MATH 350 - DISCRETE
MATHEMATICS - Permutations and
combinations
Question Bank - Set 3
Liberty University
Question 1
Question
A computer science class has 20 students, and the professor needs to form a
team of 4 students to work on a project. In how many ways can the professor
choose the team if: (a) Order does not matter? (b) Order matters?
Solution
(a) When order does not matter, we are dealing with combinations.
Step 1: Calculate the number of ways to choose a team of 4 students without
considering order.
20
4=20!
4!(20 −4)! =20 ×19 ×18 ×17
4×3×2×1= 4845.
Therefore, there are 4845 ways for the professor to choose the team when
order does not matter.
(b) When order matters, we are dealing with permutations.
Step 2: Calculate the number of ways to choose a team of 4 students con-
sidering order.
P(20,4) = 20!
(20 −4)! =20 ×19 ×18 ×17 ×16!
16! = 116280.
Therefore, there are 116280 ways for the professor to choose the team when
order matters.
Question 2
Question
A committee of 5 people is to be formed from a group of 10 students and 5
professors. If the committee must consist of at least 2 professors, how many
different committees can be formed?
Solution
Step 1: Calculate the number of ways to select 2, 3, 4, or 5 professors for the
committee.
For 2 professors out of 5: There are 5
2ways to select 2 professors from the
5 available. There are 10
3ways to select the remaining 3 members from the 10
students. So, the total number of ways to form the committee with 2 professors
is 5
2×10
3.
For 3 professors out of 5: There are 5
3ways to select 3 professors from the
5 available. There are 10
2ways to select the remaining 2 members from the 10
students. So, the total number of ways to form the committee with 3 professors
is 5
3×10
2.
For 4 professors out of 5: There are 5
4ways to select 4 professors from the
5 available. There is 10
1way to select the remaining 1 member from the 10
students. So, the total number of ways to form the committee with 4 professors
is 5
4×10
1.
For all 5 professors: There is 5
5= 1 way to select all 5 professors from the
5 available. There are 10
0= 1 way to select 0 students. So, the total number
of ways to form the committee with all 5 professors is 1.
Step 2: Add up the total number of ways from each case to find the total
number of different committees that can be formed.
Total number of committees = 5
2×10
3+5
3×10
2+5
4×10
1+ 1
Calculate the values and find the total number of different committees that
can be formed.
Question 3
Question
A committee of 5 people is to be formed from a group of 7 women and 6 men.
If the committee needs to consist of at least 2 women and 2 men, how many
different committees can be formed?
2
Solution
Step 1: Calculate the number of ways to choose 2 women from 7.
7
2=7!
2!(7 −2)! =7×6
2×1= 21
Step 2: Calculate the number of ways to choose 2 men from 6.
6
2=6!
2!(6 −2)! =6×5
2×1= 15
Step 3: Calculate the number of ways to choose the remaining person (either
man or woman) for the committee. From the remaining 5 women and 4 men,
there are 9 choices.
Step 4: Multiply the number of choices for women, men, and the remaining
person to find the total number of possible committees.
21 ×15 ×9 = 2835
Therefore, there are 2835 different committees that can be formed consisting
of at least 2 women and 2 men.
Question 4
Question
In a class of 30 students, 10 students are chosen to form a committee. If the
committee must consist of 4 seniors and 6 juniors, how many ways can this
committee be formed if there are 12 seniors and 18 juniors in total?
Solution
Step 1: Calculate the number of ways to choose 4 seniors from 12 seniors. There
are 12
4ways to choose 4 seniors out of 12.
Step 2: Calculate the number of ways to choose 6 juniors from 18 juniors.
There are 18
6ways to choose 6 juniors out of 18.
Step 3: Multiply the results from Step 1 and Step 2 to find the total number
of ways to form the committee. Therefore, the total number of ways to form
the committee is: 12
4×18
6
Question 5
Question
A committee of 5 people is to be formed from a group of 7 men and 5 women.
If the committee must have at least 2 women and at least 2 men, how many
different ways can the committee be formed?
3
Solution
Step 1: Calculate the number of ways to choose 2 men and 3 women.
7
25
3= 21 ×10 = 210
Step 2: Calculate the number of ways to choose 3 men and 2 women.
7
35
2= 35 ×10 = 350
Step 3: Add the results from Step 1 and Step 2 to get the total number of
ways to form the committee with at least 2 men and at least 2 women.
210 + 350 = 560
Therefore, there are 560 different ways to form the committee.
Question 6
Question
A committee of 5 students needs to be formed from a group of 10 students (4
males and 6 females). If the committee must consist of at least 3 females and
no more than 2 males, how many different committees can be formed?
Solution
Let’s first calculate the number of committees that consist of exactly 3 females,
1 male and 1 female, and 2 males.
Step 1: Calculate the number of committees with exactly 3 females
There are 6
3ways to choose 3 females from the 6 females, and there are 4
2
ways to choose 2 males from the 4 males. Therefore, the number of committees
with exactly 3 females is 6
3×4
2.
Step 2: Calculate the number of committees with 1 male and 4
females There are 6
4ways to choose 4 females from the 6 females, and there
are 4
1ways to choose 1 male from the 4 males. Therefore, the number of
committees with 1 male and 4 females is 6
4×4
1.
Step 3: Calculate the number of committees with 2 males There
are 6
5ways to choose 5 females from the 6 females, and there are 4
2ways to
choose 2 males from the 4 males. Therefore, the number of committees with 2
males is 6
5×4
2.
Step 4: Add the results The total number of committees that can be
formed is the sum of the committees with exactly 3 females, 1 male and 1
female, and 2 males. Therefore, the total number of committees is 6
3×4
2+
6
4×4
1+6
5×4
2.
Calculate the value to find the total number of committees.
4
Question 7
Question
A committee of 7 people is to be formed from a group of 4 men and 5 women.
If the committee must consist of at least 3 men, how many different committees
can be formed?
Solution
Step 1: Find the number of ways to choose 3 men out of 4 men. There are 4
3
ways to choose 3 men from 4 men.
Step 2: Find the number of ways to choose the remaining 4 people from the
5 women. There are 5
4ways to choose 4 women from 5 women.
Step 3: Multiply the results from Step 1 and Step 2. So the total number of
ways to form the committee with at least 3 men is: 4
3×5
4= 4 ×5 = 20
Therefore, there are 20 different committees that can be formed with at least
3 men.
Question 8
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 exactly 1 man, 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 1 man out of 8. Step 3: Multiply the
results from Steps 1 and 2 to get the total number of committees.
Step 1: To choose 2 women out of 10, we use the combination formula
C(n, k) = n!
k!(n−k)! . So, the number of ways to choose 2 women out of 10 is:
C(10,2) = 10!
2!(10 −2)! =10 ×9
2×1= 45
Step 2: To choose 1 man out of 8, we use the combination formula again.
The number of ways to choose 1 man out of 8 is:
C(8,1) = 8!
1!(8 −1)! = 8
Step 3: Multiply the results from Steps 1 and 2 to get the total number of
committees with at least 2 women and exactly 1 man:
45 ×8 = 360
Therefore, there are 360 different committees that can be formed.
5
Question 9
Question
In how many ways can 5 different rings be worn on 4 fingers if each finger can
have at most 2 rings?
Solution
Step 1: We can break this problem down into cases based on the number of
fingers with 2 rings.
Case 1: All 4 fingers have 1 ring each. In this case, we need to choose 4 rings
out of 5 to assign to the 4 fingers. This can be done in P(5,4) = 5!
(5−4)! = 5
ways.
Case 2: 3 fingers have 1 ring each, and 1 finger has 2 rings. First, we choose
the finger that will have 2 rings, which can be done in C(4,1) = 4
1= 4 ways.
Then we choose 2 rings out of 5 to be worn on that finger, which can be done
in C(5,2) = 5
2= 10 ways. Finally, we assign the remaining 3 rings to the 3
fingers, which can be done in P(3,3) = 3! = 6 ways.
Step 2: Summing up the possibilities from both cases: Total number of ways
= Case 1 + Case 2 Total number of ways = 5 + (4 ×10 ×6) Total number of
ways = 5 + 240 Total number of ways = 245
Therefore, there are 245 ways in which 5 different rings can be worn on 4
fingers if each finger can have at most 2 rings.
Question 10
Question
A committee of 5 people is to be formed from a group of 8 men and 6 women.
If the committee must have at least 2 men and 2 women, how many different
committees can be formed?
Solution
Let’s break this problem down into cases: - Case 1: 2 men and 3 women - Case
2: 3 men and 2 women - Case 3: 4 men and 1 woman - Case 4: 5 men
Let’s calculate the number of ways for each case and then sum them up to
get the total number of different committees that can be formed.
Case 1: 2 men and 3 women Step 1: Choose 2 men from 8 men: 8
2= 28
ways Step 2: Choose 3 women from 6 women: 6
3= 20 ways Step 3: Multiply
the number of ways: 28 ×20 = 560 ways
Case 2: 3 men and 2 women Step 1: Choose 3 men from 8 men: 8
3= 56
ways Step 2: Choose 2 women from 6 women: 6
2= 15 ways Step 3: Multiply
the number of ways: 56 ×15 = 840 ways
6
Case 3: 4 men and 1 woman Step 1: Choose 4 men from 8 men: 8
4= 70
ways Step 2: Choose 1 woman from 6 women: 6
1= 6 ways Step 3: Multiply
the number of ways: 70 ×6 = 420 ways
Case 4: 5 men Step 1: Choose 5 men from 8 men: 8
5= 56 ways
Therefore, the total number of different committees that can be formed is
560 + 840 + 420 + 56 = 1876 ways.
Question 11
Question
A committee of 5 people is to be formed from a group of 10 men and 8 women. If
the committee must consist of at least 2 men and 2 women, how many different
committees can be formed?
Solution
Step 1: Calculate the number of ways to choose 2 men from 10 men. There are
10
2ways to choose 2 men.
Step 2: Calculate the number of ways to choose 2 women from 8 women.
There are 8
2ways to choose 2 women.
Step 3: Calculate the number of ways to choose the remaining person (either
man or woman) for the committee. There are 8 men and 6 women remaining
from which we can choose 1 person. So, there are 8 + 6 = 14 ways to choose the
last person.
Step 4: Multiply the results from Steps 1, 2, and 3 to find the total number
of committees. There are 10
2×8
2×14 = 45 ×28 ×14 = 17640 different
committees that can be formed.
Question 12
Question
In a group of 10 friends, how many ways can we choose 3 people to form a
committee if two particular people refuse to serve on the committee together?
Solution
Step 1: Let’s first find the total number of ways we can select 3 people from a
group of 10 friends. This is given by the combination formula n
k=n!
k!(n−k)! .
In this case, n= 10 and k= 3. So, the total number of ways to select 3 people
is: 10
3=10!
3!(10 −3)!
7
=10 ×9×8
3×2×1
= 120
Step 2: Now, we need to subtract the number of ways in which the two
particular people are together. Let’s assume the two particular friends who
refuse to serve together are A and B. We can treat them as one entity and find
the number of ways to select the committee as if there are only 9 people. The
number of ways to select the committee with A and B together is:
9
2=9!
2!(9 −2)!
=9×8
2×1
= 36
Step 3: Finally, we subtract the number of ways with A and B together from
the total number of ways to get the final answer. Total ways without A and B
together = Total ways - Ways with A and B together
= 120 −36
= 84
Therefore, there are 84 ways to choose 3 people to form a committee from
a group of 10 friends if two particular people refuse to serve on the committee
together.
Question 13
Question
How many different 7-letter words can be formed using the letters of the word
”UNIVERSITY” if: 1. repetition of letters is not allowed? 2. repetition of
letters is allowed?
Solution
1. When repetition of letters is not allowed: Step 1: Count the number of
distinct letters in the word ”UNIVERSITY”. There are 8 distinct letters: U,
N, I, V, E, R, S, T.
Step 2: Choose 7 letters out of 8 for the 7-letter word. This is equivalent
to selecting a 7-element subset from an 8-element set, which can be done in 8
7
ways.
Step 3: Calculate the number of ways to arrange the selected 7 letters. Since
order matters in a word, we need to arrange the 7 selected letters in 7! ways.
8
Step 4: Multiply the results from Step 2 and Step 3 to find the total number
of different 7-letter words. The total number of different 7-letter words = 8
7×7!.
2. When repetition of letters is allowed: Step 1: Count the number of
distinct letters in the word ”UNIVERSITY”. There are 8 distinct letters: U,
N, I, V, E, R, S, T.
Step 2: For each letter, we have the option to include it or not include it in
the 7-letter word, leading to 2 choices for each of the 8 letters. Since repetition
is allowed, we need to consider each letter as many times as we want.
Step 3: Determine the total number of different 7-letter words by considering
all possible combinations of the 8 letters. The total number of different 7-letter
words = 28.
Question 14
Question
A committee of 5 people is to be formed from a group of 10 students and 5
professors. If the committee must consist of 3 students and 2 professors, how
many different committees can be formed?
Solution
Step 1: Calculate the number of ways to choose 3 students out of 10. Step 2:
Calculate the number of ways to choose 2 professors out of 5. Step 3: Mul-
tiply the results from Step 1 and Step 2 to find the total number of different
committees that can be formed.
Step 1: The number of ways to choose 3 students out of 10 is given by
the combination formula n
k=n!
k!(n−k)! . So, the number of ways to choose 3
students from 10 is 10
3=10!
3!(10−3)! =10×9×8
3×2×1= 120.
Step 2: The number of ways to choose 2 professors out of 5 is given by
the combination formula n
k=n!
k!(n−k)! . So, the number of ways to choose 2
professors from 5 is 5
2=5!
2!(5−2)! =5×4
2×1= 10.
Step 3: Now, multiply the results from Step 1 and Step 2 to find the total
number of different committees that can be formed. Total number of committees
= 120 ×10 = 1200.
