2)=7!
2!(7 −2)! =7×6
2×1= 21
15
Step 3: We need at least 2 men and 2 women in the committee, so the
remaining member must be either a man or a woman. Therefore, the remaining
member can be chosen from the remaining 6 men and 5 women in (6
1)ways if
the remaining member is a man, or in (5
1)ways if the remaining member is a
woman. (6
1)= 6 ways (if the remaining member is a man)
(5
1)= 5 ways (if the remaining member is a woman)
Step 4: To find the total number of possible committees, multiply the number
of ways for each step together: Total number of committees = 28 ×21 ×(6 + 5)
= 588 ×11
= 6468
Therefore, there are 6468 different committees that can be formed if the
committee must consist of at least 2 men and 2 women.
Question 24
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 at least 2 women, how many different committees
can be formed?
Solution
Step 1: Calculate the number of ways to select a committee with exactly 2
women and 2 men. There are (8
2)ways to select 2 women from the 8 available,
and (6
2)ways to select 2 men from the 6 available. Therefore, the number of
ways to select a committee with exactly 2 women and 2 men is given by:
(8
2)×(6
2)= 28 ×15 = 420
Step 2: Calculate the number of ways to select a committee with 3 women
and 1 man. There are (8
3)ways to select 3 women from the 8 available, and (6
1)
ways to select 1 man from the 6 available. Therefore, the number of ways to
select a committee with 3 women and 1 man is given by:
(8
3)×(6
1)= 56 ×6 = 336
Step 3: Calculate the number of ways to select a committee with all 4
members being women. There are (8
4)ways to select 4 women from the 8
16
available. Therefore, the number of ways to select a committee with all women
is given by:
(8
4)= 70
Step 4: Add the results from Steps 1, 2, and 3 to find the total number of
different committees that can be formed.
420 + 336 + 70 = 826
Therefore, there are 826 different committees that can be formed from the
group of 8 women and 6 men, where the committee must consist of at least 2
women.
Question 25
Question
A committee of 5 people is to be formed from a group of 8 men and 6 women.
How many different committees can be formed if there must be at least 2 women
on the committee?
Solution
Step 1: Calculate the number of ways to choose 2 women out of 6. There are
(6
2)= 15 ways to choose 2 women from 6.
Step 2: Calculate the number of ways to choose the remaining 3 people (men
and women). Since there must be at least 2 women on the committee, we can
have either 2 women and 3 men or all 3 women. 2 women and 3 men: (8
3)= 56
ways to choose 3 men from 8. All 3 women: 1 way.
Step 3: Calculate the total number of committees that can be formed. The
total number of committees that can be formed is the product of the choices
made in steps 1 and 2. For committees with 2 women and 3 men: 15 ×56 = 840
committees. For committees with all 3 women: 1×1 = 1 committee.
Therefore, there are a total of 840 + 1 = 841 different committees that can
be formed with at least 2 women.
17
Question 2
Question
A committee of 5 people is to be formed from a group of 10 students (5 males
and 5 females). Calculate the number of ways the committee can be formed if
it must consist of 3 males and 2 females.
Solution
Step 1: Calculate the number of ways to choose 3 males from the 5 available
males. There are (5
3)=5!
3!(5−3)! = 10 ways to choose 3 males from 5.
Step 2: Calculate the number of ways to choose 2 females from the 5 available
females. There are (5
2)=5!
2!(5−2)! = 10 ways to choose 2 females from 5.
Step 3: Multiply the results from Step 1 and Step 2 to find the total number
of ways to form the committee. Total number of ways = 10 ×10 = 100.
Therefore, there are 100 ways to form a committee of 5 people consisting of
3 males and 2 females from a group of 10 students.
Question 3
Question
In a deck of playing cards, how many ways can you choose 5 cards such that 3
are hearts and 2 are spades?
Solution
Step 1: Determine the number of ways to choose 3 hearts out of the 13 hearts
in the deck. Since there are 13 hearts in a deck of playing cards, the number of
ways to choose 3 hearts is given by the combination formula:
(13
3)=13!
3!(13 −3)! =13 ×12 ×11
3×2×1= 286
Step 2: Determine the number of ways to choose 2 spades out of the 13
spades in the deck. Similarly, the number of ways to choose 2 spades is given
by the combination formula:
(13
2)=13!
