1
Permutations and Combinations
1. Permutations
Definition
A permutation is an arrangement of items in a specific order. When the order matters, you use
permutations.
Formula
The number of permutations of 𝑛 objects taken 𝑟 at a time is given by:
𝑃(𝑛, 𝑟) =
𝑛!
(𝑛 − 𝑟)!
where 𝑛! (n factorial) is the product of all positive integers up to 𝑛.
Example 1:
How many ways can you arrange 3 out of 5 books on a shelf?
Answer
𝑃(5,3) =
5!
5 × 4 × 3 × 2 × 1 120
=
=
= 60.
(5 − 3)!
2×1
2
There are 60 ways.
2. Combinations
Definition
A combination is a selection of items where the order does not matter.
Formula
The number of combinations of 𝑛 objects taken 𝑟 at a time is given by:
𝐶(𝑛, 𝑟) =
𝑛!
𝑟! (𝑛 − 𝑟)!
Example 2:
How many ways can you choose 3 fruits from a basket of 5 different fruits?
Mr. Karim
2
Answer:
𝐶(5,3) =
5!
5×4×3×2×1
120
120
=
=
=
= 10.
3! (5 − 3)! (3 × 2 × 1)(2 × 1) 6 × 2
12
There are 10 ways.
3. Key Points to Remember
•
Factorial 𝑛!:
𝑛! = 𝑛 × (𝑛 − 1) × (𝑛 − 2) × ⋯ × 1
For example, 4! = 4 × 3 × 2 × 1 = 24.
•
When to use permutations vs. combinations:
•
o Use permutations when the arrangement or order is important.
o Use combinations when the order does not matter.
Zero and One Factorial:
0! = 1
•
and
1! = 1.
Practice Problems:
Try solving these problems to reinforce your understanding:
o Permutation Example: How many different 4-digit codes can be made using the
digits 1 to 9 without repetition?
o Combination Example: In a class of 20 students, how many ways can you
choose a committee of 4 students?
Understanding these fundamentals will help you solve a variety of problems in IGCSE
mathematics related to arrangements and selections.
Here are plenty of examples to help you master Permutations and Combinations for IGCSE!
1. Permutations (Order Matters)
Formula:
𝑃(𝑛, 𝑟) =
𝑛!
(𝑛 − 𝑟)!
Mr. Karim
3
(Used when order is important.)
Example 3: Arranging Books on a Shelf
How many ways can you arrange 4 books on a shelf from a selection of 7 books?
Answer:
𝑃(7,4) =
7!
7! 7 × 6 × 5 × 4 × 3!
= =
= 7 × 6 × 5 × 4 = 840
(7 − 4)! 3!
3!
Answer: 840 ways.
Example 4: Creating a 3-Letter Code
How many 3-letter codes can be formed using the letters A, B, C, D, E, if no letter is repeated?
Answer:
𝑃(5,3) =
5!
5! 5 × 4 × 3 × 2!
= =
= 5 × 4 × 3 = 60
(5 − 3)! 2!
2!
Answer: 60 codes.
Example 5: Arranging People in a Line
How many ways can 6 people sit in a row?
Answer:
𝑃(6,6) =
6!
6!
= = 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720
(6 − 6)! 0!
Answer: 720 ways.
Example 6: Passwords with Numbers
How many 4-digit PIN codes can be made using the digits 1-9, without repetition?
Answer:
𝑃(9,4) =
9!
9!
= = 9 × 8 × 7 × 6 = 3024
(9 − 4)! 5!
Answer: 3,024 PIN codes.
Mr. Karim
4
Example 7: Racing Positions
In a race with 10 runners, how many ways can we decide gold, silver, and bronze medalists?
Answer:
𝑃(10,3) =
10!
10!
=
= 10 × 9 × 8 = 720
(10 − 3)!
7!
Answer: 720 ways.
2. Combinations (Order Doesn’t Matter)
Formula:
𝐶(𝑛, 𝑟) =
𝑛!
𝑟! (𝑛 − 𝑟)!
(Used when order is not important.)
Example 8: Selecting a Committee
A school needs to choose 3 students from a class of 10 to form a committee. How many ways
can they be selected?
Answer:
𝐶(10,3) =
10!
10!
10 × 9 × 8
=
=
= 120
3! (10 − 3)! 3! × 7!
3×2×1
Answer: 120 ways.
Example 9: Choosing Ice Cream Flavors
If an ice cream shop has 8 flavors and you can choose 3 flavors for your sundae, how many
different combinations can you make?
Answer:
𝐶(8,3) =
8!