Therefore, there are 1200 different committees that can be formed consisting
of 3 students and 2 professors.
Question 15
Question
A committee of 5 people is to be formed from a group of 8 men and 5 women. If
the committee must consist of at least 2 men and 2 women, how many different
committees can be formed?
9
Solution
Step 1: First, we will calculate the number of ways to choose 2 men and 3
women for the committee. We can choose 2 men from 8 in 8
2ways, and 3
women from 5 in 5
3ways. Therefore, the number of ways to choose 2 men and
3 women is 8
2×5
3.
Step 2: Next, we will calculate the number of ways to choose 3 men and
2 women for the committee. We can choose 3 men from 8 in 8
3ways, and 2
women from 5 in 5
2ways. Therefore, the number of ways to choose 3 men and
2 women is 8
3×5
2.
Step 3: Finally, we will add the results from Step 1 and Step 2 to get the total
number of committees that can be formed with at least 2 men and 2 women.
The total number of committees = 8
2×5
3+8
3×5
2.
Question 16
Question
A committee of 5 people is chosen from a group of 8 men and 6 women. What
is the probability that the committee consists of at least 2 men?
Solution
To find the probability that the committee consists of at least 2 men, we need
to calculate the total number of ways to choose a committee and the number of
ways to choose a committee with at least 2 men.
Step 1: Calculate the Total Number of Ways to Choose a Com-
mittee To calculate the total number of ways to choose a committee of 5 people
from 14 (8 men and 6 women), we use the combination formula.
The total number of ways to choose a committee of 5 from 14 is:
14
5=14!
5!(14 −5)! =14!
5! ·9! = 2002
Step 2: Calculate the Number of Ways to Choose a Committee
with at Least 2 Men We can choose a committee with at least 2 men in two
ways: 1. Choose 2 men and 3 women. 2. Choose 3 men and 2 women.
1. Choose 2 men and 3 women: Number of ways to choose 2 men from 8:
8
2=8!
2!(8 −2)! = 28
Number of ways to choose 3 women from 6:
6
3=6!
3!(6 −3)! = 20
10
The total number of ways to choose a committee with 2 men and 3 women:
28 ×20 = 560
2. Choose 3 men and 2 women: Number of ways to choose 3 men from 8:
8
3=8!
3!(8 −3)! = 56
Number of ways to choose 2 women from 6:
6
2=6!
2!(6 −2)! = 15
The total number of ways to choose a committee with 3 men and 2 women:
56 ×15 = 840
Step 3: Calculate the Probability The probability that the committee
consists of at least 2 men is the ratio of the number of ways to choose a committee
with at least 2 men to the total number of ways to choose a committee.
The probability is:
P(at least 2 men) = 560 + 840
2002 =1400
2002 =700
1001 ≈0.699
Question 17
Question
In how many ways can you form a 7-digit number using the digits 0, 1, 2, 3, 4,
5, 6, 7, 8, 9 without repetition, such that the number is divisible by 5?
Solution
Step 1: Counting the total number of 7-digit numbers without repetition Since
we are forming a 7-digit number without repetition using the digits 0-9, the total
number of ways to do this is given by the permutation formula P(n, r) = n!
(n−r)! ,
where nis the total number of elements and ris the number of elements to be
selected. For this case, n= 10 (digits 0-9) and r= 7 (7-digit number). Thus,
the total number of ways to form a 7-digit number without repetition is:
P(10,7) = 10!
(10 −7)! =10!
3! =10 ×9×8×7×6×5×4
6= 30,240
Step 2: Counting the number of ways the number is divisible by 5 For a
number to be divisible by 5, the units digit must be either 0 or 5. If the units
digit is 0, then there are 9 choices for the first digit (excluding 0) and 8 choices for
the remaining digits (excluding the first digit and 0). This gives 9×8! numbers.
11
If the units digit is 5, then there are 8 choices for the first digit (excluding 0 and
5) and 8 choices for the remaining digits. This gives 8 ×8! numbers. Therefore,
the total number of 7-digit numbers that are divisible by 5 is:
9×8! + 8 ×8! = 9 ×8! + 8! = 10 ×8! = 28,800
Step 3: Calculating the number of ways the number is not divisible by 5 The
number of 7-digit numbers that are not divisible by 5 is given by:
Total number of 7-digit numbers without repetition−number of 7-digit numbers divisible by 5 = 30,240−28,800 = 1,440
Therefore, there are 1,440 ways to form a 7-digit number using the digits
0-9 without repetition, such that the number is not divisible by 5.
Question 18
Question
A committee of 5 people is to be formed from a group of 8 students and 5 pro-
fessors. If the committee must have at least 2 students and at least 1 professor,
how many different committees can be formed?
Solution
Step 1: Calculate the number of committees with only students or only profes-
sors.
Let’s first find the number of committees with only students: - There are 8
students to choose from. - We need to select 5 students for the committee. The
number of committees with only students is 8
5.
Similarly, we find the number of committees with only professors: - There are
5 professors to choose from. - We need to select 5 professors for the committee.
The number of committees with only professors is 5
5.
Step 2: Calculate the total number of committees with at least 2 students
and at least 1 professor.
To find the total number of committees with at least 2 students and at least
1 professor, we will subtract the number of committees with only students or
only professors from the total number of committees.
Total number of committees = 13
5(choosing 5 people from 13 total)
We subtract the number of committees with only students and only profes-
sors: 13
5−8
5−5
5
Step 3: Calculate the final answer.
Now, we calculate the expression: 13
5−8
5−5
5=13!
5!(13−5)! −8!
5!(8−5)! −
5!
5!(5−5)!
Solving the expression, we get: 13!
5!8! −8!
5!3! −5!
5!0! = 1287 −56 −1 = 1230
Therefore, there are 1230 different committees that can be formed.
12
Question 19
Question
How many ways are there to form a committee of 5 people from a group of 8
women and 5 men if the committee must have at least 2 women and at least 2
men?
Solution
Step 1: Calculate the number of ways to choose 2 women and 3 men first.
There are 8
2ways to choose 2 women from the 8 women, and 5
3ways to
choose 3 men from the 5 men.
Therefore, the number of ways to choose 2 women and 3 men is 8
2×5
3.
Step 2: Calculate the number of ways to choose 3 women and 2 men next.
Similarly, there are 8
3ways to choose 3 women and 5
2ways to choose 2
men.
Therefore, the number of ways to choose 3 women and 2 men is 8
3×5
2.
Step 3: Calculate the total number of ways to form the committee.
To find the total number of ways, we add the number of ways from Step 1
and Step 2:
Total ways = 8
2×5
3+8
3×5
2.
Computing this expression gives us the final answer.
Question 20
Question
A committee of 5 people is to be formed from a group of 10 students and 5
professors.
(a)
In how many ways can the committee be formed if it contains at least 2 profes-
sors?
(b)
In how many ways can the committee be formed if the committee can have at
most 1 professor?
13
Solution
(a)
Step 1: Find the number of ways to choose 2 professors out of 5 professors and
3 students out of 10 students.
5
2×10
3= 10 ×120 = 1200
Step 2: Find the number of ways to choose 3 professors out of 5 professors
and 2 students out of 10 students.
5
3×10
2= 10 ×45 = 450
Step 3: Find the total number of ways to form the committee with at least
2 professors by adding the results from Step 1 and Step 2.
1200 + 450 = 1650
(b)
Step 1: Find the number of ways to choose 1 professor out of 5 professors and
4 students out of 10 students.
5
1×10
4= 5 ×210 = 1050
Step 2: Find the number of ways to choose 0 professors out of 5 professors
and 5 students out of 10 students.
5
0×10
5= 1 ×252 = 252
Step 3: Find the total number of ways to form the committee with at most
1 professor by adding the results from Step 1 and Step 2.
1050 + 252 = 1302
Question 21
Question
In how many ways can you arrange the letters in the word ”MISSISSIPPI” such
that no two S’s are adjacent?
14
Solution
Step 1: Let’s first calculate the total number of arrangements of the letters in
the word ”MISSISSIPPI”. The word has a total of 11 letters, with the following
counts: - 4 M’s - 4 I’s - 2 S’s - 1 P
Therefore, the total number of arrangements is 11!/(4! ×4! ×2! ×1!).
Step 2: Next, let’s calculate the number of arrangements where the S’s are
always together. Treat the two S’s as a single entity. This reduces the total
number of letters to arrange to 10 (M, I, S, P x 4, SS). The total number of
arrangements with the S’s together is 10!/(4! ×4! ×2!).
Step 3: Now, we need to subtract the number of arrangements where the S’s
are always together from the total number of arrangements to find the number of
arrangements where no two S’s are adjacent. Therefore, the number of arrange-
ments where no two S’s are adjacent is 11!/(4! ×4! ×2! ×1!) −10!/(4! ×4! ×2!).
Step 4: Calculating this expression gives us the final answer.
Question 22
Question
A board of directors consists of 8 members, including 5 women and 3 men. A
committee of 3 members is to be formed from the board of directors. What is
the probability that the committee consists of 2 women and 1 man?
Solution
Step 1: Find the total number of ways to form a committee of 3 members from
the 8 members on the board. There are 8
3=8!
3!(8−3)! = 56 ways to choose a
committee of 3 members from a group of 8.
Step 2: Find the number of ways to form a committee with 2 women and 1
man. There are 5
23
1=5!
2!(5−2)! ×3!
1!(3−1)! = 30 ways to choose 2 women from
5 and 1 man from 3.
Step 3: Calculate the probability of selecting a committee with 2 women
and 1 man. The probability is given by the ratio of successful outcomes to total
outcomes:
P(2 women, 1 man) = Number of ways to form committee with 2 women and 1 man
Total number of ways to form a committee =30
56 =15
28 ≈0.5357
Therefore, the probability that the committee consists of 2 women and 1
man is approximately 0.5357 or 53.57
15
Question 23
Question
How many ways are there to select 5 books from a collection of 12 different
books to take on a trip if one book is a dictionary and must be included in the
selection?
Solution
Let’s approach this problem by first selecting the dictionary and then selecting
the remaining 4 books.
Step 1: Selecting the dictionary Since the dictionary must be included,
there is only 1 way to select it.
Step 2: Selecting the remaining 4 books We have 11 books left to
choose from (since the dictionary has already been chosen), and we need to
select 4 more books. This can be done in 11
4ways using combinations.
Therefore, the total number of ways to select 5 books (including the dictio-
nary) from the collection of 12 different books is 1 ×11
4= 1 ×11!
4!7! = 3300.
Hence, there are 3300 ways to select the 5 books for the trip.
Question 24
Question
Suppose a family with 6 children is going on a vacation and they have 8 different
board games to play. If each child can choose one game to bring on the trip,
how many ways are there for the children to select the games so that no game
is played by more than one child?
Solution
Step 1: Since each child can choose one game and no game can be played by more
than one child, this problem involves finding the number of ways to distribute
8 different games to 6 children.
Step 2: This is a classic example of a problem that can be solved using
the concept of permutations. The number of ways to distribute 8 games to 6
children without any restrictions is 8P6, which is given by the formula:
nP r =n!
(n−r)!
Step 3: Substituting n= 8 and r= 6 into the formula, we get:
8P6 = 8!
(8 −6)! =8×7×6×5×4×3×2
2×1= 20160
Step 4: Therefore, there are 20,160 ways for the children to select the games
so that no game is played by more than one child.
16
Question 25
Question
A committee of 5 people is to be formed from a group of 10 men and 8 women.
In how many ways can the committee be formed if it must consist of 3 men and
2 women?
Solution
To solve this problem, we can calculate the number of ways to choose 3 men
from 10 and 2 women from 8, and then multiply the results together since these
choices are independent.
Step 1: Calculate the number of ways to choose 3 men from 10 The
number of ways to choose 3 men from 10 is given by the combination formula:
10
3=10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
Step 2: Calculate the number of ways to choose 2 women from 8
Similarly, the number of ways to choose 2 women from 8 is given by:
8
2=8!
2!(8 −2)! =8×7
2×1= 28
Step 3: Multiply the results to find the total number of ways
Finally, to find the total number of ways the committee can be formed, we
multiply the two results:
120 ×28 = 3360
Therefore, there are 3360 ways to form a committee consisting of 3 men and
2 women from the given group.
Question 26
Question
A committee of 4 members is to be formed from a group of 10 people. If 3 of
the people are hostile towards each other and refuse to serve on the committee
together, how many different committees can be formed?
Solution
Step 1: First, let’s consider the total number of ways to choose 4 people out of
10. This is simply the number of combinations of 10 items taken 4 at a time:
10
4=10!
4!(10 −4)! = 210
17
Step 2: Next, we need to subtract the number of committees that include
all 3 hostile people. Since the 3 hostile people cannot be on the committee
together, we choose 1 out of the 3 and then fill the remaining 3 spots from the
remaining 7 amiable people. So the number of committees that include all 3
hostile people is:
3
1×7
3= 3 ×35 = 105
Step 3: Finally, we subtract the number of committees with all 3 hostile
people from the total number of committees:
210 −105 = 105
Therefore, there are 105 different committees that can be formed from the
group of 10 people where 3 are hostile towards each other.
Question 27
Question
A pianist is selecting 4 pieces to play in a piano recital from a repertoire of 10
pieces. However, she cannot play more than 2 pieces by a particular composer.
How many ways can she choose her program?
Solution
Step 1: Calculate the total number of ways she can select 4 pieces out of 10.
Step 2: Calculate the number of ways she can choose the program without any
restrictions. Step 3: Calculate the number of ways she can choose the program
with the restriction on no more than 2 pieces by a composer. Step 4: Subtract
the result from Step 3 from the result in Step 2 to find the final answer.
Step 1: The total number of ways she can choose 4 pieces out of 10 is given
by the combination formula.
10
4=10!