2!(13 −2)! =13 ×12
2×1= 78
Step 3: Determine the total number of ways to choose 5 cards (3 hearts
and 2 spades). To determine the total number of ways to choose 5 cards with
3 hearts and 2 spades, you need to multiply the number of ways to choose 3
hearts and 2 spades.
Total ways =(13
3)×(13
2)= 286 ×78 = 22236
2
Therefore, there are 22,236 ways to choose 5 cards from a deck of playing
cards such that 3 are hearts and 2 are spades.
Question 4
Question
In a group of 12 students, how many ways are there to form a committee of
4 students if the committee must include exactly 1 male student and 1 female
student?
Solution
To solve this problem, we can break it down into two steps: first, selecting one
male student and one female student for the committee, and then selecting the
remaining two students from the remaining group.
Step 1: Selecting one male student and one female student Since
there are 12 students in total, the number of ways to select 1 male student out
of 6 male students is (6
1)= 6 ways. Similarly, the number of ways to select 1
female student out of 6 female students is (6
1)= 6 ways.
So, the total number of ways to select one male student and one female
student is 6×6 = 36 ways.
Step 2: Selecting the remaining two students After selecting one male
student and one female student, there are 10 students remaining. We need to
select 2 students from this group.
The number of ways to select 2 students out of 10 students is (10
2)=
10!
2!(10−2)! = 45 ways.
Combining both steps To find the total number of ways to form the
committee of 4 students with exactly 1 male and 1 female, we multiply the
number of ways in Step 1 and Step 2:
Total number of ways = 36 ×45 = 1620 ways.
Therefore, there are 1620 ways to form a committee of 4 students with
exactly 1 male student and 1 female student from a group of 12 students.
Question 5
Question
In how many ways can you arrange the letters of the word ”UNIVERSITY”
such that no two vowels are adjacent?
Solution
Step 1: Calculate the total number of ways to arrange the letters of the word
”UNIVERSITY”. There are 10 letters in the word ”UNIVERSITY”, but the
3
letter ”I” appears twice. Therefore, the total number of ways to arrange the
letters is 10!
2! .
Step 2: Calculate the number of ways to arrange the vowels (U, I, E) such
that no two vowels are adjacent. Consider the vowels (U, I, E) as one group.
This group can be arranged in 3! ways. Within this group, the vowels I can be
arranged in 2! ways.
Step 3: Calculate the number of ways to arrange the consonants (N, V, R,
S, T, Y). There are 6 consonants left to be arranged after grouping the vowels.
These can be arranged in 6! ways.
Step 4: Calculate the total number of ways to arrange the letters such that
no two vowels are adjacent. The number of ways to arrange the vowels without
adjacency is the product of the number of ways to arrange the vowels (Step 2)
and the number of ways to arrange the consonants (Step 3).
Therefore, the total number of ways to arrange the letters of the word ”UNI-
VERSITY” such that no two vowels are adjacent is
3! ×2! ×6! = 6 ×2×720 = 8640.
Question 6
Question
In a group of 10 friends, how many ways can we select a committee of 3 members
and a subcommittee of 2 members from the same group?
Solution
Step 1: Calculate the number of ways to select a committee of 3 members from
10 friends. This can be calculated using the combination formula (n
k)=n!
k!(n−k)! ,
where nis the total number of friends and kis the number of members in the
committee.
Number of ways to select committee of 3 members =(10
3)=10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
Step 2: Calculate the number of ways to select a subcommittee of 2 members
from the same group. This can be calculated in the same way as in Step 1.
Number of ways to select subcommittee of 2 members =(10
2)=10!
2!(10 −2)! =10 ×9
2×1= 45
Step 3: Multiply the results from Step 1 and Step 2 to find the total number
of ways to select a committee of 3 members and a subcommittee of 2 members
from the group of 10 friends.
Total number of ways to select both committees = 120 ×45 = 5400
Therefore, there are 5400 ways to select a committee of 3 members and a
subcommittee of 2 members from a group of 10 friends.
4
Question 7
Question
A committee of 5 people is to be chosen from a group of 10 students. How
many different committees can be formed if 2 particular students refuse to serve
together on the same committee?
Solution
Step 1: First, we find the total number of ways to choose a committee of 5
people from 10 students.