8!
8×7×6
=
=
= 56
3! (8 − 3)! 3! × 5! 3 × 2 × 1
Answer: 56 sundaes.
Mr. Karim
5
Example 10: Selecting a Football Team
A coach must choose 5 players from 12 for a match. How many possible teams can be formed?
Answer:
𝐶(12,5) =
12!
12!
12 × 11 × 10 × 9 × 8
=
=
= 792
5! (12 − 5)! 5! × 7!
5×4×3×2×1
Answer: 792 teams.
Example 11: Lottery Ticket Selection
A lottery requires selecting 6 numbers from 49. How many different lottery tickets are possible?
Answer:
𝐶(49,6) =
49!
49!
49 × 48 × 47 × 46 × 45 × 44
=
=
= 13,983,816
6! (49 − 6)! 6! × 43!
6×5×4×3×2×1
Answer: 13,983,816 different tickets!
Example 12: Forming a Study Group
A teacher wants to select 4 students from a group of 15 for a special project. How many ways
can they be selected?
Answer:
𝐶(15,4) =
15!
15!
15 × 14 × 13 × 12
=
=
= 1365
4! (15 − 4)! 4! × 11!
4×3×2×1
Answer: 1,365 ways.
3. Permutations vs. Combinations – Key Differences
Situation
Choosing team members (order doesn’t matter)
Permutation
𝑃(𝑛, 𝑟)
Combination
𝐶(𝑛, 𝑟)
Arranging books on a shelf (order matters)
Choosing a committee (order doesn’t matter)
Ranking students in a competition (order
matters)
Mr. Karim
6
Situation
Choosing ice cream flavors (order doesn’t
matter)
Permutation
𝑃(𝑛, 𝑟)
Combination
𝐶(𝑛, 𝑟)
4. Summary & Tips
Use permutations when order is important (e.g., seating arrangements, races,
passwords).
Use combinations when order doesn’t matter (e.g., committees, teams, lottery
selections).
Factorial shortcuts:
•
•
•
0! = 1
𝑛! = 𝑛 × (𝑛 − 1) × (𝑛 − 2) × ⋯ × 1
𝐶(𝑛, 𝑟) is always smaller than 𝑃(𝑛, 𝑟) since permutations consider order.
Mr. Karim
7
Let's review some past IGCSE questions on Permutations
and Combinations.
Question 1:
There are 720 different 5-digit numbers that can be formed using the digits 1, 2, 3, 5, 7 and 8, if
each digit may be used only once in any number.
How many of these 5-digit numbers are even and greater than 30 000?
[4]
Answer:
To solve this problem, we'll work with the following conditions:
1. Even number: The number must end in an even digit. In our set {1,2,3,5,7,8}, the even
digits are 2 and 8.
2. Greater than 30,000: For a 5-digit number to be greater than 30,000, the first digit must
be at least 3. Since our available digits are 1, 2, 3, 5, 7, and 8, the first digit must be one
of {3,5,7,8}.
We will handle the counting in two separate cases based on the last digit.
Case 1: Last Digit is 2
•
•
•
•
Last digit: Fixed as 2.
First digit: Must be chosen from {3,5,7,8} (4 choices).
After choosing the first and last digits, there are 4 digits remaining.
Middle three digits: The number of arrangements is the permutation of 4 digits taken 3
at a time:
𝑃(4,3) = 4 × 3 × 2 = 24.
Thus, the total number for this case is:
4 (first digit choices) × 24 (arrangements) = 96.
Case 2: Last Digit is 8
•
•
•
Last digit: Fixed as 8.
First digit: Must be chosen from {3,5,7} because 8 is already used as the last digit (3
choices).
After choosing the first and last digits, there are 4 digits remaining.
Mr. Karim
8
•
Middle three digits: Again, can be arranged in 𝑃(4,3) = 24 ways.
Thus, the total number for this case is:
3 (first digit choices) × 24 (arrangements) = 72.
Final Count
Adding the two cases together:
96 (Case 1) + 72 (Case 2) = 168.
Thus, the number of 5-digit numbers that are even and greater than 30,000 is:
168 .
Question 2:
The number of combinations of n items taken 3 at a time is 6 times the number of combinations
of n items taken 2 at a time.
Find the value of the constant n.