4!(10 −4)! =10 ×9×8×7
4×3×2×1= 210
Step 2: If there were no restrictions, she could choose any 4 pieces out of
10, so the number of ways would be 10
4= 210.
Step 3: Now, we calculate the number of ways she can choose the program
while ensuring no more than 2 pieces by a composer. Let’s consider the cases
separately: - If she chooses 4 pieces from different composers: 6
4×4! = 15×24 =
360 ways. - If she chooses 3 pieces from one composer and 1 from another:
2
1×6
1×4
3×3!×2! = 12×6×4×6 = 1728 ways. - If she chooses 2 pieces from
one composer and 2 from another: 2
2×6
2×2!×2!×2! = 1×15×2×2×2 = 120
ways.
Total number of ways with the restriction = 360 + 1728 + 120 = 2208.
18
Step 4: The number of ways she can choose her program with the restriction
is 2208, so the final answer is:
210 −2208 = −1998
Therefore, there are 1998 ways for the pianist to choose her program under
the given restrictions.
Question 28
Question
How many ways are there to choose a committee of 5 people from a group of 10
people, where 3 of them must be chosen and 2 of them must not be together?
Solution
Step 1: First, we will calculate the number of ways to choose 3 people out of 10.
We will use the combination formula n
k=n!
k!(n−k)! where n is the total number
of people and k is the number of people to be chosen.
10
3=10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
So, there are 120 ways to choose 3 people out of 10.
Step 2: Next, we will calculate the number of ways to choose 2 people out
of 7 (remaining 7 people after choosing 3).
7
2=7!
2!(7 −2)! =7×6
2×1= 21
So, there are 21 ways to choose 2 people out of the remaining 7.
Step 3: Since 2 of the 5 members must not be together, we need to subtract
the number of ways in which both of them are together. To choose 2 people out
of 3 (chosen initially) in a group of 10, we have 1 way.
Step 4: Now, we will calculate the total number of ways to form the com-
mittee by combining the results from steps 1, 2, and 3.
The total number of ways to select a committee of 5 people with the given
conditions is: 120 ×21 −1 = 2520 −1 = 2519
Therefore, there are 2519 ways to choose a committee of 5 people from a
group of 10 people, where 3 of them must be chosen and 2 of them must not be
together.
19
Question 29
Question
A committee of 5 people is to be formed from a group of 10 men and 6 women. If
the committee must consist of at least 2 women, how many different committees
can be formed?
Solution
Step 1: Calculate the number of committees with exactly 2 women.
To form a committee with exactly 2 women, we choose 2 women from the
6 available women and 3 men from the 10 available men. This can be done in
6
2·10
3ways.
6
2·10
3=6!
2!4! ·10!
3!7! = 15 ·120 = 1800.
Therefore, there are 1800 ways to form a committee with exactly 2 women.
Step 2: Calculate the number of committees with exactly 3 women.
To form a committee with exactly 3 women, we choose 3 women from the
6 available women and 2 men from the 10 available men. This can be done in
6
3·10
2ways.
6
3·10
2=6!
3!3! ·10!
2!8! = 20 ·45 = 900.
Therefore, there are 900 ways to form a committee with exactly 3 women.
Step 3: Calculate the total number of committees with at least 2 women.
The total number of committees with at least 2 women is the sum of the
committees with exactly 2 women and exactly 3 women.
Total = 1800 (exactly 2 women) + 900 (exactly 3 women) = 2700.
Therefore, there are 2700 different committees that can be formed with at
least 2 women.
Question 30
Question
In a group of 10 students, how many ways can we form a committee consisting
of a president, a vice-president, and a secretary?
Solution
Step 1: To find the number of ways to choose the president, we have 10 options.
Step 2: After choosing the president, there are 9 students remaining to
choose from for the vice-president.
Step 3: Finally, after choosing the president and vice-president, there are 8
students left to choose from for the secretary.
20
Step 4: To find the total number of ways to form the committee, we multiply
the number of ways for each position:
10 ×9×8 = 720
Therefore, there are 720 ways to form a committee consisting of a president,
a vice-president, and a secretary from a group of 10 students.
Question 31
Question
How many ways are there to arrange the letters in the word ”MISSISSIPPI”
such that no two S’s are adjacent?
Solution
Step 1: Let’s first consider the total number of ways we can arrange the letters
in ”MISSISSIPPI” without any restrictions.
Step 2: The word ”MISSISSIPPI” has 11 letters in total, with the following
counts: - 1 M - 4 I’s - 4 S’s - 2 P’s
Step 3: The total number of ways to arrange the 11 letters is given by the
multinomial coefficient 11!
1!×4!×4!×2! .
Step 4: Now, let’s consider the number of ways in which the S’s are adjacent.
Step 5: Treat the four S’s as a single entity. This reduces the problem to
arranging the letters ”M, SS, I, I, I, I, P, P” which can be done in 8!
1!×4!×2! ways.
Step 6: There are 5 ways to position the S’s within the SS block. - For
example, ”SSMISSISSIPI” has the four S’s separated.
Step 7: The total number of ways in which the four S’s are adjacent is
5×8!
1!×4!×2! .
Step 8: Finally, subtract the number of ways in which the S’s are adjacent
from the total number of ways to get the final result.
Step 9: The number of ways to arrange the letters in ”MISSISSIPPI” such
that no two S’s are adjacent is 11!
1!×4!×4!×2! −5×8!
1!×4!×2! .
Question 32
Question
A committee of 5 people is to be formed from a group of 8 men and 4 women. If
the committee must consist of at least 3 men and 1 woman, how many different
committees can be formed?
21
Solution
To find the number of different committees that can be formed, we need to
consider the number of ways to choose the men and women separately.
Step 1: Calculate the number of ways to choose at least 3 men out of 8.
Choose 3 men out of 8: 8
3=8!
3!(8−3)! = 56
Choose 4 men out of 8: 8
4=8!
4!(8−4)! = 70
Choose 5 men out of 8: 8
5=8!
5!(8−5)! = 56
Step 2: Calculate the number of ways to choose 1 woman out of 4.
Choose 1 woman out of 4: 4
1=4!
1!(4−1)! = 4
Step 3: Multiply the number of ways to choose the men and women to get
the total number of different committees.
(56 + 70 + 56) ×4 = 182 ×4 = 728
Therefore, there are 728 different committees that can be formed.
Question 33
Question
A committee of 5 people is to be formed from a group of 10 men and 5 women.
How many ways can the committee be formed if there must be at least 3 women
on the committee?
Solution
To determine the number of ways a committee can be formed if there must be
at least 3 women on the committee, we need to consider different cases:
Case 1: 3 women and 2 men on the committee
Choose 3 women out of 5: 5
3= 10 ways
Choose 2 men out of 10: 10
2= 45 ways
Step 1: Calculate the number of ways for Case 1:
10 ×45 = 450 ways
Case 2: 4 women and 1 man on the committee
Choose 4 women out of 5: 5
4= 5 ways
Choose 1 man out of 10: 10
1= 10 ways
22
Step 2: Calculate the number of ways for Case 2:
5×10 = 50 ways
Case 3: 5 women on the committee
Choose all 5 women out of 5: 5
5= 1 way
Step 3: Calculate the number of ways for Case 3:
1×1 = 1 way
Finally, add the number of ways from each case to get the total number of
ways the committee can be formed:
450 + 50 + 1 = 501 ways
Therefore, there are 501 ways to form a committee of 5 people with at least
3 women.
Question 34
Question
Let’s say you have 5 blue balls, 4 red balls, and 3 green balls in a bag. If you
randomly draw 6 balls from the bag without replacement, what is the probability
that you end up with exactly 3 blue balls, 2 red balls, and 1 green ball?
Solution
Step 1: Determine the total number of ways to choose 6 balls out of 12. There
are a total of 12 balls in the bag, so we need to calculate the total number of
ways to choose 6 balls out of 12. This can be done using combinations, denoted
as n
k, where n is the total number of items and k is the number of items to
choose.
Total ways to choose 6 balls out of 12 = 12
6
Step 2: Determine the number of ways to choose 3 blue balls, 2 red balls,
and 1 green ball. To calculate the number of ways to choose 3 blue balls out of
5, 2 red balls out of 4, and 1 green ball out of 3, we can use combinations for
each color.
Ways to choose 3 blue balls = 5
3
Ways to choose 2 red balls = 4
2
Ways to choose 1 green ball = 3
1
23
Step 3: Calculate the total number of successful outcomes. The total number
of successful outcomes is the product of the number of ways to choose each color.
Total successful outcomes = 5
3×4
2×3
1
Step 4: Calculate the probability. The probability of getting exactly 3 blue
balls, 2 red balls, and 1 green ball is the ratio of the total successful outcomes
to the total ways to choose 6 balls out of 12.
Probability = Total successful outcomes
Total ways to choose 6 balls out of 12 =5
3×4
2×3
1
12
6
Question 35
Question
A committee of 5 people must be formed from a group of 10 women and 5 men.
If at least 2 women must be on the committee, how many different committees
can be formed?
Solution
Step 1: Calculate the number of ways to select a committee with exactly 2, 3,
4, or 5 women.
For exactly 2 women: - Select 2 women from 10: 10
2- Select 3 people
(2 women and 1 man) from the remaining 8 people: 8
3- Multiply the two
selections: 10
2×8
3
For exactly 3 women: - Select 3 women from 10: 10
3- Select 2 people
(3 women and 2 men) from the remaining 8 people: 8
2- Multiply the two
selections: 10
3×8
2
For exactly 4 women: - Select 4 women from 10: 10
4- Select 1 person
(4 women and 1 man) from the remaining 8 people: 8
1- Multiply the two
selections: 10
4×8
1
For exactly 5 women: - Select 5 women from 10: 10
5- Since all committee
members are women, we are done: 10
5
Step 2: Add up the possibilities from Step 1 to get the total number of
different committees that can be formed. - Total number of committees =
10
2×8
3+10
3×8
2+10
4×8
1+10
5
24
Question 2
Question
A committee of 5 people is to be formed from a group of 10 students and 5
professors. If the committee must consist of at least 2 professors, how many
different committees can be formed?
Solution
Step 1: Calculate the number of ways to select 2, 3, 4, or 5 professors for the
committee.
For 2 professors out of 5: There are 5
2ways to select 2 professors from the
5 available. There are 10
3ways to select the remaining 3 members from the 10
students. So, the total number of ways to form the committee with 2 professors
is 5
2×10
3.
For 3 professors out of 5: There are 5
3ways to select 3 professors from the
5 available. There are 10
2ways to select the remaining 2 members from the 10
students. So, the total number of ways to form the committee with 3 professors
is 5
3×10
2.
For 4 professors out of 5: There are 5
4ways to select 4 professors from the
5 available. There is 10
1way to select the remaining 1 member from the 10
students. So, the total number of ways to form the committee with 4 professors
is 5
4×10
1.
For all 5 professors: There is 5
5= 1 way to select all 5 professors from the
5 available. There are 10
0= 1 way to select 0 students. So, the total number
of ways to form the committee with all 5 professors is 1.
Step 2: Add up the total number of ways from each case to find the total
number of different committees that can be formed.
Total number of committees = 5
2×10
3+5
3×10
2+5
4×10
1+ 1
Calculate the values and find the total number of different committees that
can be formed.
Question 3
Question
A committee of 5 people is to be formed from a group of 7 women and 6 men.
If the committee needs to consist of at least 2 women and 2 men, how many
different committees can be formed?
2
Solution
Step 1: Calculate the number of ways to choose 2 women from 7.
7
2=7!
2!(7 −2)! =7×6
2×1= 21
Step 2: Calculate the number of ways to choose 2 men from 6.
6
2=6!
2!(6 −2)! =6×5
2×1= 15
Step 3: Calculate the number of ways to choose the remaining person (either
man or woman) for the committee. From the remaining 5 women and 4 men,
there are 9 choices.
Step 4: Multiply the number of choices for women, men, and the remaining
person to find the total number of possible committees.
21 ×15 ×9 = 2835
Therefore, there are 2835 different committees that can be formed consisting
of at least 2 women and 2 men.
Question 4
Question
In a class of 30 students, 10 students are chosen to form a committee. If the
committee must consist of 4 seniors and 6 juniors, how many ways can this
committee be formed if there are 12 seniors and 18 juniors in total?
Solution
Step 1: Calculate the number of ways to choose 4 seniors from 12 seniors. There
are 12
4ways to choose 4 seniors out of 12.
Step 2: Calculate the number of ways to choose 6 juniors from 18 juniors.
There are 18
6ways to choose 6 juniors out of 18.
Step 3: Multiply the results from Step 1 and Step 2 to find the total number
of ways to form the committee. Therefore, the total number of ways to form
the committee is: 12
4×18
6
Question 5
Question
A committee of 5 people is to be formed from a group of 7 men and 5 women.
If the committee must have at least 2 women and at least 2 men, how many
different ways can the committee be formed?
3
Solution
Step 1: Calculate the number of ways to choose 2 men and 3 women.
7
25
3= 21 ×10 = 210
Step 2: Calculate the number of ways to choose 3 men and 2 women.
7
35
2= 35 ×10 = 350
Step 3: Add the results from Step 1 and Step 2 to get the total number of
ways to form the committee with at least 2 men and at least 2 women.
210 + 350 = 560
Therefore, there are 560 different ways to form the committee.
Question 6
Question
A committee of 5 students needs to be formed from a group of 10 students (4
males and 6 females). If the committee must consist of at least 3 females and
no more than 2 males, how many different committees can be formed?
Solution
Let’s first calculate the number of committees that consist of exactly 3 females,
1 male and 1 female, and 2 males.
Step 1: Calculate the number of committees with exactly 3 females
There are 6
3ways to choose 3 females from the 6 females, and there are 4
2
ways to choose 2 males from the 4 males. Therefore, the number of committees
with exactly 3 females is 6
3×4
2.