(10
5)=10!
5!(10 −5)! =10 ×9×8×7×6
5×4×3×2×1= 252
Step 2: Next, we find the number of ways in which the 2 particular students
can serve together on the same committee. Since the 2 students must be on
the same committee, we treat them as a single unit. So, there are 8 remaining
students from which to choose 3 more to accompany the pair.
(8
3)=8!
3!(8 −3)! =8×7×6
3×2×1= 56
Step 3: Finally, we subtract the number of committees where the 2 particular
students are together from the total number of committees to find the number
of committees where they are not together.
252 −56 = 196
Therefore, there are 196 different committees that can be formed if the 2
particular students refuse to serve together on the same committee.
Question 8
Question
In a group of 12 people, how many ways can we choose a committee of 3 people
to serve as president, vice president, and secretary?
Solution
Step 1: To find the number of ways to select a president, we have 12 options.
Step 2: After selecting the president, there are 11 remaining people for the vice
president position. Step 3: Finally, for the secretary position, there are 10
remaining people. Step 4: To find the total number of ways to choose the
committee, we multiply the number of choices for each position:
12 ×11 ×10 = 1320
5
So, there are 1320 ways to choose a committee of 3 people to serve as president,
vice president, and secretary from a group of 12 people.
Question 9
Question
In a group of 10 students, how many ways can we form a committee of 3 students
and a subcommittee of 2 students from that committee?
Solution
To determine the number of ways we can form a committee of 3 students and
a subcommittee of 2 students from that committee, we’ll use combinatorial
analysis:
Step 1: Calculate the number of ways to choose a committee of 3 students
out of 10. We can do this by using the combination formula: (n
r)=n!
r!(n−r)! .
So, the number of ways to choose a committee of 3 students from a group of 10
is: (10
3)=10!
3!(10−3)! =10∗9∗8
3∗2∗1= 120 ways.
Step 2: Calculate the number of ways to choose a subcommittee of 2 students
from the committee of 3. Similarly, using the combination formula, the number
of ways to select 2 students from a committee of 3 is: (3
2)=3!
2!(3−2)! =3
2= 3
ways.
Step 3: Multiply the results from Step 1 and Step 2 to find the total number
of ways. The total number of ways to form a committee of 3 students and a
subcommittee of 2 students from the initial group of 10 students is: 120×3 = 360
ways.
Therefore, there are 360 different ways to form a committee of 3 students
and a subcommittee of 2 students from a group of 10 students.
Question 10
Question
In a mathematics competition, there are 6 multiple-choice questions with 4
answer choices each. If a student guesses randomly on every question, what is
the probability that the student gets at least 5 questions correct?
Solution
Step 1: Calculate the total number of possible outcomes when guessing ran-
domly on every question.
Since there are 4 answer choices for each of the 6 questions, the total number
of possible outcomes is 46.
Step 2: Calculate the number of ways to get exactly 5 questions correct.
6
To get exactly 5 questions correct, the student must guess the correct answer
for 5 questions and the incorrect answer for the remaining 1 question. There
are (6
5)ways to choose the 5 questions to answer correctly, and for each of these
combinations, there is only 1 way to answer the remaining question incorrectly.
Therefore, the number of ways to get exactly 5 questions correct is (6
5)×1
= 6 ways.
Step 3: Calculate the number of ways to get all 6 questions correct.
To get all 6 questions correct, the student must guess the correct answer for
all 6 questions. There is only 1 way to do this.
Step 4: Calculate the probability of getting at least 5 questions correct.
The probability of getting at least 5 questions correct is the sum of the
probabilities of getting exactly 5 questions correct and getting all 6 questions
correct, divided by the total number of possible outcomes.
The probability is 6+1
46=7
4096.
Question 11
Question
A committee of 5 people is to be formed from a group of 6 men and 4 women.
In how many ways can the committee be formed if it must contain at least 2
men and 2 women?
Solution
Step 1: Calculate the number of ways to select exactly 2 men and 2 women for
the committee. There are (6
2)ways to choose 2 out of 6 men, and (4
2)ways to
choose 2 out of 4 women. Therefore, the number of ways to select exactly 2 men
and 2 women is (6
2)×(4
2).