[4]
Answer:
We are given the equation based on combinations:
𝐶(𝑛, 3) = 6 × 𝐶(𝑛, 2)
Step 1: Write the Formulas for Combinations
Recall that:
𝐶(𝑛, 3) =
𝑛(𝑛 − 1)(𝑛 − 2)
6
and
𝐶(𝑛, 2) =
𝑛(𝑛 − 1)
2
Step 2: Set Up the Equation
Substitute the formulas into the given equation:
𝑛(𝑛 − 1)(𝑛 − 2)
𝑛(𝑛 − 1)
= 6×
6
2
Mr. Karim
9
Step 3: Simplify the Equation
1. Multiply both sides by 6 to eliminate the denominator on the left:
𝑛(𝑛 − 1)
𝑛(𝑛 − 1)(𝑛 − 2) = 6 ×
×6
2
Alternatively, simplify the right-hand side step-by-step:
𝑛(𝑛 − 1)
6×
= 3𝑛(𝑛 − 1)
2
So the equation becomes:
𝑛(𝑛 − 1)(𝑛 − 2)
= 3𝑛(𝑛 − 1)
6
2. Multiply both sides by 6:
𝑛(𝑛 − 1)(𝑛 − 2) = 18𝑛(𝑛 − 1)
Step 4: Solve for 𝑛
Assuming 𝑛(𝑛 − 1) ≠ 0 (which is true for 𝑛 > 1), we can divide both sides by 𝑛(𝑛 − 1):
𝑛 − 2 = 18
𝑛 = 18 + 2 = 20
Step 5: Verification
•
Calculate 𝐶(20,3):
𝐶(20,3) =
•
20 × 19 × 18 6840
=
= 1140
6
6
Calculate 𝐶(20,2):
𝐶(20,2) =
•
20 × 19 380
=
= 190
2
2
Check:
1140 = 6 × 190
(True)
Thus, the value of 𝑛 is:
20
Question 3:
The number of combinations of n items taken 3 at a time is 92n. Find the value of the constant n.
[4]
Answer:
Mr. Karim
10
We are given that the number of combinations of 𝑛 items taken 3 at a time is equal to 92𝑛. This
means:
𝐶(𝑛, 3) = 92𝑛.
Step 1: Write the Combination Formula
Recall the formula for combinations:
𝐶(𝑛, 3) =
𝑛(𝑛 − 1)(𝑛 − 2)
.
6
Substitute this into the equation:
𝑛(𝑛 − 1)(𝑛 − 2)
= 92𝑛.
6
Step 2: Simplify the Equation
Assuming 𝑛 ≠ 0, divide both sides by 𝑛:
(𝑛 − 1)(𝑛 − 2)
= 92.
6
Multiply both sides by 6 to eliminate the denominator:
(𝑛 − 1)(𝑛 − 2) = 552.
Step 3: Solve the Quadratic Equation
Expand the left-hand side:
𝑛2 − 3𝑛 + 2 = 552.
Subtract 552 from both sides:
𝑛2 − 3𝑛 − 550 = 0.
Now, solve this quadratic equation using the quadratic formula:
𝑛=
−𝑏 ± √𝑏 2 − 4𝑎𝑐
,
2𝑎
where 𝑎 = 1, 𝑏 = −3, and 𝑐 = −550. Calculate the discriminant:
𝛥 = (−3)2 − 4(1)(−550) = 9 + 2200 = 2209.
Since √2209 = 47, we have:
𝑛=
3 ± 47
.
2
This gives two solutions:
Mr. Karim
11
•
𝑛=
•
𝑛=
3+47
2
3−47
2
50
= 2 = 25,
=
−44
2
= −22.
Since 𝑛 must be positive, we discard 𝑛 = −22.
Final Answer
Thus, the value of 𝑛 is:
25 .
Question 4:
There are 360 different 4-digit numbers that can be formed using the digits 2, 3, 5, 7, 8 and 9, if
each digit may be used only once in any number.
How many of these numbers are odd and greater than 7000?
[4]
Answer:
We need to form 4-digit numbers from the digits {2,3,5,7,8,9} (each used at most once) that are:
1. Odd (last digit must be odd), and
2. Greater than 7000 (first digit must be 7 or higher).
Step 1: Determine the Eligible Digits for the First and Last Places
•
•
Odd digits available: 3,5,7,9.
For the number to be greater than 7000: The first digit must be at least 7. From our
set, the eligible first digits are: 7,8,9.
Step 2: Break into Cases Based on the First Digit
Case 1: First Digit is 7
• First digit: 7 (fixed).
• Last digit (must be odd): Options are from {3,5,7,9}, but 7 is already used. So available
choices: 3,5,9.
Choices = 3.
• Remaining digits: After using the first and last digits, there are 6 − 2 = 4 digits left.
• Middle two positions: These can be filled in 𝑃(4,2) = 4 × 3 = 12 ways.