Step 2: Calculate the number of committees with 1 male and 4
females There are 6
4ways to choose 4 females from the 6 females, and there
are 4
1ways to choose 1 male from the 4 males. Therefore, the number of
committees with 1 male and 4 females is 6
4×4
1.
Step 3: Calculate the number of committees with 2 males There
are 6
5ways to choose 5 females from the 6 females, and there are 4
2ways to
choose 2 males from the 4 males. Therefore, the number of committees with 2
males is 6
5×4
2.
Step 4: Add the results The total number of committees that can be
formed is the sum of the committees with exactly 3 females, 1 male and 1
female, and 2 males. Therefore, the total number of committees is 6
3×4
2+
6
4×4
1+6
5×4
2.
Calculate the value to find the total number of committees.
4
Question 7
Question
A committee of 7 people is to be formed from a group of 4 men and 5 women.
If the committee must consist of at least 3 men, how many different committees
can be formed?
Solution
Step 1: Find the number of ways to choose 3 men out of 4 men. There are 4
3
ways to choose 3 men from 4 men.
Step 2: Find the number of ways to choose the remaining 4 people from the
5 women. There are 5
4ways to choose 4 women from 5 women.
Step 3: Multiply the results from Step 1 and Step 2. So the total number of
ways to form the committee with at least 3 men is: 4
3×5
4= 4 ×5 = 20
Therefore, there are 20 different committees that can be formed with at least
3 men.
Question 8
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 exactly 1 man, 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 1 man out of 8. Step 3: Multiply the
results from Steps 1 and 2 to get the total number of committees.
Step 1: To choose 2 women out of 10, we use the combination formula
C(n, k) = n!
k!(n−k)! . So, the number of ways to choose 2 women out of 10 is:
C(10,2) = 10!
2!(10 −2)! =10 ×9
2×1= 45
Step 2: To choose 1 man out of 8, we use the combination formula again.
The number of ways to choose 1 man out of 8 is:
C(8,1) = 8!
1!(8 −1)! = 8
Step 3: Multiply the results from Steps 1 and 2 to get the total number of
committees with at least 2 women and exactly 1 man:
45 ×8 = 360
Therefore, there are 360 different committees that can be formed.
5
Question 9
Question
In how many ways can 5 different rings be worn on 4 fingers if each finger can
have at most 2 rings?
Solution
Step 1: We can break this problem down into cases based on the number of
fingers with 2 rings.
Case 1: All 4 fingers have 1 ring each. In this case, we need to choose 4 rings
out of 5 to assign to the 4 fingers. This can be done in P(5,4) = 5!
(5−4)! = 5
ways.
Case 2: 3 fingers have 1 ring each, and 1 finger has 2 rings. First, we choose
the finger that will have 2 rings, which can be done in C(4,1) = 4
1= 4 ways.
Then we choose 2 rings out of 5 to be worn on that finger, which can be done
in C(5,2) = 5
2= 10 ways. Finally, we assign the remaining 3 rings to the 3
fingers, which can be done in P(3,3) = 3! = 6 ways.
Step 2: Summing up the possibilities from both cases: Total number of ways
= Case 1 + Case 2 Total number of ways = 5 + (4 ×10 ×6) Total number of
ways = 5 + 240 Total number of ways = 245
Therefore, there are 245 ways in which 5 different rings can be worn on 4
fingers if each finger can have at most 2 rings.
Question 10
Question
A committee of 5 people is to be formed from a group of 8 men and 6 women.
If the committee must have at least 2 men and 2 women, how many different
committees can be formed?
Solution
Let’s break this problem down into cases: - Case 1: 2 men and 3 women - Case
2: 3 men and 2 women - Case 3: 4 men and 1 woman - Case 4: 5 men
Let’s calculate the number of ways for each case and then sum them up to
get the total number of different committees that can be formed.
Case 1: 2 men and 3 women Step 1: Choose 2 men from 8 men: 8
2= 28
ways Step 2: Choose 3 women from 6 women: 6
3= 20 ways Step 3: Multiply
the number of ways: 28 ×20 = 560 ways
Case 2: 3 men and 2 women Step 1: Choose 3 men from 8 men: 8
3= 56
ways Step 2: Choose 2 women from 6 women: 6
2= 15 ways Step 3: Multiply
the number of ways: 56 ×15 = 840 ways
6
Case 3: 4 men and 1 woman Step 1: Choose 4 men from 8 men: 8
4= 70
ways Step 2: Choose 1 woman from 6 women: 6
1= 6 ways Step 3: Multiply
the number of ways: 70 ×6 = 420 ways
Case 4: 5 men Step 1: Choose 5 men from 8 men: 8
5= 56 ways
Therefore, the total number of different committees that can be formed is
560 + 840 + 420 + 56 = 1876 ways.
Question 11
Question
A committee of 5 people is to be formed from a group of 10 men and 8 women. If
the committee must consist of at least 2 men and 2 women, how many different
committees can be formed?
Solution
Step 1: Calculate the number of ways to choose 2 men from 10 men. There are
10
2ways to choose 2 men.
Step 2: Calculate the number of ways to choose 2 women from 8 women.
There are 8
2ways to choose 2 women.
Step 3: Calculate the number of ways to choose the remaining person (either
man or woman) for the committee. There are 8 men and 6 women remaining
from which we can choose 1 person. So, there are 8 + 6 = 14 ways to choose the
last person.
Step 4: Multiply the results from Steps 1, 2, and 3 to find the total number
of committees. There are 10
2×8
2×14 = 45 ×28 ×14 = 17640 different
committees that can be formed.
Question 12
Question
In a group of 10 friends, how many ways can we choose 3 people to form a
committee if two particular people refuse to serve on the committee together?
Solution
Step 1: Let’s first find the total number of ways we can select 3 people from a
group of 10 friends. This is given by the combination formula n
k=n!
k!(n−k)! .
In this case, n= 10 and k= 3. So, the total number of ways to select 3 people
is: 10
3=10!
3!(10 −3)!
7
=10 ×9×8
3×2×1
= 120
Step 2: Now, we need to subtract the number of ways in which the two
particular people are together. Let’s assume the two particular friends who
refuse to serve together are A and B. We can treat them as one entity and find
the number of ways to select the committee as if there are only 9 people. The
number of ways to select the committee with A and B together is:
9
2=9!
2!(9 −2)!
=9×8
2×1
= 36
Step 3: Finally, we subtract the number of ways with A and B together from
the total number of ways to get the final answer. Total ways without A and B
together = Total ways - Ways with A and B together
= 120 −36
= 84
Therefore, there are 84 ways to choose 3 people to form a committee from
a group of 10 friends if two particular people refuse to serve on the committee
together.
Question 13
Question
How many different 7-letter words can be formed using the letters of the word
”UNIVERSITY” if: 1. repetition of letters is not allowed? 2. repetition of
letters is allowed?
Solution
1. When repetition of letters is not allowed: Step 1: Count the number of
distinct letters in the word ”UNIVERSITY”. There are 8 distinct letters: U,
N, I, V, E, R, S, T.
Step 2: Choose 7 letters out of 8 for the 7-letter word. This is equivalent
to selecting a 7-element subset from an 8-element set, which can be done in 8
7
ways.
Step 3: Calculate the number of ways to arrange the selected 7 letters. Since
order matters in a word, we need to arrange the 7 selected letters in 7! ways.
8
Step 4: Multiply the results from Step 2 and Step 3 to find the total number
of different 7-letter words. The total number of different 7-letter words = 8
7×7!.
2. When repetition of letters is allowed: Step 1: Count the number of
distinct letters in the word ”UNIVERSITY”. There are 8 distinct letters: U,
N, I, V, E, R, S, T.
Step 2: For each letter, we have the option to include it or not include it in
the 7-letter word, leading to 2 choices for each of the 8 letters. Since repetition
is allowed, we need to consider each letter as many times as we want.
Step 3: Determine the total number of different 7-letter words by considering
all possible combinations of the 8 letters. The total number of different 7-letter
words = 28.
Question 14
Question
A committee of 5 people is to be formed from a group of 10 students and 5
professors. If the committee must consist of 3 students and 2 professors, how
many different committees can be formed?
Solution
Step 1: Calculate the number of ways to choose 3 students out of 10. Step 2:
Calculate the number of ways to choose 2 professors out of 5. Step 3: Mul-
tiply the results from Step 1 and Step 2 to find the total number of different
committees that can be formed.
Step 1: The number of ways to choose 3 students out of 10 is given by
the combination formula n
k=n!
k!(n−k)! . So, the number of ways to choose 3
students from 10 is 10
3=10!
3!(10−3)! =10×9×8
3×2×1= 120.
Step 2: The number of ways to choose 2 professors out of 5 is given by
the combination formula n
k=n!
k!(n−k)! . So, the number of ways to choose 2
professors from 5 is 5
2=5!
2!(5−2)! =5×4
2×1= 10.
Step 3: Now, multiply the results from Step 1 and Step 2 to find the total
number of different committees that can be formed. Total number of committees
= 120 ×10 = 1200.
Therefore, there are 1200 different committees that can be formed consisting
of 3 students and 2 professors.
Question 15
Question
A committee of 5 people is to be formed from a group of 8 men and 5 women. If
the committee must consist of at least 2 men and 2 women, how many different
committees can be formed?
9
Solution
Step 1: First, we will calculate the number of ways to choose 2 men and 3
women for the committee. We can choose 2 men from 8 in 8
2ways, and 3
women from 5 in 5
3ways. Therefore, the number of ways to choose 2 men and
3 women is 8
2×5
3.
Step 2: Next, we will calculate the number of ways to choose 3 men and
2 women for the committee. We can choose 3 men from 8 in 8
3ways, and 2
women from 5 in 5
2ways. Therefore, the number of ways to choose 3 men and
2 women is 8
3×5
2.
Step 3: Finally, we will add the results from Step 1 and Step 2 to get the total
number of committees that can be formed with at least 2 men and 2 women.
The total number of committees = 8
2×5
3+8
3×5
2.
Question 16
Question
A committee of 5 people is chosen from a group of 8 men and 6 women. What
is the probability that the committee consists of at least 2 men?
Solution
To find the probability that the committee consists of at least 2 men, we need
to calculate the total number of ways to choose a committee and the number of
ways to choose a committee with at least 2 men.
Step 1: Calculate the Total Number of Ways to Choose a Com-
mittee To calculate the total number of ways to choose a committee of 5 people
from 14 (8 men and 6 women), we use the combination formula.
The total number of ways to choose a committee of 5 from 14 is:
14
5=14!
5!(14 −5)! =14!
5! ·9! = 2002
Step 2: Calculate the Number of Ways to Choose a Committee
with at Least 2 Men We can choose a committee with at least 2 men in two
ways: 1. Choose 2 men and 3 women. 2. Choose 3 men and 2 women.
1. Choose 2 men and 3 women: Number of ways to choose 2 men from 8:
8
2=8!
2!(8 −2)! = 28
Number of ways to choose 3 women from 6:
6
3=6!
3!(6 −3)! = 20
10
The total number of ways to choose a committee with 2 men and 3 women:
28 ×20 = 560
2. Choose 3 men and 2 women: Number of ways to choose 3 men from 8:
8
3=8!
3!(8 −3)! = 56
Number of ways to choose 2 women from 6:
6
2=6!
2!(6 −2)! = 15
The total number of ways to choose a committee with 3 men and 2 women:
56 ×15 = 840
Step 3: Calculate the Probability The probability that the committee
consists of at least 2 men is the ratio of the number of ways to choose a committee
with at least 2 men to the total number of ways to choose a committee.
The probability is:
P(at least 2 men) = 560 + 840
2002 =1400
2002 =700
1001 ≈0.699
Question 17
Question
In how many ways can you form a 7-digit number using the digits 0, 1, 2, 3, 4,
5, 6, 7, 8, 9 without repetition, such that the number is divisible by 5?
Solution
Step 1: Counting the total number of 7-digit numbers without repetition Since
we are forming a 7-digit number without repetition using the digits 0-9, the total
number of ways to do this is given by the permutation formula P(n, r) = n!
(n−r)! ,
where nis the total number of elements and ris the number of elements to be
selected. For this case, n= 10 (digits 0-9) and r= 7 (7-digit number). Thus,
the total number of ways to form a 7-digit number without repetition is:
P(10,7) = 10!
(10 −7)! =10!
3! =10 ×9×8×7×6×5×4
6= 30,240
Step 2: Counting the number of ways the number is divisible by 5 For a
number to be divisible by 5, the units digit must be either 0 or 5. If the units
digit is 0, then there are 9 choices for the first digit (excluding 0) and 8 choices for
the remaining digits (excluding the first digit and 0). This gives 9×8! numbers.
11
If the units digit is 5, then there are 8 choices for the first digit (excluding 0 and
5) and 8 choices for the remaining digits. This gives 8 ×8! numbers. Therefore,
the total number of 7-digit numbers that are divisible by 5 is:
9×8! + 8 ×8! = 9 ×8! + 8! = 10 ×8! = 28,800
Step 3: Calculating the number of ways the number is not divisible by 5 The
number of 7-digit numbers that are not divisible by 5 is given by:
Total number of 7-digit numbers without repetition−number of 7-digit numbers divisible by 5 = 30,240−28,800 = 1,440
Therefore, there are 1,440 ways to form a 7-digit number using the digits
0-9 without repetition, such that the number is not divisible by 5.
Question 18
Question
A committee of 5 people is to be formed from a group of 8 students and 5 pro-
fessors. If the committee must have at least 2 students and at least 1 professor,
how many different committees can be formed?
Solution
Step 1: Calculate the number of committees with only students or only profes-
sors.
Let’s first find the number of committees with only students: - There are 8
students to choose from. - We need to select 5 students for the committee. The
number of committees with only students is 8
5.
Similarly, we find the number of committees with only professors: - There are
5 professors to choose from. - We need to select 5 professors for the committee.