Step 2: Calculate the number of ways to select 3 men and 2 women for the
committee. There are (6
3)ways to choose 3 out of 6 men, and (4
2)ways to choose
2 out of 4 women. Therefore, the number of ways to select 3 men and 2 women
is (6
3)×(4
2).
Step 3: Calculate the number of ways to select 4 men and 1 woman for the
committee. There are (6
4)ways to choose 4 out of 6 men, and (4
1)ways to choose
1 out of 4 women. Therefore, the number of ways to select 4 men and 1 woman
is (6
4)×(4
1).
Step 4: Add the results from Step 1, Step 2, and Step 3 to find the total
number of ways to form the committee. Total number of ways = (6
2)×(4
2)+
(6
3)×(4
2)+(6
4)×(4
1).
Finally, calculate the total number of ways to form the committee by eval-
uating the expression.
7
Question 12
Question
In how many ways can a committee of 5 people be selected from a group of 10
people if 2 of them refuse to serve together?
Solution
Step 1: First, we find the total number of ways to select a committee of 5 people
from a group of 10 people. This can be done using the combination formula:
(n
r)=n!
r!(n−r)!
Total ways to select a committee of 5 people =(10
5)
=10!
5!(10 −5)!
=10 ×9×8×7×6
5×4×3×2×1
= 252
So, there are 252 total ways to select a committee of 5 people from a group
of 10 people.
Step 2: Next, we find the number of ways to select a committee of 5 people in
which the 2 people who refuse to serve together are selected. This can be done
by considering those two people as one unit, and then selecting the remaining
3 people from the group of 8 people left.
Ways to select the 2 people who refuse to serve together = 1 (considered as one unit)
Ways to select the remaining 3 people from the group of 8 people left =(8
3)
=8!
3!(8 −3)!
=8×7×6
3×2×1
= 56
Therefore, there are 56 ways to select a committee of 5 people where the 2
people who refuse to serve together are selected.
Step 3: Finally, we subtract the number of ways to select a committee where
the 2 people who refuse to serve together are selected from the total number of
ways to select a committee to find the number of ways to select a committee
where the 2 people do not serve together.
8
Ways to select a committee where the 2 people do not serve together =Total ways −Ways with the 2 people selected together
= 252 −56
= 196
Therefore, there are 196 ways to select a committee of 5 people from a group
of 10 people such that the 2 people who refuse to serve together are not both
selected.
Question 13
Question
In how many ways can you arrange the letters in the word ”STATISTICS” if no
two vowels can be next to each other?
Solution
Step 1: Count the total number of ways to arrange the letters in ”STATISTICS”.
There are 10 letters in the word ”STATISTICS”, but the letter ’S’ appears 3
times and the letter ’T’ appears 3 times. So, the total number of ways to arrange
the letters is 10!
3!3! .
Step 2: Count the number of ways when the vowels are grouped together.
Let’s consider the group of vowels ’AII’, which is treated as a single unit. So,
we have 6 units to consider: - AII - S - T - T - S - C - T - S - T - C Now, the
number of ways to arrange these 6 units is 6!.
Step 3: Count the number of ways when the vowels are treated as distinct
letters. Now we will consider the vowels ’A’, ’I’, ’I’ as distinct and not grouped
together. So, we have 7 units to consider: - A - I - I - S - T - S - T - S - T - C
Now, the number of ways to arrange these 7 units is 7!.
Step 4: Calculating the total number of ways when no two vowels can be
next to each other. The total number of ways can be calculated as total ways -
ways when the vowels are grouped together + ways when the vowels are treated
as distinct. So, the total number of ways =10!
3!3! −6! + 7!.
Question 14
Question
A committee of 5 people is to be formed from a group of 8 men and 6 women. If
the committee must include at least 2 women, how many different committees
can be formed?
9
Solution
Step 1: Calculate the number of ways to form a committee with 2, 3, 4, or 5
women.
Let’s consider the following cases: - Case 1: Committee with 2 women and
3 men There are (6
2)ways to choose 2 women from 6, and (8
3)ways to choose 3
men from 8. The total number of committees in this case is (6
2)×(8
3).
- Case 2: Committee with 3 women and 2 men There are (6
3)ways to choose
3 women from 6, and (8
2)ways to choose 2 men from 8. The total number of
committees in this case is (6
3)×(8
2).