Mr. Karim
12
Numbers for Case 1 = 3 × 12 = 36.
Case 2: First Digit is 8
• First digit: 8 (fixed).
• Last digit (must be odd): Options are from {3,5,7,9} (all available since 8 is even).
Choices = 4.
• Remaining digits: After using the first and last digits, there are 4 digits left.
• Middle two positions: 𝑃(4,2) = 4 × 3 = 12 ways.
Numbers for Case 2 = 4 × 12 = 48.
Case 3: First Digit is 9
• First digit: 9 (fixed).
• Last digit (must be odd): Options are from {3,5,7,9}, but 9 is already used. So available
choices: 3,5,7.
Choices = 3.
• Remaining digits: After using the first and last digits, there are 4 digits left.
• Middle two positions: 𝑃(4,2) = 12 ways.
Numbers for Case 3 = 3 × 12 = 36.
Step 3: Add Up All the Cases
Total numbers = 36(Case 1) + 48(Case 2) + 36(Case 3) = 120.
Thus, the total number of 4-digit numbers that are odd and greater than 7000 is:
120
Question 5:
A team of 8 players is to be chosen from 6 girls and 8 boys. Find the number of different ways
the team may be chosen if.
a) there are no restrictions,
b) all the girls are in the team,
c) at least 1 girl is in the team
[4]
Mr. Karim
13
Answer:
We have a total of 6 girls and 8 boys, so 14 players in all. We need to choose an 8-player team
under various conditions.
(
No Restrictions
a) The total number of ways to choose any 8 players from 14 is given by:
(
14
14!
)=
= 3003.
8
8! 6!
(b) All the Girls Are in the Team
If all 6 girls must be on the team, then we have 6 spots taken by the girls, and the remaining 2
spots must be filled by boys from the 8 available. The number of ways to choose 2 boys is:
8
8!
( )=
= 28.
2
2! 6!
(c) At Least 1 Girl Is in the Team
To count the number of teams with at least one girl, we can use the complement principle. First,
find the total number of teams (without any restrictions) and then subtract the number of teams
that have no girls (i.e., all boys).
14
•
Total teams (no restrictions): ( 8 ) = 3003.
•
Teams with no girls (all boys): Since there are 8 boys, choosing 8 out of 8 gives:
8
( ) = 1.
8
Thus, the number of teams with at least one girl is:
3003 − 1 = 3002.
Final Answers:
•
•
•
(a) 3003
(b) 28
(c) 3002
Mr. Karim
14
Question 6:
A group of people is to be selected from 5 women and 3 men.
a) Calculate the number of different groups of 4 people that have exactly 3 women.
b) Calculate the number of different groups of at most 4 people where the number of women
is the same as the number of men.
[4]
Answer:
We are given a group of 5 women and 3 men, and we need to solve for different groups under
two conditions:
(a) Groups of 4 People with Exactly 3 Women
To form a group of 4 people with exactly 3 women, we need to select:
•
•
3 women from the 5 women, and
1 man from the 3 men.
The number of ways to choose 3 women from 5 is:
5
5!
5×4
( )=
=
= 10.
3
3! (5 − 3)! 2 × 1
The number of ways to choose 1 man from 3 is:
3
( ) = 3.
1
Thus, the total number of ways to form the group is:
5
3
( ) × ( ) = 10 × 3 = 30.
3
1
So, the number of different groups of 4 people with exactly 3 women is:
30.
Mr. Karim
15
(b) Groups of At Most 4 People Where the Number of Women is the Same as the
Number of Men
For this part, we are selecting groups of up to 4 people, where the number of women is equal to
the number of men. This means we are looking for groups where the number of women and the
number of men are equal.
•
Case 1: Group size is 2 (1 woman and 1 man):
5
o Choose 1 woman from 5 women: (1) = 5.
3
o Choose 1 man from 3 men: (1) = 3.
The number of ways to choose this group is:
5
3
(1) × (1) = 5 × 3 = 15.
•
Case 2: Group size is 4 (2 women and 2 men):
5
5×4
o Choose 2 women from 5 women: (2) = 2×1 = 10.
3
o Choose 2 men from 3 men: (2) = 3.
The number of ways to choose this group is:
5
3
(2) × (2) = 10 × 3 = 30.
Total Number of Groups
The total number of groups of at most 4 people where the number of women is equal to the
number of men is the sum of the cases:
15 + 30 = 45.
So, the number of different groups of at most 4 people where the number of women is the same
as the number of men is:
45.