The number of committees with only professors is 5
5.
Step 2: Calculate the total number of committees with at least 2 students
and at least 1 professor.
To find the total number of committees with at least 2 students and at least
1 professor, we will subtract the number of committees with only students or
only professors from the total number of committees.
Total number of committees = 13
5(choosing 5 people from 13 total)
We subtract the number of committees with only students and only profes-
sors: 13
5−8
5−5
5
Step 3: Calculate the final answer.
Now, we calculate the expression: 13
5−8
5−5
5=13!
5!(13−5)! −8!
5!(8−5)! −
5!
5!(5−5)!
Solving the expression, we get: 13!
5!8! −8!
5!3! −5!
5!0! = 1287 −56 −1 = 1230
Therefore, there are 1230 different committees that can be formed.
12
Question 19
Question
How many ways are there to form a committee of 5 people from a group of 8
women and 5 men if the committee must have at least 2 women and at least 2
men?
Solution
Step 1: Calculate the number of ways to choose 2 women and 3 men first.
There are 8
2ways to choose 2 women from the 8 women, and 5
3ways to
choose 3 men from the 5 men.
Therefore, the number of ways to choose 2 women and 3 men is 8
2×5
3.
Step 2: Calculate the number of ways to choose 3 women and 2 men next.
Similarly, there are 8
3ways to choose 3 women and 5
2ways to choose 2
men.
Therefore, the number of ways to choose 3 women and 2 men is 8
3×5
2.
Step 3: Calculate the total number of ways to form the committee.
To find the total number of ways, we add the number of ways from Step 1
and Step 2:
Total ways = 8
2×5
3+8
3×5
2.
Computing this expression gives us the final answer.
Question 20
Question
A committee of 5 people is to be formed from a group of 10 students and 5
professors.
(a)
In how many ways can the committee be formed if it contains at least 2 profes-
sors?
(b)
In how many ways can the committee be formed if the committee can have at
most 1 professor?
13
Solution
(a)
Step 1: Find the number of ways to choose 2 professors out of 5 professors and
3 students out of 10 students.
5
2×10
3= 10 ×120 = 1200
Step 2: Find the number of ways to choose 3 professors out of 5 professors
and 2 students out of 10 students.
5
3×10
2= 10 ×45 = 450
Step 3: Find the total number of ways to form the committee with at least
2 professors by adding the results from Step 1 and Step 2.
1200 + 450 = 1650
(b)
Step 1: Find the number of ways to choose 1 professor out of 5 professors and
4 students out of 10 students.
5
1×10
4= 5 ×210 = 1050
Step 2: Find the number of ways to choose 0 professors out of 5 professors
and 5 students out of 10 students.
5
0×10
5= 1 ×252 = 252
Step 3: Find the total number of ways to form the committee with at most
1 professor by adding the results from Step 1 and Step 2.
1050 + 252 = 1302
Question 21
Question
In how many ways can you arrange the letters in the word ”MISSISSIPPI” such
that no two S’s are adjacent?
14
Solution
Step 1: Let’s first calculate the total number of arrangements of the letters in
the word ”MISSISSIPPI”. The word has a total of 11 letters, with the following
counts: - 4 M’s - 4 I’s - 2 S’s - 1 P
Therefore, the total number of arrangements is 11!/(4! ×4! ×2! ×1!).
Step 2: Next, let’s calculate the number of arrangements where the S’s are
always together. Treat the two S’s as a single entity. This reduces the total
number of letters to arrange to 10 (M, I, S, P x 4, SS). The total number of
arrangements with the S’s together is 10!/(4! ×4! ×2!).
Step 3: Now, we need to subtract the number of arrangements where the S’s
are always together from the total number of arrangements to find the number of
arrangements where no two S’s are adjacent. Therefore, the number of arrange-
ments where no two S’s are adjacent is 11!/(4! ×4! ×2! ×1!) −10!/(4! ×4! ×2!).
Step 4: Calculating this expression gives us the final answer.
Question 22
Question
A board of directors consists of 8 members, including 5 women and 3 men. A
committee of 3 members is to be formed from the board of directors. What is
the probability that the committee consists of 2 women and 1 man?
Solution
Step 1: Find the total number of ways to form a committee of 3 members from
the 8 members on the board. There are 8
3=8!
3!(8−3)! = 56 ways to choose a
committee of 3 members from a group of 8.
Step 2: Find the number of ways to form a committee with 2 women and 1
man. There are 5
23
1=5!
2!(5−2)! ×3!
1!(3−1)! = 30 ways to choose 2 women from
5 and 1 man from 3.
Step 3: Calculate the probability of selecting a committee with 2 women
and 1 man. The probability is given by the ratio of successful outcomes to total
outcomes:
P(2 women, 1 man) = Number of ways to form committee with 2 women and 1 man
Total number of ways to form a committee =30
56 =15
28 ≈0.5357
Therefore, the probability that the committee consists of 2 women and 1
man is approximately 0.5357 or 53.57
15
Question 23
Question
How many ways are there to select 5 books from a collection of 12 different
books to take on a trip if one book is a dictionary and must be included in the
selection?
Solution
Let’s approach this problem by first selecting the dictionary and then selecting
the remaining 4 books.
Step 1: Selecting the dictionary Since the dictionary must be included,
there is only 1 way to select it.
Step 2: Selecting the remaining 4 books We have 11 books left to
choose from (since the dictionary has already been chosen), and we need to
select 4 more books. This can be done in 11
4ways using combinations.
Therefore, the total number of ways to select 5 books (including the dictio-
nary) from the collection of 12 different books is 1 ×11
4= 1 ×11!
4!7! = 3300.
Hence, there are 3300 ways to select the 5 books for the trip.
Question 24
Question
Suppose a family with 6 children is going on a vacation and they have 8 different
board games to play. If each child can choose one game to bring on the trip,
how many ways are there for the children to select the games so that no game
is played by more than one child?
Solution
Step 1: Since each child can choose one game and no game can be played by more
than one child, this problem involves finding the number of ways to distribute
8 different games to 6 children.
Step 2: This is a classic example of a problem that can be solved using
the concept of permutations. The number of ways to distribute 8 games to 6
children without any restrictions is 8P6, which is given by the formula:
nP r =n!
(n−r)!
Step 3: Substituting n= 8 and r= 6 into the formula, we get:
8P6 = 8!
(8 −6)! =8×7×6×5×4×3×2
2×1= 20160
Step 4: Therefore, there are 20,160 ways for the children to select the games
so that no game is played by more than one child.
16
Question 25
Question
A committee of 5 people is to be formed from a group of 10 men and 8 women.
In how many ways can the committee be formed if it must consist of 3 men and
2 women?
Solution
To solve this problem, we can calculate the number of ways to choose 3 men
from 10 and 2 women from 8, and then multiply the results together since these
choices are independent.
Step 1: Calculate the number of ways to choose 3 men from 10 The
number of ways to choose 3 men from 10 is given by the combination formula:
10
3=10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
Step 2: Calculate the number of ways to choose 2 women from 8
Similarly, the number of ways to choose 2 women from 8 is given by:
8
2=8!
2!(8 −2)! =8×7
2×1= 28
Step 3: Multiply the results to find the total number of ways
Finally, to find the total number of ways the committee can be formed, we
multiply the two results:
120 ×28 = 3360
Therefore, there are 3360 ways to form a committee consisting of 3 men and
2 women from the given group.
Question 26
Question
A committee of 4 members is to be formed from a group of 10 people. If 3 of
the people are hostile towards each other and refuse to serve on the committee
together, how many different committees can be formed?
Solution
Step 1: First, let’s consider the total number of ways to choose 4 people out of
10. This is simply the number of combinations of 10 items taken 4 at a time:
10
4=10!
4!(10 −4)! = 210
17
Step 2: Next, we need to subtract the number of committees that include
all 3 hostile people. Since the 3 hostile people cannot be on the committee
together, we choose 1 out of the 3 and then fill the remaining 3 spots from the
remaining 7 amiable people. So the number of committees that include all 3
hostile people is:
3
1×7
3= 3 ×35 = 105
Step 3: Finally, we subtract the number of committees with all 3 hostile
people from the total number of committees:
210 −105 = 105
Therefore, there are 105 different committees that can be formed from the
group of 10 people where 3 are hostile towards each other.
Question 27
Question
A pianist is selecting 4 pieces to play in a piano recital from a repertoire of 10
pieces. However, she cannot play more than 2 pieces by a particular composer.
How many ways can she choose her program?
Solution
Step 1: Calculate the total number of ways she can select 4 pieces out of 10.
Step 2: Calculate the number of ways she can choose the program without any
restrictions. Step 3: Calculate the number of ways she can choose the program
with the restriction on no more than 2 pieces by a composer. Step 4: Subtract
the result from Step 3 from the result in Step 2 to find the final answer.
Step 1: The total number of ways she can choose 4 pieces out of 10 is given
by the combination formula.
10
4=10!
4!(10 −4)! =10 ×9×8×7
4×3×2×1= 210
Step 2: If there were no restrictions, she could choose any 4 pieces out of
10, so the number of ways would be 10
4= 210.
Step 3: Now, we calculate the number of ways she can choose the program
while ensuring no more than 2 pieces by a composer. Let’s consider the cases
separately: - If she chooses 4 pieces from different composers: 6
4×4! = 15×24 =
360 ways. - If she chooses 3 pieces from one composer and 1 from another:
2
1×6
1×4
3×3!×2! = 12×6×4×6 = 1728 ways. - If she chooses 2 pieces from
one composer and 2 from another: 2
2×6
2×2!×2!×2! = 1×15×2×2×2 = 120
ways.
Total number of ways with the restriction = 360 + 1728 + 120 = 2208.
18
Step 4: The number of ways she can choose her program with the restriction
is 2208, so the final answer is:
210 −2208 = −1998
Therefore, there are 1998 ways for the pianist to choose her program under
the given restrictions.
Question 28
Question
How many ways are there to choose a committee of 5 people from a group of 10
people, where 3 of them must be chosen and 2 of them must not be together?
Solution
Step 1: First, we will calculate the number of ways to choose 3 people out of 10.
We will use the combination formula n
k=n!
k!(n−k)! where n is the total number
of people and k is the number of people to be chosen.
10
3=10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
So, there are 120 ways to choose 3 people out of 10.
Step 2: Next, we will calculate the number of ways to choose 2 people out
of 7 (remaining 7 people after choosing 3).
7
2=7!
2!(7 −2)! =7×6
2×1= 21
So, there are 21 ways to choose 2 people out of the remaining 7.
Step 3: Since 2 of the 5 members must not be together, we need to subtract
the number of ways in which both of them are together. To choose 2 people out
of 3 (chosen initially) in a group of 10, we have 1 way.
Step 4: Now, we will calculate the total number of ways to form the com-
mittee by combining the results from steps 1, 2, and 3.
The total number of ways to select a committee of 5 people with the given
conditions is: 120 ×21 −1 = 2520 −1 = 2519
Therefore, there are 2519 ways to choose a committee of 5 people from a
group of 10 people, where 3 of them must be chosen and 2 of them must not be
together.
19
Question 29
Question
A committee of 5 people is to be formed from a group of 10 men and 6 women. If
the committee must consist of at least 2 women, how many different committees
can be formed?
Solution
Step 1: Calculate the number of committees with exactly 2 women.
To form a committee with exactly 2 women, we choose 2 women from the
6 available women and 3 men from the 10 available men. This can be done in
6
2·10
3ways.
6
2·10
3=6!
2!4! ·10!
3!7! = 15 ·120 = 1800.
Therefore, there are 1800 ways to form a committee with exactly 2 women.
Step 2: Calculate the number of committees with exactly 3 women.
To form a committee with exactly 3 women, we choose 3 women from the
6 available women and 2 men from the 10 available men. This can be done in
6
3·10
2ways.
6
3·10
2=6!
3!3! ·10!
2!8! = 20 ·45 = 900.
Therefore, there are 900 ways to form a committee with exactly 3 women.
Step 3: Calculate the total number of committees with at least 2 women.
The total number of committees with at least 2 women is the sum of the
committees with exactly 2 women and exactly 3 women.
Total = 1800 (exactly 2 women) + 900 (exactly 3 women) = 2700.
Therefore, there are 2700 different committees that can be formed with at
least 2 women.
Question 30
Question
In a group of 10 students, how many ways can we form a committee consisting
of a president, a vice-president, and a secretary?
Solution
Step 1: To find the number of ways to choose the president, we have 10 options.
Step 2: After choosing the president, there are 9 students remaining to
choose from for the vice-president.
Step 3: Finally, after choosing the president and vice-president, there are 8
students left to choose from for the secretary.
20
Step 4: To find the total number of ways to form the committee, we multiply
the number of ways for each position:
10 ×9×8 = 720
Therefore, there are 720 ways to form a committee consisting of a president,
a vice-president, and a secretary from a group of 10 students.
Question 31
Question
How many ways are there to arrange the letters in the word ”MISSISSIPPI”
such that no two S’s are adjacent?
Solution
Step 1: Let’s first consider the total number of ways we can arrange the letters
in ”MISSISSIPPI” without any restrictions.
Step 2: The word ”MISSISSIPPI” has 11 letters in total, with the following
counts: - 1 M - 4 I’s - 4 S’s - 2 P’s
Step 3: The total number of ways to arrange the 11 letters is given by the
multinomial coefficient 11!
1!×4!×4!×2! .
Step 4: Now, let’s consider the number of ways in which the S’s are adjacent.
Step 5: Treat the four S’s as a single entity. This reduces the problem to
arranging the letters ”M, SS, I, I, I, I, P, P” which can be done in 8!
1!×4!×2! ways.
Step 6: There are 5 ways to position the S’s within the SS block. - For
example, ”SSMISSISSIPI” has the four S’s separated.
Step 7: The total number of ways in which the four S’s are adjacent is
5×8!
1!×4!×2! .