- Case 3: Committee with 4 women and 1 man There are (6
4)ways to choose
4 women from 6, and (8
1)ways to choose 1 man from 8. The total number of
committees in this case is (6
4)×(8
1).
- Case 4: Committee with 5 women and 0 men There is only (6
5)way to
choose all 5 women from 6. The total number of committees in this case is (6
5).
Step 2: Calculate the total number of committees by summing up the results
from the different cases.
The total number of committees is:
(6
2)×(8
3)+(6
3)×(8
2)+(6
4)×(8
1)+(6
5)
= 15 ×56 + 20 ×28 + 15 ×8+6
= 840 + 560 + 120 + 6
= 1526
Therefore, there are 1526 different committees that can be formed with at
least 2 women.
Question 15
Question
In a deck of 52 playing cards, how many different ways can you choose 5 cards
if you want exactly 2 of them to be aces?
Solution
Step 1: Calculate the number of ways to choose 2 aces out of the 4 available.
- Since there are 4 aces in a deck of 52 cards, the number of ways to choose 2
aces is given by (4
2)=4!
2!(4−2)! = 6.
Step 2: Calculate the number of ways to choose the remaining 3 non-ace
cards from the 48 remaining cards (52 total cards minus the 4 aces). - The
number of ways to choose 3 non-ace cards from the remaining 48 cards is (48
3)=
48!
3!(48−3)! = 17,296.
10
Step 3: Multiply the results from Step 1 and Step 2 to determine the total
number of ways to choose 5 cards with exactly 2 aces. - The total number of
ways to choose 5 cards with exactly 2 aces is 6×17,296 = 103,776.
Therefore, there are 103,776 different ways to choose 5 cards from a deck of
52 cards if exactly 2 of them are aces.
Question 16
Question
A committee of 5 people is to be formed from a group of 7 men and 4 women.
In how many ways can the committee be formed if it must contain at least 2
men and 2 women?
Solution
Step 1: Calculate the number of ways to choose 2 men and 3 women. Step 2:
Calculate the number of ways to choose 3 men and 2 women. Step 3: Add the
results from Step 1 and Step 2 to find the total number of ways to form the
committee.
Step 1: To choose 2 men from 7, we use the combination formula (n
k)=
n!
k!(n−k)! . Therefore, the number of ways to choose 2 men from 7 is:
(7
2)=7!
2!(7 −2)! =7×6
2×1= 21
To choose 3 women from 4, we use the combination formula. So, the number
of ways to choose 3 women from 4 is:
(4
3)=4!
3!(4 −3)! =4
1= 4
Therefore, the total number of ways to choose 2 men and 3 women is 21×4 =
84 ways.
Step 2: To choose 3 men from 7, the number of ways is:
(7
3)=7!
3!(7 −3)! =7×6×5
3×2×1= 35
To choose 2 women from 4, the number of ways is:
(4
2)=4!
2!(4 −2)! =4×3
2×1= 6
Therefore, the total number of ways to choose 3 men and 2 women is 35×6 =
210 ways.
Step 3: Finally, the total number of ways to form the committee with at
least 2 men and 2 women is the sum of the results from Step 1 and Step 2:
84 + 210 = 294 ways.
11
Question 17
Question
A committee of 5 people is to be formed from a group of 10 students (5 boys
and 5 girls). If the committee must consist of exactly 3 girls and 2 boys, how
many different ways can the committee be formed?
Solution
Step 1: Calculate the number of ways to select 3 girls from a group of 5 girls.
There are (5
3)ways to select 3 girls from 5 girls. (5
3)=5!
3!(5−3)! =5×4×3
3×2×1= 10
ways.
Step 2: Calculate the number of ways to select 2 boys from a group of 5 boys.
There are (5
2)ways to select 2 boys from 5 boys. (5
2)=5!
2!(5−2)! =5×4
2×1= 10
ways.
Step 3: Multiply the number of ways in Step 1 and Step 2 to find the total
number of ways to form the committee. Total number of ways = (5
3)×(5
2)=
10 ×10 = 100.
Therefore, there are 100 different ways to form a committee with exactly 3
girls and 2 boys from a group of 10 students.
Question 18
Question
In a group of 10 students, how many ways can we choose a committee of 4
students if 2 of them refuse to serve together?