Mr. Karim
16
Question 7:
6 books are to be chosen from 8 different books.
Find the number of different selections of 6 books that could be made.
[2]
Answer:
Selecting 6 Books from 8
Since we are selecting 6 books out of 8 without regard to order, we use the combination formula:
8
8!
8×7
( )=
=
= 28.
6
6! (8 − 6)! 2 × 1
Thus, the number of different selections of 6 books is:
28 .
Question 8:
A quiz team of 6 children is to be chosen from a class of 8 boys and 10 girls. Find the number of
ways of choosing the team if
a) there are no restrictions,
b) there are more boys than girls in the team.
[5]
Answer:
We are given a class of 8 boys and 10 girls (18 children in total). We need to select a 6-member
quiz team under different conditions.
(a) No Restrictions
Since we are simply choosing 6 children out of 18 without any restrictions, we use the
combination formula:
(
18
18!
)=
6
6! (18 − 6)!
Mr. Karim
17
Calculating:
(
18
18 × 17 × 16 × 15 × 14 × 13
)=
= 18564.
6
6×5×4×3×2×1
So, the number of ways to choose the team with no restrictions is:
18564 .
(b) More Boys Than Girls in the Team
This means the number of boys must be greater than the number of girls in the selected team
of 6.
So, we consider the possible distributions of boys and girls:
Boys
4
5
6
Girls
2
1
0
Total Team Size
6
6
6
Case 1: 4 Boys, 2 Girls
• Ways to choose 4 boys from 8:
8
8!
8×7×6×5
( )=
=
= 70.
4
4! (8 − 4)! 4 × 3 × 2 × 1
• Ways to choose 2 girls from 10:
10
10!
10 × 9
( )=
=
= 45.
2
2! (10 − 2)!
2×1
Total for this case:
70 × 45 = 3150.
Case 2: 5 Boys, 1 Girl
• Ways to choose 5 boys from 8:
8
8!
8×7×6
( )=
=
= 56.
5
5! (8 − 5)! 3 × 2 × 1
• Ways to choose 1 girl from 10:
10
( ) = 10.
1
Total for this case:
56 × 10 = 560.
Mr. Karim
18
Case 3: 6 Boys, 0 Girls
• Ways to choose 6 boys from 8:
8
8!
8×7
( )=
=
= 28.
6
6! (8 − 6)! 2 × 1
• Ways to choose 0 girls from 10:
10
( ) = 1.
0
Total for this case:
28 × 1 = 28.
Final Total (Sum of All Cases)
3150 + 560 + 28 = 3738.
Thus, the number of ways to choose a team where there are more boys than girls is:
3738 .
Question 9:
a) How many even numbers less than 500 can be formed using the digits 1, 2, 3, 4 and 5?
Each digit may be used only once in any number.
[4]
b) A committee of 8 people is to be chosen from 7 men and 5 women. Find the number of
different committees that could be selected if
i.
the committee contains at least 3 men and at least 3 women,
[4]
ii.
the oldest man or the oldest woman, but not both, must be included in the
committee.
[2]
[Total: 10]
Answer:
Mr. Karim
19
Let's solve each part step by step.
(a) Counting Even Numbers Less Than 500
We are forming even numbers less than 500 using the digits 1, 2, 3, 4, and 5, with each digit
used only once.
One digit even number: 2,4 (two choices)
Arrangement: 2
Two digits even number:
•
First digit: 1,2,3,4 (four choices)
•
Last digit: 2,4 (two choices)
Arrangement: 4 × 2 = 8
Three digits even number:
Step 1: Identify Conditions
• The number must be even, so the last digit must be 2 or 4.
• The number must be less than 500, so the first digit must be 1, 2, 3 or 4.
Case 1: Last digit = 2
•
•
The first digit choices (must be 1, 3 or 4 because 2 is already used): 3 choices.
The second digit choices (must be one of the remaining 3 digits): 3 choices.
Total for this case:
3 × 3 = 9.
Case 2: Last digit = 4
•
•
The first digit choices (must be 1, 2, or 3 because 4 is already used): 3 choices.
The second digit choices (must be one of the remaining 3 digits): 3 choices.
Total for this case:
3 × 3 = 9.
Total Even Numbers Less Than 500
Adding both cases:
Mr. Karim
20
2 + 8 + 9 + 9 = 28.
Thus, the number of even numbers less than 500 that can be formed is:
28 .
(b) Selecting an 8-Person Committee from 7 Men and 5 Women
We are choosing 8 people from 7 men and 5 women under different conditions.