Step 8: Finally, subtract the number of ways in which the S’s are adjacent
from the total number of ways to get the final result.
Step 9: The number of ways to arrange the letters in ”MISSISSIPPI” such
that no two S’s are adjacent is 11!
1!×4!×4!×2! −5×8!
1!×4!×2! .
Question 32
Question
A committee of 5 people is to be formed from a group of 8 men and 4 women. If
the committee must consist of at least 3 men and 1 woman, how many different
committees can be formed?
21
Solution
To find the number of different committees that can be formed, we need to
consider the number of ways to choose the men and women separately.
Step 1: Calculate the number of ways to choose at least 3 men out of 8.
Choose 3 men out of 8: 8
3=8!
3!(8−3)! = 56
Choose 4 men out of 8: 8
4=8!
4!(8−4)! = 70
Choose 5 men out of 8: 8
5=8!
5!(8−5)! = 56
Step 2: Calculate the number of ways to choose 1 woman out of 4.
Choose 1 woman out of 4: 4
1=4!
1!(4−1)! = 4
Step 3: Multiply the number of ways to choose the men and women to get
the total number of different committees.
(56 + 70 + 56) ×4 = 182 ×4 = 728
Therefore, there are 728 different committees that can be formed.
Question 33
Question
A committee of 5 people is to be formed from a group of 10 men and 5 women.
How many ways can the committee be formed if there must be at least 3 women
on the committee?
Solution
To determine the number of ways a committee can be formed if there must be
at least 3 women on the committee, we need to consider different cases:
Case 1: 3 women and 2 men on the committee
Choose 3 women out of 5: 5
3= 10 ways
Choose 2 men out of 10: 10
2= 45 ways
Step 1: Calculate the number of ways for Case 1:
10 ×45 = 450 ways
Case 2: 4 women and 1 man on the committee
Choose 4 women out of 5: 5
4= 5 ways
Choose 1 man out of 10: 10
1= 10 ways
22
Step 2: Calculate the number of ways for Case 2:
5×10 = 50 ways
Case 3: 5 women on the committee
Choose all 5 women out of 5: 5
5= 1 way
Step 3: Calculate the number of ways for Case 3:
1×1 = 1 way
Finally, add the number of ways from each case to get the total number of
ways the committee can be formed:
450 + 50 + 1 = 501 ways
Therefore, there are 501 ways to form a committee of 5 people with at least
3 women.
Question 34
Question
Let’s say you have 5 blue balls, 4 red balls, and 3 green balls in a bag. If you
randomly draw 6 balls from the bag without replacement, what is the probability
that you end up with exactly 3 blue balls, 2 red balls, and 1 green ball?
Solution
Step 1: Determine the total number of ways to choose 6 balls out of 12. There
are a total of 12 balls in the bag, so we need to calculate the total number of
ways to choose 6 balls out of 12. This can be done using combinations, denoted
as n
k, where n is the total number of items and k is the number of items to
choose.
Total ways to choose 6 balls out of 12 = 12
6
Step 2: Determine the number of ways to choose 3 blue balls, 2 red balls,
and 1 green ball. To calculate the number of ways to choose 3 blue balls out of
5, 2 red balls out of 4, and 1 green ball out of 3, we can use combinations for
each color.
Ways to choose 3 blue balls = 5
3
Ways to choose 2 red balls = 4
2
Ways to choose 1 green ball = 3
1
23
Step 3: Calculate the total number of successful outcomes. The total number
of successful outcomes is the product of the number of ways to choose each color.
Total successful outcomes = 5
3×4
2×3
1
Step 4: Calculate the probability. The probability of getting exactly 3 blue
balls, 2 red balls, and 1 green ball is the ratio of the total successful outcomes
to the total ways to choose 6 balls out of 12.
Probability = Total successful outcomes
Total ways to choose 6 balls out of 12 =5
3×4
2×3
1
12
6
Question 35
Question
A committee of 5 people must be formed from a group of 10 women and 5 men.
If at least 2 women must be on the committee, how many different committees
can be formed?
Solution
Step 1: Calculate the number of ways to select a committee with exactly 2, 3,
4, or 5 women.
For exactly 2 women: - Select 2 women from 10: 10
2- Select 3 people
(2 women and 1 man) from the remaining 8 people: 8
3- Multiply the two
selections: 10
2×8
3
For exactly 3 women: - Select 3 women from 10: 10
3- Select 2 people
(3 women and 2 men) from the remaining 8 people: 8
2- Multiply the two
selections: 10
3×8
2
For exactly 4 women: - Select 4 women from 10: 10
4- Select 1 person
(4 women and 1 man) from the remaining 8 people: 8
1- Multiply the two
selections: 10
4×8
1
For exactly 5 women: - Select 5 women from 10: 10
5- Since all committee
members are women, we are done: 10
5
Step 2: Add up the possibilities from Step 1 to get the total number of
different committees that can be formed. - Total number of committees =
10
2×8
3+10
3×8
2+10
4×8
1+10
5
24
Question 2
Question
A committee of 5 people is to be formed from a group of 10 students and 5
professors. If the committee must consist of at least 2 professors, how many
different committees can be formed?
Solution
Step 1: Calculate the number of ways to select 2, 3, 4, or 5 professors for the
committee.
For 2 professors out of 5: There are 5
2ways to select 2 professors from the
5 available. There are 10
3ways to select the remaining 3 members from the 10
students. So, the total number of ways to form the committee with 2 professors
is 5
2×10
3.
For 3 professors out of 5: There are 5
3ways to select 3 professors from the
5 available. There are 10
2ways to select the remaining 2 members from the 10
students. So, the total number of ways to form the committee with 3 professors
is 5
3×10
2.
For 4 professors out of 5: There are 5
4ways to select 4 professors from the
5 available. There is 10
1way to select the remaining 1 member from the 10
students. So, the total number of ways to form the committee with 4 professors
is 5
4×10
1.
For all 5 professors: There is 5
5= 1 way to select all 5 professors from the
5 available. There are 10
0= 1 way to select 0 students. So, the total number
of ways to form the committee with all 5 professors is 1.
Step 2: Add up the total number of ways from each case to find the total
number of different committees that can be formed.
Total number of committees = 5
2×10
3+5
3×10
2+5
4×10
1+ 1
Calculate the values and find the total number of different committees that
can be formed.
Question 3
Question
A committee of 5 people is to be formed from a group of 7 women and 6 men.
If the committee needs to consist of at least 2 women and 2 men, how many
different committees can be formed?
2
Solution
Step 1: Calculate the number of ways to choose 2 women from 7.
7
2=7!
2!(7 −2)! =7×6
2×1= 21
Step 2: Calculate the number of ways to choose 2 men from 6.
6
2=6!
2!(6 −2)! =6×5
2×1= 15
Step 3: Calculate the number of ways to choose the remaining person (either
man or woman) for the committee. From the remaining 5 women and 4 men,
there are 9 choices.
Step 4: Multiply the number of choices for women, men, and the remaining
person to find the total number of possible committees.
21 ×15 ×9 = 2835
Therefore, there are 2835 different committees that can be formed consisting
of at least 2 women and 2 men.
Question 4
Question
In a class of 30 students, 10 students are chosen to form a committee. If the
committee must consist of 4 seniors and 6 juniors, how many ways can this
committee be formed if there are 12 seniors and 18 juniors in total?
Solution
Step 1: Calculate the number of ways to choose 4 seniors from 12 seniors. There
are 12
4ways to choose 4 seniors out of 12.
Step 2: Calculate the number of ways to choose 6 juniors from 18 juniors.
There are 18
6ways to choose 6 juniors out of 18.
Step 3: Multiply the results from Step 1 and Step 2 to find the total number
of ways to form the committee. Therefore, the total number of ways to form
the committee is: 12
4×18
6
Question 5
Question
A committee of 5 people is to be formed from a group of 7 men and 5 women.
If the committee must have at least 2 women and at least 2 men, how many
different ways can the committee be formed?
3
Solution
Step 1: Calculate the number of ways to choose 2 men and 3 women.
7
25
3= 21 ×10 = 210
Step 2: Calculate the number of ways to choose 3 men and 2 women.
7
35
2= 35 ×10 = 350
Step 3: Add the results from Step 1 and Step 2 to get the total number of
ways to form the committee with at least 2 men and at least 2 women.
210 + 350 = 560
Therefore, there are 560 different ways to form the committee.
Question 6
Question
A committee of 5 students needs to be formed from a group of 10 students (4
males and 6 females). If the committee must consist of at least 3 females and
no more than 2 males, how many different committees can be formed?
Solution
Let’s first calculate the number of committees that consist of exactly 3 females,
1 male and 1 female, and 2 males.
Step 1: Calculate the number of committees with exactly 3 females
There are 6
3ways to choose 3 females from the 6 females, and there are 4
2
ways to choose 2 males from the 4 males. Therefore, the number of committees
with exactly 3 females is 6
3×4
2.
Step 2: Calculate the number of committees with 1 male and 4
females There are 6
4ways to choose 4 females from the 6 females, and there
are 4
1ways to choose 1 male from the 4 males. Therefore, the number of
committees with 1 male and 4 females is 6
4×4
1.
Step 3: Calculate the number of committees with 2 males There
are 6
5ways to choose 5 females from the 6 females, and there are 4
2ways to
choose 2 males from the 4 males. Therefore, the number of committees with 2
males is 6
5×4
2.
Step 4: Add the results The total number of committees that can be
formed is the sum of the committees with exactly 3 females, 1 male and 1
female, and 2 males. Therefore, the total number of committees is 6
3×4
2+
6
4×4
1+6
5×4
2.
Calculate the value to find the total number of committees.
4
Question 7
Question
A committee of 7 people is to be formed from a group of 4 men and 5 women.
If the committee must consist of at least 3 men, how many different committees
can be formed?
Solution
Step 1: Find the number of ways to choose 3 men out of 4 men. There are 4
3
ways to choose 3 men from 4 men.
Step 2: Find the number of ways to choose the remaining 4 people from the
5 women. There are 5
4ways to choose 4 women from 5 women.
Step 3: Multiply the results from Step 1 and Step 2. So the total number of
ways to form the committee with at least 3 men is: 4
3×5
4= 4 ×5 = 20
Therefore, there are 20 different committees that can be formed with at least
3 men.
Question 8
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 exactly 1 man, 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 1 man out of 8. Step 3: Multiply the
results from Steps 1 and 2 to get the total number of committees.
Step 1: To choose 2 women out of 10, we use the combination formula
C(n, k) = n!
k!(n−k)! . So, the number of ways to choose 2 women out of 10 is:
C(10,2) = 10!
2!(10 −2)! =10 ×9
2×1= 45
Step 2: To choose 1 man out of 8, we use the combination formula again.
The number of ways to choose 1 man out of 8 is:
C(8,1) = 8!
1!(8 −1)! = 8
Step 3: Multiply the results from Steps 1 and 2 to get the total number of
committees with at least 2 women and exactly 1 man:
45 ×8 = 360
Therefore, there are 360 different committees that can be formed.
5
Question 9
Question
In how many ways can 5 different rings be worn on 4 fingers if each finger can
have at most 2 rings?
Solution
Step 1: We can break this problem down into cases based on the number of
fingers with 2 rings.
Case 1: All 4 fingers have 1 ring each. In this case, we need to choose 4 rings
out of 5 to assign to the 4 fingers. This can be done in P(5,4) = 5!
(5−4)! = 5
ways.
Case 2: 3 fingers have 1 ring each, and 1 finger has 2 rings. First, we choose
the finger that will have 2 rings, which can be done in C(4,1) = 4
1= 4 ways.
Then we choose 2 rings out of 5 to be worn on that finger, which can be done
in C(5,2) = 5
2= 10 ways. Finally, we assign the remaining 3 rings to the 3
fingers, which can be done in P(3,3) = 3! = 6 ways.
Step 2: Summing up the possibilities from both cases: Total number of ways
= Case 1 + Case 2 Total number of ways = 5 + (4 ×10 ×6) Total number of
ways = 5 + 240 Total number of ways = 245
Therefore, there are 245 ways in which 5 different rings can be worn on 4
fingers if each finger can have at most 2 rings.
Question 10
Question
A committee of 5 people is to be formed from a group of 8 men and 6 women.
If the committee must have at least 2 men and 2 women, how many different
committees can be formed?
Solution
Let’s break this problem down into cases: - Case 1: 2 men and 3 women - Case
2: 3 men and 2 women - Case 3: 4 men and 1 woman - Case 4: 5 men
Let’s calculate the number of ways for each case and then sum them up to
get the total number of different committees that can be formed.
Case 1: 2 men and 3 women Step 1: Choose 2 men from 8 men: 8
2= 28
ways Step 2: Choose 3 women from 6 women: 6
3= 20 ways Step 3: Multiply
the number of ways: 28 ×20 = 560 ways
Case 2: 3 men and 2 women Step 1: Choose 3 men from 8 men: 8
3= 56
ways Step 2: Choose 2 women from 6 women: 6
2= 15 ways Step 3: Multiply
the number of ways: 56 ×15 = 840 ways
6
Case 3: 4 men and 1 woman Step 1: Choose 4 men from 8 men: 8
4= 70
ways Step 2: Choose 1 woman from 6 women: 6
1= 6 ways Step 3: Multiply
the number of ways: 70 ×6 = 420 ways
Case 4: 5 men Step 1: Choose 5 men from 8 men: 8
5= 56 ways
Therefore, the total number of different committees that can be formed is
560 + 840 + 420 + 56 = 1876 ways.
Question 11
Question
A committee of 5 people is to be formed from a group of 10 men and 8 women. If
the committee must consist of at least 2 men and 2 women, how many different
committees can be formed?
Solution
Step 1: Calculate the number of ways to choose 2 men from 10 men. There are
10
2ways to choose 2 men.
Step 2: Calculate the number of ways to choose 2 women from 8 women.