Solution
Step 1: First, we find the total number of ways to choose a committee of
4 students from a group of 10 students. Since the order of selection does
not matter and repetitions are not allowed, we use the combination formula:
C(n, r) = n!
r!(n−r)! . Here, n= 10 and r= 4. The total number of ways is:
C(10,4) = 10!
4!(10 −4)! =10!
4!6! = 210
Step 2: Next, we find the number of ways to choose a committee of 4 stu-
dents such that the 2 students who refuse to serve together are included in the
committee. There are 8 students left to choose from, and we need to select 2
more students to complete the committee. Using the combination formula, the
number of ways to choose 2 students from 8 is:
C(8,2) = 8!
2!(8 −2)! =8!
2!6! = 28
12
Step 3: Finally, we calculate the number of ways to choose a committee of
4 students where the 2 students who refuse to serve together are not included.
This can be done by subtracting the number of ways to choose a committee
with the 2 students from the total number of ways to choose a committee:
210 −28 = 182
Therefore, there are 182 ways to choose a committee of 4 students from a
group of 10 students if 2 of them refuse to serve together.
Question 19
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 3 men and exactly 2 women, how many
different committees can be formed?
Solution
Step 1: Calculate the number of ways to choose 3 men out of 7. There are (7
3)
ways to choose 3 men from a group of 7.
Step 2: Calculate the number of ways to choose 2 women out of 6. There
are (6
2)ways to choose 2 women from a group of 6.
Step 3: Multiply the results of Step 1 and Step 2 to determine the total
number of committees that can be formed with at least 3 men and exactly 2
women. (7
3)×(6
2)=7!
3!4! ×6!
2!4!
Step 4: Simplify the expression. 7!
3!4! ×6!
2!4! =7×6×5
3×2×1×6×5
2×1
Step 5: Compute the final answer. 7×6×5
3×2×1×6×5
2×1= 35 ×15 = 525
Therefore, there are 525 different committees that can be formed with at
least 3 men and exactly 2 women.
Question 20
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 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 10. Step 2:
Calculate the number of ways to choose 2 men out of 8. Step 3: Calculate
the number of ways to choose the remaining person (can be either a man or a
13
woman). Step 4: Multiply the results from Steps 1, 2, and 3 to find the total
number of committees.
Step 1: There are (10
2)ways to choose 2 women out of 10. (10
2)=10!
2!(10−2)! =
10×9
2×1= 45
Step 2: There are (8
2)ways to choose 2 men out of 8. (8
2)=8!
2!(8−2)! =8×7
2×1=
28
Step 3: Since the committee must consist of at least 2 women and 2 men, the
remaining person can be either a man or a woman. So, there are 10 women and
8 men left to choose from. There are 10 + 8 = 18 ways to choose the remaining
person.
Step 4: Multiply the results from Steps 1, 2, and 3 to find the total number
of committees: Total number of committees = 45 ×28 ×18 = 22680
Therefore, there are 22,680 different committees that can be formed with at
least 2 women and 2 men.
Question 21
Question
A committee of 5 people is to be formed from a group of 10 men and 8 women.
How many ways can the committee be formed if at least 2 men and 2 women
must be included?
Solution
Step 1: Calculate the number of ways to choose 2 men and 2 women out of 10
men and 8 women.
There are (10
2)ways to choose 2 men from the 10 men, and (8
2)ways to
choose 2 women from the 8 women.
So, the number of ways to choose 2 men and 2 women is (10
2)×(8
2).
Step 2: Calculate the number of ways to choose the remaining person on the
committee.
After selecting 2 men and 2 women, there are 6 people left to choose from -
8 men and 6 women.
We need to choose 1 more person to complete the committee, so there are
(14
1)ways to choose 1 more person.
Step 3: Calculate the total number of ways the committee can be formed.
To find the total number of ways the committee can be formed, multiply the
number of ways from Step 1 and Step 2.
Therefore, the total number of ways the committee can be formed if at least
2 men and 2 women must be included is:
(10
2)×(8
2)×(14
1)
14
Question 22
Question
In a class of 30 students, how many ways can a group of 5 students be selected
to form a project team if two particular students, Alice and Bob, must both be
on the team?
Solution
To solve this problem, we can use the concept of combinations, denoted by (n
k),
which represents the number of ways to choose kobjects from a set of nobjects
without regard to order.