(i) The Committee Contains at Least 3 Men and at Least 3 Women
This means the possible distributions of men and women in the committee are:
Men
3
4
5
Women
5
4
3
Total Team Size
8
8
8
We compute each case separately.
Case 1: 3 Men, 5 Women
• Choose 3 men from 7:
•
7
7!
7×6×5
( )=
=
= 35.
3
3! (7 − 3)! 3 × 2 × 1
Choose 5 women from 5 (must take all):
5
( ) = 1.
5
Total for this case:
35 × 1 = 35.
Case 2: 4 Men, 4 Women
• Choose 4 men from 7:
•
7
7!
7×6×5
( )=
=
= 35.
4
4! (7 − 4)! 3 × 2 × 1
Choose 4 women from 5:
5
5!
5
( )=
= = 5.
4
4! (5 − 4)! 1
Total for this case:
Mr. Karim
21
35 × 5 = 175.
Case 3: 5 Men, 3 Women
• Choose 5 men from 7:
•
7
7!
7×6
( )=
=
= 21.
5
5! (7 − 5)! 2 × 1
Choose 3 women from 5:
5
5!
5×4
( )=
=
= 10.
3
3! (5 − 3)! 2 × 1
Total for this case:
21 × 10 = 210.
Total Committees with at Least 3 Men and 3 Women
Adding all cases:
35 + 175 + 210 = 420.
Thus, the number of ways to select a committee with at least 3 men and 3 women is:
420 .
(ii) The Oldest Man or the Oldest Woman, but Not Both, Must Be Included
•
Step 1: We fix either The oldest Man or The oldest Women, but not both.
•
Step 2: We choose the remaining 7 committee members from the remaining 10,
Case 1: The oldest Man is in the committee (so we pick 7 more, out of 10 remaining. Because,
there is no restriction given for the remaining 7 committee members)
10
10!
( )=
= 120.
7
7! (10 − 7)!
Case 2: The oldest Man is in the committee (so we pick 7 more out of 10 remaining)
10
10!
( )=
= 120.
7
7! (10 − 7)!
Total number of committees satisfying the condition:
120 + 120 = 240 .
Mr. Karim
22
Question 10:
A. A team of 7 people is to be chosen from 5 women and 7 men. Calculate the number of
different ways in which this can be done if
I.
there are no restrictions,
II.
the team is to contain more women than men.
[4]
B. (i) How many different 4-digit numbers, less than 5000, can be formed using 4 of the 6
digits 1, 2, 3, 4, 5 and 6 if no digit can be used more than once?
(ii) How many of these 4-digit numbers are divisible by 5?
[4]
[Total: 8]
Answer:
Let's go step by step.
(a) Choosing a Team of 7 People from 5 Women and 7 Men
We have 5 women and 7 men, and we need to choose 7 people.
(i) No Restrictions
We simply choose 7 people from the total of 12:
12
12!
12 × 11 × 10 × 9 × 8
( )=
=
= 792.
7
7! (12 − 7)!
5×4×3×2×1
Thus, the total number of ways to form the team without restrictions is:
792 .
Mr. Karim
23
(ii) The Team Must Contain More Women than Men
This means we need at least 4 women (since 3 women and 4 men would not satisfy this
condition).
Possible distributions:
Women Men Total Team
4
3
7
5
2
7
We calculate each case separately.
Case 1: 4 Women, 3 Men
• Choose 4 women from 5:
5
5!
5
( )=
= = 5.
4
4! (5 − 4)! 1
•
Choose 3 men from 7:
7
7!
7×6×5
( )=
=
= 35.
3
3! (7 − 3)! 3 × 2 × 1
Total ways for this case:
5 × 35 = 175.
Case 2: 5 Women, 2 Men
• Choose 5 women from 5 (all must be selected):
5
( ) = 1.
5
•
Choose 2 men from 7:
7
7!
7×6
( )=
=
= 21.
2
2! (7 − 2)! 2 × 1
Total ways for this case:
1 × 21 = 21.
Total Ways for More Women than Men
175 + 21 = 196.
Mr. Karim
24
Thus, the number of ways to form the team with more women than men is:
196 .
(b) Forming 4-Digit Numbers from 1, 2, 3, 4, 5, and 6
We must form 4-digit numbers using 4 different digits from the set {1, 2, 3, 4, 5, 6}.
(i) Numbers Less Than 5000
• The first digit must be 1, 2, 3, or 4 (to be less than 5000).
• The remaining 3 digits can be any of the remaining 5 digits.
• The number arrangement matters (since they are numbers), so we use permutations.