There are 8
2ways to choose 2 women.
Step 3: Calculate the number of ways to choose the remaining person (either
man or woman) for the committee. There are 8 men and 6 women remaining
from which we can choose 1 person. So, there are 8 + 6 = 14 ways to choose the
last person.
Step 4: Multiply the results from Steps 1, 2, and 3 to find the total number
of committees. There are 10
2×8
2×14 = 45 ×28 ×14 = 17640 different
committees that can be formed.
Question 12
Question
In a group of 10 friends, how many ways can we choose 3 people to form a
committee if two particular people refuse to serve on the committee together?
Solution
Step 1: Let’s first find the total number of ways we can select 3 people from a
group of 10 friends. This is given by the combination formula n
k=n!
k!(n−k)! .
In this case, n= 10 and k= 3. So, the total number of ways to select 3 people
is: 10
3=10!
3!(10 −3)!
7
=10 ×9×8
3×2×1
= 120
Step 2: Now, we need to subtract the number of ways in which the two
particular people are together. Let’s assume the two particular friends who
refuse to serve together are A and B. We can treat them as one entity and find
the number of ways to select the committee as if there are only 9 people. The
number of ways to select the committee with A and B together is:
9
2=9!
2!(9 −2)!
=9×8
2×1
= 36
Step 3: Finally, we subtract the number of ways with A and B together from
the total number of ways to get the final answer. Total ways without A and B
together = Total ways - Ways with A and B together
= 120 −36
= 84
Therefore, there are 84 ways to choose 3 people to form a committee from
a group of 10 friends if two particular people refuse to serve on the committee
together.
Question 13
Question
How many different 7-letter words can be formed using the letters of the word
”UNIVERSITY” if: 1. repetition of letters is not allowed? 2. repetition of
letters is allowed?
Solution
1. When repetition of letters is not allowed: Step 1: Count the number of
distinct letters in the word ”UNIVERSITY”. There are 8 distinct letters: U,
N, I, V, E, R, S, T.
Step 2: Choose 7 letters out of 8 for the 7-letter word. This is equivalent
to selecting a 7-element subset from an 8-element set, which can be done in 8
7
ways.
Step 3: Calculate the number of ways to arrange the selected 7 letters. Since
order matters in a word, we need to arrange the 7 selected letters in 7! ways.
8
Step 4: Multiply the results from Step 2 and Step 3 to find the total number
of different 7-letter words. The total number of different 7-letter words = 8
7×7!.
2. When repetition of letters is allowed: Step 1: Count the number of
distinct letters in the word ”UNIVERSITY”. There are 8 distinct letters: U,
N, I, V, E, R, S, T.
Step 2: For each letter, we have the option to include it or not include it in
the 7-letter word, leading to 2 choices for each of the 8 letters. Since repetition
is allowed, we need to consider each letter as many times as we want.
Step 3: Determine the total number of different 7-letter words by considering
all possible combinations of the 8 letters. The total number of different 7-letter
words = 28.
Question 14
Question
A committee of 5 people is to be formed from a group of 10 students and 5
professors. If the committee must consist of 3 students and 2 professors, how
many different committees can be formed?
Solution
Step 1: Calculate the number of ways to choose 3 students out of 10. Step 2:
Calculate the number of ways to choose 2 professors out of 5. Step 3: Mul-
tiply the results from Step 1 and Step 2 to find the total number of different
committees that can be formed.
Step 1: The number of ways to choose 3 students out of 10 is given by
the combination formula n
k=n!
k!(n−k)! . So, the number of ways to choose 3
students from 10 is 10
3=10!
3!(10−3)! =10×9×8
3×2×1= 120.
Step 2: The number of ways to choose 2 professors out of 5 is given by
the combination formula n
k=n!
k!(n−k)! . So, the number of ways to choose 2
professors from 5 is 5
2=5!
2!(5−2)! =5×4
2×1= 10.
Step 3: Now, multiply the results from Step 1 and Step 2 to find the total
number of different committees that can be formed. Total number of committees
= 120 ×10 = 1200.
Therefore, there are 1200 different committees that can be formed consisting
of 3 students and 2 professors.
Question 15
Question
A committee of 5 people is to be formed from a group of 8 men and 5 women. If
the committee must consist of at least 2 men and 2 women, how many different
committees can be formed?
9
Solution
Step 1: First, we will calculate the number of ways to choose 2 men and 3
women for the committee. We can choose 2 men from 8 in 8
2ways, and 3
women from 5 in 5
3ways. Therefore, the number of ways to choose 2 men and
3 women is 8
2×5
3.
Step 2: Next, we will calculate the number of ways to choose 3 men and
2 women for the committee. We can choose 3 men from 8 in 8
3ways, and 2
women from 5 in 5
2ways. Therefore, the number of ways to choose 3 men and
2 women is 8
3×5
2.
Step 3: Finally, we will add the results from Step 1 and Step 2 to get the total
number of committees that can be formed with at least 2 men and 2 women.
The total number of committees = 8
2×5
3+8
3×5
2.
Question 16
Question
A committee of 5 people is chosen from a group of 8 men and 6 women. What
is the probability that the committee consists of at least 2 men?
Solution
To find the probability that the committee consists of at least 2 men, we need
to calculate the total number of ways to choose a committee and the number of
ways to choose a committee with at least 2 men.
Step 1: Calculate the Total Number of Ways to Choose a Com-
mittee To calculate the total number of ways to choose a committee of 5 people
from 14 (8 men and 6 women), we use the combination formula.
The total number of ways to choose a committee of 5 from 14 is:
14
5=14!
5!(14 −5)! =14!
5! ·9! = 2002
Step 2: Calculate the Number of Ways to Choose a Committee
with at Least 2 Men We can choose a committee with at least 2 men in two
ways: 1. Choose 2 men and 3 women. 2. Choose 3 men and 2 women.
1. Choose 2 men and 3 women: Number of ways to choose 2 men from 8:
8
2=8!
2!(8 −2)! = 28
Number of ways to choose 3 women from 6:
6
3=6!
3!(6 −3)! = 20
10
The total number of ways to choose a committee with 2 men and 3 women:
28 ×20 = 560
2. Choose 3 men and 2 women: Number of ways to choose 3 men from 8:
8
3=8!
3!(8 −3)! = 56
Number of ways to choose 2 women from 6:
6
2=6!
2!(6 −2)! = 15
The total number of ways to choose a committee with 3 men and 2 women:
56 ×15 = 840
Step 3: Calculate the Probability The probability that the committee
consists of at least 2 men is the ratio of the number of ways to choose a committee
with at least 2 men to the total number of ways to choose a committee.
The probability is:
P(at least 2 men) = 560 + 840
2002 =1400
2002 =700
1001 ≈0.699
Question 17
Question
In how many ways can you form a 7-digit number using the digits 0, 1, 2, 3, 4,
5, 6, 7, 8, 9 without repetition, such that the number is divisible by 5?
Solution
Step 1: Counting the total number of 7-digit numbers without repetition Since
we are forming a 7-digit number without repetition using the digits 0-9, the total
number of ways to do this is given by the permutation formula P(n, r) = n!
(n−r)! ,
where nis the total number of elements and ris the number of elements to be
selected. For this case, n= 10 (digits 0-9) and r= 7 (7-digit number). Thus,
the total number of ways to form a 7-digit number without repetition is:
P(10,7) = 10!
(10 −7)! =10!
3! =10 ×9×8×7×6×5×4
6= 30,240
Step 2: Counting the number of ways the number is divisible by 5 For a
number to be divisible by 5, the units digit must be either 0 or 5. If the units
digit is 0, then there are 9 choices for the first digit (excluding 0) and 8 choices for
the remaining digits (excluding the first digit and 0). This gives 9×8! numbers.
11
If the units digit is 5, then there are 8 choices for the first digit (excluding 0 and
5) and 8 choices for the remaining digits. This gives 8 ×8! numbers. Therefore,
the total number of 7-digit numbers that are divisible by 5 is:
9×8! + 8 ×8! = 9 ×8! + 8! = 10 ×8! = 28,800
Step 3: Calculating the number of ways the number is not divisible by 5 The
number of 7-digit numbers that are not divisible by 5 is given by:
Total number of 7-digit numbers without repetition−number of 7-digit numbers divisible by 5 = 30,240−28,800 = 1,440
Therefore, there are 1,440 ways to form a 7-digit number using the digits
0-9 without repetition, such that the number is not divisible by 5.
Question 18
Question
A committee of 5 people is to be formed from a group of 8 students and 5 pro-
fessors. If the committee must have at least 2 students and at least 1 professor,
how many different committees can be formed?
Solution
Step 1: Calculate the number of committees with only students or only profes-
sors.
Let’s first find the number of committees with only students: - There are 8
students to choose from. - We need to select 5 students for the committee. The
number of committees with only students is 8
5.
Similarly, we find the number of committees with only professors: - There are
5 professors to choose from. - We need to select 5 professors for the committee.
The number of committees with only professors is 5
5.
Step 2: Calculate the total number of committees with at least 2 students
and at least 1 professor.
To find the total number of committees with at least 2 students and at least
1 professor, we will subtract the number of committees with only students or
only professors from the total number of committees.
Total number of committees = 13
5(choosing 5 people from 13 total)
We subtract the number of committees with only students and only profes-
sors: 13
5−8
5−5
5
Step 3: Calculate the final answer.
Now, we calculate the expression: 13
5−8
5−5
5=13!
5!(13−5)! −8!
5!(8−5)! −
5!
5!(5−5)!
Solving the expression, we get: 13!
5!8! −8!
5!3! −5!
5!0! = 1287 −56 −1 = 1230
Therefore, there are 1230 different committees that can be formed.
12
Question 19
Question
How many ways are there to form a committee of 5 people from a group of 8
women and 5 men if the committee must have at least 2 women and at least 2
men?
Solution
Step 1: Calculate the number of ways to choose 2 women and 3 men first.
There are 8
2ways to choose 2 women from the 8 women, and 5
3ways to
choose 3 men from the 5 men.
Therefore, the number of ways to choose 2 women and 3 men is 8
2×5
3.
Step 2: Calculate the number of ways to choose 3 women and 2 men next.
Similarly, there are 8
3ways to choose 3 women and 5
2ways to choose 2
men.
Therefore, the number of ways to choose 3 women and 2 men is 8
3×5
2.
Step 3: Calculate the total number of ways to form the committee.
To find the total number of ways, we add the number of ways from Step 1
and Step 2:
Total ways = 8
2×5
3+8
3×5
2.
Computing this expression gives us the final answer.
Question 20
Question
A committee of 5 people is to be formed from a group of 10 students and 5
professors.
(a)
In how many ways can the committee be formed if it contains at least 2 profes-
sors?
(b)
In how many ways can the committee be formed if the committee can have at
most 1 professor?
13
Solution
(a)
Step 1: Find the number of ways to choose 2 professors out of 5 professors and
3 students out of 10 students.
5
2×10
3= 10 ×120 = 1200
Step 2: Find the number of ways to choose 3 professors out of 5 professors
and 2 students out of 10 students.
5
3×10
2= 10 ×45 = 450
Step 3: Find the total number of ways to form the committee with at least
2 professors by adding the results from Step 1 and Step 2.
1200 + 450 = 1650
(b)
Step 1: Find the number of ways to choose 1 professor out of 5 professors and
4 students out of 10 students.
5
1×10
4= 5 ×210 = 1050
Step 2: Find the number of ways to choose 0 professors out of 5 professors
and 5 students out of 10 students.
5
0×10
5= 1 ×252 = 252
Step 3: Find the total number of ways to form the committee with at most
1 professor by adding the results from Step 1 and Step 2.
1050 + 252 = 1302
Question 21
Question
In how many ways can you arrange the letters in the word ”MISSISSIPPI” such
that no two S’s are adjacent?
14
Solution
Step 1: Let’s first calculate the total number of arrangements of the letters in
the word ”MISSISSIPPI”. The word has a total of 11 letters, with the following
counts: - 4 M’s - 4 I’s - 2 S’s - 1 P
Therefore, the total number of arrangements is 11!/(4! ×4! ×2! ×1!).
Step 2: Next, let’s calculate the number of arrangements where the S’s are
always together. Treat the two S’s as a single entity. This reduces the total
number of letters to arrange to 10 (M, I, S, P x 4, SS). The total number of
arrangements with the S’s together is 10!/(4! ×4! ×2!).
Step 3: Now, we need to subtract the number of arrangements where the S’s
are always together from the total number of arrangements to find the number of
arrangements where no two S’s are adjacent. Therefore, the number of arrange-
ments where no two S’s are adjacent is 11!/(4! ×4! ×2! ×1!) −10!/(4! ×4! ×2!).
Step 4: Calculating this expression gives us the final answer.
Question 22
Question
A board of directors consists of 8 members, including 5 women and 3 men. A
committee of 3 members is to be formed from the board of directors. What is
the probability that the committee consists of 2 women and 1 man?
Solution
Step 1: Find the total number of ways to form a committee of 3 members from
the 8 members on the board. There are 8
3=8!
3!(8−3)! = 56 ways to choose a
committee of 3 members from a group of 8.
Step 2: Find the number of ways to form a committee with 2 women and 1
man. There are 5
23
1=5!
2!(5−2)! ×3!
1!(3−1)! = 30 ways to choose 2 women from
5 and 1 man from 3.
Step 3: Calculate the probability of selecting a committee with 2 women
and 1 man. The probability is given by the ratio of successful outcomes to total
outcomes:
P(2 women, 1 man) = Number of ways to form committee with 2 women and 1 man
Total number of ways to form a committee =30
56 =15
28 ≈0.5357
Therefore, the probability that the committee consists of 2 women and 1
man is approximately 0.5357 or 53.57
15
Question 23
Question
How many ways are there to select 5 books from a collection of 12 different
books to take on a trip if one book is a dictionary and must be included in the
selection?