Step 1: First, we select Alice and Bob to be on the team. This leaves us
with the task of selecting 3 more students from the remaining 28 students.
Step 2: The number of ways to select 3 students from 28 is given by (28
3).
Step 3: To find the total number of ways to form the project team with
Alice and Bob on it, we multiply the results from Step 2 and 3.
Step 4: Therefore, the total number of ways to select a group of 5 students
with Alice and Bob on the team is (28
3)=28!
3!(28−3)! =28×27×26
3×2×1= 3276.
Hence, there are 3276 ways to select a project team of 5 students where both
Alice and Bob must be included.
Question 23
Question
A committee of 5 members is to be formed from a group of 8 men and 7 women.
How many different committees can be formed if the committee must consist of
at least 2 men and 2 women?
Solution
Step 1: Calculate the number of ways to choose 2 men from 8 men. Step
2: Calculate the number of ways to choose 2 women from 7 women. Step 3:
Calculate the number of ways to choose the remaining member (either man or
woman) to complete the committee. Step 4: Use the multiplication principle to
find the total number of possible committees.
Step 1: There are (8
2)ways to choose 2 men from 8 men.
(8
2)=8!
2!(8 −2)! =8×7
2×1= 28
Step 2: There are (7
2)ways to choose 2 women from 7 women.
(7
2)=7!
2!(7 −2)! =7×6
2×1= 21
15
Step 3: We need at least 2 men and 2 women in the committee, so the
remaining member must be either a man or a woman. Therefore, the remaining
member can be chosen from the remaining 6 men and 5 women in (6
1)ways if
the remaining member is a man, or in (5
1)ways if the remaining member is a
woman. (6
1)= 6 ways (if the remaining member is a man)
(5
1)= 5 ways (if the remaining member is a woman)
Step 4: To find the total number of possible committees, multiply the number
of ways for each step together: Total number of committees = 28 ×21 ×(6 + 5)
= 588 ×11
= 6468
Therefore, there are 6468 different committees that can be formed if the
committee must consist of at least 2 men and 2 women.
Question 24
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 at least 2 women, how many different committees
can be formed?
Solution
Step 1: Calculate the number of ways to select a committee with exactly 2
women and 2 men. There are (8
2)ways to select 2 women from the 8 available,
and (6
2)ways to select 2 men from the 6 available. Therefore, the number of
ways to select a committee with exactly 2 women and 2 men is given by:
(8
2)×(6
2)= 28 ×15 = 420
Step 2: Calculate the number of ways to select a committee with 3 women
and 1 man. There are (8
3)ways to select 3 women from the 8 available, and (6
1)
ways to select 1 man from the 6 available. Therefore, the number of ways to
select a committee with 3 women and 1 man is given by:
(8
3)×(6
1)= 56 ×6 = 336
Step 3: Calculate the number of ways to select a committee with all 4
members being women. There are (8
4)ways to select 4 women from the 8
16
available. Therefore, the number of ways to select a committee with all women
is given by:
(8
4)= 70
Step 4: Add the results from Steps 1, 2, and 3 to find the total number of
different committees that can be formed.
420 + 336 + 70 = 826
Therefore, there are 826 different committees that can be formed from the
group of 8 women and 6 men, where the committee must consist of at least 2
women.
Question 25
Question
A committee of 5 people is to be formed from a group of 8 men and 6 women.
How many different committees can be formed if there must be at least 2 women
on the committee?
Solution
Step 1: Calculate the number of ways to choose 2 women out of 6. There are
(6
2)= 15 ways to choose 2 women from 6.
Step 2: Calculate the number of ways to choose the remaining 3 people (men
and women). Since there must be at least 2 women on the committee, we can
have either 2 women and 3 men or all 3 women. 2 women and 3 men: (8
3)= 56
ways to choose 3 men from 8. All 3 women: 1 way.
Step 3: Calculate the total number of committees that can be formed. The
total number of committees that can be formed is the product of the choices
made in steps 1 and 2. For committees with 2 women and 3 men: 15 ×56 = 840
committees. For committees with all 3 women: 1×1 = 1 committee.
Therefore, there are a total of 840 + 1 = 841 different committees that can
be formed with at least 2 women.
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