Step 1: Choose the First Digit
Choices: 1, 2, 3, or 4 → 4 choices.
Step 2: Arrange 3 More Digits
We now have 5 remaining digits, and we must select and arrange 3 of them:
𝑃(5,3) =
5!
5×4×3
=
= 60.
(5 − 3)!
1
Total Numbers Less Than 5000
4 × 60 = 240.
Thus, the number of 4-digit numbers less than 5000 is:
240 .
(ii) Numbers Divisible by 5
A number is divisible by 5 if its last digit is 0 or 5. Since 0 is not available, the last digit must
be 5.
Step 1: Choose the First Digit (Less than 5000)
The first digit must be 1, 2, 3, or 4 → 4 choices.
Step 2: Choose and Arrange 2 More Digits
Now, we need to select 2 more digits from the remaining 4 digits (since 5 is already chosen as
the last digit). The number arrangement matters.
Ways to select and arrange 2 digits:
Mr. Karim
25
𝑃(4,2) =
4!
4×3
=
= 12.
(4 − 2)!
1
Total Numbers Divisible by 5
4 × 12 = 48.
Thus, the number of 4-digit numbers divisible by 5 is:
48 .
Question 11:
a)
(i)
Find how many different 4-digit numbers can be formed using the digits 1, 2, 3, 4,
5 and 6 if no digit is repeated.
(ii)
How many of the 4-digit numbers found in part (i) are greater than 6000? (iii)
How many of the 4-digit numbers found in part (i) are greater than 6000 and are
odd?
[2]
b) A quiz team of 10 players is to be chosen from a class of 8 boys and 12 girls.
i.
Find the number of different teams that can be chosen if the team has to have equal
numbers of girls and boys.
ii.
Find the number of different teams that can be chosen if the team has to include the
youngest and oldest boy and the youngest and oldest girl.
[5]
[Total: 8]
Answer:
Let's go step by step.
(a) Forming 4-digit numbers from {1, 2, 3, 4, 5, 6} with no repetition
We are given 6 digits: 1, 2, 3, 4, 5, 6, and we need to form 4-digit numbers without repeating
any digit.
Mr. Karim
26
(i) Total number of different 4-digit numbers
We must select 4 digits and arrange them. Since order matters, we use permutations:
𝑃(6,4) =
6!
6! 6 × 5 × 4 × 3
= =
= 360.
(6 − 4)! 2!
1
Thus, the total number of 4-digit numbers is:
360 .
(ii) 4-digit numbers greater than 6000
For a number to be greater than 6000, its first digit must be 6.
•
•
The first digit is fixed as 6 → 1 choice.
The remaining 3 digits can be selected from {1, 2, 3, 4, 5}, and arranged in any order.
𝑃(5,3) =
5!
5×4×3
=
= 60.
(5 − 3)!
1
Thus, the number of 4-digit numbers greater than 6000 is:
60 .
(iii) 4-digit numbers greater than 6000 and odd
For a number to be odd, its last digit must be 1, 3, or 5.
•
•
•
The first digit is 6 → 1 choice.
The last digit must be 1, 3, or 5 → 3 choices.
The remaining 2 digits must be chosen from the remaining 4 digits ({2, 3, 4, 5} or {1, 2,
4, 5}, depending on the last digit).
Ways to arrange these 2 digits:
𝑃(4,2) =
4!
4×3
=
= 12.
(4 − 2)!
1
Total numbers satisfying both conditions:
1 × 3 × 12 = 36.
Thus, the number of 4-digit numbers greater than 6000 and odd is:
36 .
Mr. Karim
27
(b) Choosing a Quiz Team of 10 Players from 8 Boys and 12 Girls
We need to form a 10-player team from 8 boys and 12 girls.
(i) Team must have equal numbers of girls and boys
Since the team must have equal numbers of boys and girls, we must select 5 boys and 5 girls.
•
Choose 5 boys from 8:
8
8!
8×7×6
( )=
=
= 56.
5
5! (8 − 5)! 3 × 2 × 1
•
Choose 5 girls from 12:
12
12!
12 × 11 × 10 × 9 × 8
( )=
=
= 792.
5
5! (12 − 5)!
5×4×3×2×1
Total ways:
56 × 792 = 44352.
Thus, the number of teams with equal boys and girls is:
44,352 .
(ii) Team must include the youngest and oldest boy and the youngest and
oldest girl
Since these 4 players must be included in the team, we fix these 4 players and choose the
remaining 6 players from the remaining 6 boys and 10 girls.
•
•
We now have 6 boys and 10 girls remaining.
We must choose 6 more players from this 16.