Solution
Let’s approach this problem by first selecting the dictionary and then selecting
the remaining 4 books.
Step 1: Selecting the dictionary Since the dictionary must be included,
there is only 1 way to select it.
Step 2: Selecting the remaining 4 books We have 11 books left to
choose from (since the dictionary has already been chosen), and we need to
select 4 more books. This can be done in 11
4ways using combinations.
Therefore, the total number of ways to select 5 books (including the dictio-
nary) from the collection of 12 different books is 1 ×11
4= 1 ×11!
4!7! = 3300.
Hence, there are 3300 ways to select the 5 books for the trip.
Question 24
Question
Suppose a family with 6 children is going on a vacation and they have 8 different
board games to play. If each child can choose one game to bring on the trip,
how many ways are there for the children to select the games so that no game
is played by more than one child?
Solution
Step 1: Since each child can choose one game and no game can be played by more
than one child, this problem involves finding the number of ways to distribute
8 different games to 6 children.
Step 2: This is a classic example of a problem that can be solved using
the concept of permutations. The number of ways to distribute 8 games to 6
children without any restrictions is 8P6, which is given by the formula:
nP r =n!
(n−r)!
Step 3: Substituting n= 8 and r= 6 into the formula, we get:
8P6 = 8!
(8 −6)! =8×7×6×5×4×3×2
2×1= 20160
Step 4: Therefore, there are 20,160 ways for the children to select the games
so that no game is played by more than one child.
16
Question 25
Question
A committee of 5 people is to be formed from a group of 10 men and 8 women.
In how many ways can the committee be formed if it must consist of 3 men and
2 women?
Solution
To solve this problem, we can calculate the number of ways to choose 3 men
from 10 and 2 women from 8, and then multiply the results together since these
choices are independent.
Step 1: Calculate the number of ways to choose 3 men from 10 The
number of ways to choose 3 men from 10 is given by the combination formula:
10
3=10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
Step 2: Calculate the number of ways to choose 2 women from 8
Similarly, the number of ways to choose 2 women from 8 is given by:
8
2=8!
2!(8 −2)! =8×7
2×1= 28
Step 3: Multiply the results to find the total number of ways
Finally, to find the total number of ways the committee can be formed, we
multiply the two results:
120 ×28 = 3360
Therefore, there are 3360 ways to form a committee consisting of 3 men and
2 women from the given group.
Question 26
Question
A committee of 4 members is to be formed from a group of 10 people. If 3 of
the people are hostile towards each other and refuse to serve on the committee
together, how many different committees can be formed?
Solution
Step 1: First, let’s consider the total number of ways to choose 4 people out of
10. This is simply the number of combinations of 10 items taken 4 at a time:
10
4=10!
4!(10 −4)! = 210
17
Step 2: Next, we need to subtract the number of committees that include
all 3 hostile people. Since the 3 hostile people cannot be on the committee
together, we choose 1 out of the 3 and then fill the remaining 3 spots from the
remaining 7 amiable people. So the number of committees that include all 3
hostile people is:
3
1×7
3= 3 ×35 = 105
Step 3: Finally, we subtract the number of committees with all 3 hostile
people from the total number of committees:
210 −105 = 105
Therefore, there are 105 different committees that can be formed from the
group of 10 people where 3 are hostile towards each other.
Question 27
Question
A pianist is selecting 4 pieces to play in a piano recital from a repertoire of 10
pieces. However, she cannot play more than 2 pieces by a particular composer.
How many ways can she choose her program?
Solution
Step 1: Calculate the total number of ways she can select 4 pieces out of 10.
Step 2: Calculate the number of ways she can choose the program without any
restrictions. Step 3: Calculate the number of ways she can choose the program
with the restriction on no more than 2 pieces by a composer. Step 4: Subtract
the result from Step 3 from the result in Step 2 to find the final answer.
Step 1: The total number of ways she can choose 4 pieces out of 10 is given
by the combination formula.
10
4=10!
4!(10 −4)! =10 ×9×8×7
4×3×2×1= 210
Step 2: If there were no restrictions, she could choose any 4 pieces out of
10, so the number of ways would be 10
4= 210.
Step 3: Now, we calculate the number of ways she can choose the program
while ensuring no more than 2 pieces by a composer. Let’s consider the cases
separately: - If she chooses 4 pieces from different composers: 6
4×4! = 15×24 =
360 ways. - If she chooses 3 pieces from one composer and 1 from another:
2
1×6
1×4
3×3!×2! = 12×6×4×6 = 1728 ways. - If she chooses 2 pieces from
one composer and 2 from another: 2
2×6
2×2!×2!×2! = 1×15×2×2×2 = 120
ways.
Total number of ways with the restriction = 360 + 1728 + 120 = 2208.
18
Step 4: The number of ways she can choose her program with the restriction
is 2208, so the final answer is:
210 −2208 = −1998
Therefore, there are 1998 ways for the pianist to choose her program under
the given restrictions.
Question 28
Question
How many ways are there to choose a committee of 5 people from a group of 10
people, where 3 of them must be chosen and 2 of them must not be together?
Solution
Step 1: First, we will calculate the number of ways to choose 3 people out of 10.
We will use the combination formula n
k=n!
k!(n−k)! where n is the total number
of people and k is the number of people to be chosen.
10
3=10!
3!(10 −3)! =10 ×9×8
3×2×1= 120
So, there are 120 ways to choose 3 people out of 10.
Step 2: Next, we will calculate the number of ways to choose 2 people out
of 7 (remaining 7 people after choosing 3).
7
2=7!
2!(7 −2)! =7×6
2×1= 21
So, there are 21 ways to choose 2 people out of the remaining 7.
Step 3: Since 2 of the 5 members must not be together, we need to subtract
the number of ways in which both of them are together. To choose 2 people out
of 3 (chosen initially) in a group of 10, we have 1 way.
Step 4: Now, we will calculate the total number of ways to form the com-
mittee by combining the results from steps 1, 2, and 3.
The total number of ways to select a committee of 5 people with the given
conditions is: 120 ×21 −1 = 2520 −1 = 2519
Therefore, there are 2519 ways to choose a committee of 5 people from a
group of 10 people, where 3 of them must be chosen and 2 of them must not be
together.
19
Question 29
Question
A committee of 5 people is to be formed from a group of 10 men and 6 women. If
the committee must consist of at least 2 women, how many different committees
can be formed?
Solution
Step 1: Calculate the number of committees with exactly 2 women.
To form a committee with exactly 2 women, we choose 2 women from the
6 available women and 3 men from the 10 available men. This can be done in
6
2·10
3ways.
6
2·10
3=6!
2!4! ·10!
3!7! = 15 ·120 = 1800.
Therefore, there are 1800 ways to form a committee with exactly 2 women.
Step 2: Calculate the number of committees with exactly 3 women.
To form a committee with exactly 3 women, we choose 3 women from the
6 available women and 2 men from the 10 available men. This can be done in
6
3·10
2ways.
6
3·10
2=6!
3!3! ·10!
2!8! = 20 ·45 = 900.
Therefore, there are 900 ways to form a committee with exactly 3 women.
Step 3: Calculate the total number of committees with at least 2 women.
The total number of committees with at least 2 women is the sum of the
committees with exactly 2 women and exactly 3 women.
Total = 1800 (exactly 2 women) + 900 (exactly 3 women) = 2700.
Therefore, there are 2700 different committees that can be formed with at
least 2 women.
Question 30
Question
In a group of 10 students, how many ways can we form a committee consisting
of a president, a vice-president, and a secretary?
Solution
Step 1: To find the number of ways to choose the president, we have 10 options.
Step 2: After choosing the president, there are 9 students remaining to
choose from for the vice-president.
Step 3: Finally, after choosing the president and vice-president, there are 8
students left to choose from for the secretary.
20
Step 4: To find the total number of ways to form the committee, we multiply
the number of ways for each position:
10 ×9×8 = 720
Therefore, there are 720 ways to form a committee consisting of a president,
a vice-president, and a secretary from a group of 10 students.
Question 31
Question
How many ways are there to arrange the letters in the word ”MISSISSIPPI”
such that no two S’s are adjacent?
Solution
Step 1: Let’s first consider the total number of ways we can arrange the letters
in ”MISSISSIPPI” without any restrictions.
Step 2: The word ”MISSISSIPPI” has 11 letters in total, with the following
counts: - 1 M - 4 I’s - 4 S’s - 2 P’s
Step 3: The total number of ways to arrange the 11 letters is given by the
multinomial coefficient 11!
1!×4!×4!×2! .
Step 4: Now, let’s consider the number of ways in which the S’s are adjacent.
Step 5: Treat the four S’s as a single entity. This reduces the problem to
arranging the letters ”M, SS, I, I, I, I, P, P” which can be done in 8!
1!×4!×2! ways.
Step 6: There are 5 ways to position the S’s within the SS block. - For
example, ”SSMISSISSIPI” has the four S’s separated.
Step 7: The total number of ways in which the four S’s are adjacent is
5×8!
1!×4!×2! .
Step 8: Finally, subtract the number of ways in which the S’s are adjacent
from the total number of ways to get the final result.
Step 9: The number of ways to arrange the letters in ”MISSISSIPPI” such
that no two S’s are adjacent is 11!
1!×4!×4!×2! −5×8!
1!×4!×2! .
Question 32
Question
A committee of 5 people is to be formed from a group of 8 men and 4 women. If
the committee must consist of at least 3 men and 1 woman, how many different
committees can be formed?
21
Solution
To find the number of different committees that can be formed, we need to
consider the number of ways to choose the men and women separately.
Step 1: Calculate the number of ways to choose at least 3 men out of 8.
Choose 3 men out of 8: 8
3=8!
3!(8−3)! = 56
Choose 4 men out of 8: 8
4=8!
4!(8−4)! = 70
Choose 5 men out of 8: 8
5=8!
5!(8−5)! = 56
Step 2: Calculate the number of ways to choose 1 woman out of 4.
Choose 1 woman out of 4: 4
1=4!
1!(4−1)! = 4
Step 3: Multiply the number of ways to choose the men and women to get
the total number of different committees.
(56 + 70 + 56) ×4 = 182 ×4 = 728
Therefore, there are 728 different committees that can be formed.
Question 33
Question
A committee of 5 people is to be formed from a group of 10 men and 5 women.
How many ways can the committee be formed if there must be at least 3 women
on the committee?
Solution
To determine the number of ways a committee can be formed if there must be
at least 3 women on the committee, we need to consider different cases:
Case 1: 3 women and 2 men on the committee
Choose 3 women out of 5: 5
3= 10 ways
Choose 2 men out of 10: 10
2= 45 ways
Step 1: Calculate the number of ways for Case 1:
10 ×45 = 450 ways
Case 2: 4 women and 1 man on the committee
Choose 4 women out of 5: 5
4= 5 ways
Choose 1 man out of 10: 10
1= 10 ways
22
Step 2: Calculate the number of ways for Case 2:
5×10 = 50 ways
Case 3: 5 women on the committee
Choose all 5 women out of 5: 5
5= 1 way
Step 3: Calculate the number of ways for Case 3:
1×1 = 1 way
Finally, add the number of ways from each case to get the total number of
ways the committee can be formed:
450 + 50 + 1 = 501 ways
Therefore, there are 501 ways to form a committee of 5 people with at least
3 women.
Question 34
Question
Let’s say you have 5 blue balls, 4 red balls, and 3 green balls in a bag. If you
randomly draw 6 balls from the bag without replacement, what is the probability
that you end up with exactly 3 blue balls, 2 red balls, and 1 green ball?
Solution
Step 1: Determine the total number of ways to choose 6 balls out of 12. There
are a total of 12 balls in the bag, so we need to calculate the total number of
ways to choose 6 balls out of 12. This can be done using combinations, denoted
as n
k, where n is the total number of items and k is the number of items to
choose.
Total ways to choose 6 balls out of 12 = 12
6
Step 2: Determine the number of ways to choose 3 blue balls, 2 red balls,
and 1 green ball. To calculate the number of ways to choose 3 blue balls out of
5, 2 red balls out of 4, and 1 green ball out of 3, we can use combinations for
each color.
Ways to choose 3 blue balls = 5
3
Ways to choose 2 red balls = 4
2
Ways to choose 1 green ball = 3
1
23
Step 3: Calculate the total number of successful outcomes. The total number
of successful outcomes is the product of the number of ways to choose each color.
Total successful outcomes = 5
3×4
2×3
1
Step 4: Calculate the probability. The probability of getting exactly 3 blue
balls, 2 red balls, and 1 green ball is the ratio of the total successful outcomes
to the total ways to choose 6 balls out of 12.
Probability = Total successful outcomes
Total ways to choose 6 balls out of 12 =5
3×4
2×3
1
12
6
Question 35
Question
A committee of 5 people must be formed from a group of 10 women and 5 men.
If at least 2 women must be on the committee, how many different committees
can be formed?
Solution
Step 1: Calculate the number of ways to select a committee with exactly 2, 3,
4, or 5 women.
For exactly 2 women: - Select 2 women from 10: 10
2- Select 3 people
(2 women and 1 man) from the remaining 8 people: 8
3- Multiply the two
selections: 10
2×8
3
For exactly 3 women: - Select 3 women from 10: 10
3- Select 2 people
(3 women and 2 men) from the remaining 8 people: 8
2- Multiply the two
selections: 10
3×8
2
For exactly 4 women: - Select 4 women from 10: 10
4- Select 1 person
(4 women and 1 man) from the remaining 8 people: 8
1- Multiply the two
selections: 10
4×8
1
For exactly 5 women: - Select 5 women from 10: 10
5- Since all committee
members are women, we are done: 10
5
Step 2: Add up the possibilities from Step 1 to get the total number of
different committees that can be formed. - Total number of committees =
10
2×8
3+10
3×8
2+10
4×8
1+10
5
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