(
16
16!
16 × 15 × 14 × 13 × 12 × 11
)=
=
= 8008.
6
6! (16 − 6)!
6×5×4×3×2×1
Thus, the number of teams that must include the youngest and oldest boy and girl is:
8,008 .
Mr. Karim
28
Question 12:
A committee of four is to be selected from 7 men and 5 women. Find the number of different
committees that could be selected if
(a) there are no restrictions,
(b) there must be two male and two female members.
A brother and sister, Ken and Betty, are among the 7 men and 5 women.
(c) Find how many different committees of four could be selected so that there are two male
and two female members which must include either Ken or Betty but not both.
[7]
Answer:
Let's solve each part step by step.
(a) No restrictions
We need to select 4 people from a total of 7 men + 5 women = 12 people.
Since order does not matter, we use combinations:
(
12
12!
12 × 11 × 10 × 9
)=
=
= 495.
4
4! (12 − 4)!
4×3×2×1
Thus, the number of different committees with no restrictions is:
495 .
(b) Committee must have 2 men and 2 women
We now select:
•
2 men from 7:
7
7!
7×6
( )=
=
= 21.
2
2! (7 − 2)! 2 × 1
•
2 women from 5:
5
5!
5×4
( )=
=
= 10.
2
2! (5 − 2)! 2 × 1
Total ways:
Mr. Karim
29
21 × 10 = 210.
Thus, the number of different committees with 2 men and 2 women is:
210 .
(c) Committee must have 2 men and 2 women, and must include either Ken or
Betty but not both
•
•
Step 1: We fix either Ken (M) or Betty (F), but not both.
Step 2: We choose the remaining 1 man and 2 women if Ken is chosen, or 2 men and 1
woman if Betty is chosen.
Case 1: Ken is in the committee (so we pick 1 more man and 2 women)
• Choose 1 more man from the remaining 6 men:
6
( ) = 6.
1
•
Choose 2 women from the remaining 4 women:
4
4!
4×3
( )=
=
= 6.
2
2! (4 − 2)! 2 × 1
Total for this case:
6 × 6 = 36.
Case 2: Betty is in the committee (so we pick 2 men and 1 more woman)
• Choose 2 men from the remaining 6 men:
6
6!
6×5
( )=
=
= 15.
2
2! (6 − 2)! 2 × 1
•
Choose 1 more woman from the remaining 4 women:
4
( ) = 4.
1
Total for this case:
15 × 4 = 60.
Total number of committees satisfying the condition:
36 + 60 = 96 .
Question 13:
Mr. Karim
30
There are twenty numbered balls in a bag. Two of the balls are numbered 0, six are numbered 1,
five are numbered 2 and seven are numbered 3, as shown in the table below.
Number on Ball 0 1 2 3
Frequency
2 6 5 7
Four of these balls are chosen at random, without replacement. Calculate the number of ways
this can be done so that
a. the four balls all have the same number,
b. the four balls all have different numbers,
[5]
Answer:
Let's go step by step for each part.
Given Data
We have 20 balls with the following distribution:
Number on Ball 0 1 2 3
Frequency
2 6 5 7
We need to select 4 balls randomly.
(a) The four balls all have the same number
For this case, we must pick 4 balls of the same number. We check whether we have at least 4
balls for each number:
•
•
•
•
Number 0 → Only 2 balls →
Not possible.
Number 1 → 6 balls →
Choose 4 from 6:
6
6!
6×5
( )=
=
= 15.
4
4! (6 − 4)! 2 × 1
Number 2 → 5 balls →
Choose 4 from 5:
5
5!
5
( )=
= = 5.
4
4! (5 − 4)! 1
Number 3 → 7 balls →
Choose 4 from 7:
7
7!
7×6×5
( )=
=
= 35.
4
4! (7 − 4)! 3 × 2 × 1
Mr. Karim
31
Total ways:
15 + 5 + 35 = 55 .
(b) The four balls all have different numbers
We must select one ball from each category (0, 1, 2, 3).
•
•
•
•
Choose 1 ball from 2 balls numbered 0:
2
( ) = 2.
1
Choose 1 ball from 6 balls numbered 1:
6
( ) = 6.
1
Choose 1 ball from 5 balls numbered 2:
5
( ) = 5.
1
Choose 1 ball from 7 balls numbered 3:
7
( ) = 7.
1
Total ways:
2 × 6 × 5 × 7 = 420.
Thus, the number of ways to select four balls with different numbers is:
420 .
Question 14:
Eight books are to be arranged on a shelf. There are 4 mathematics books, 3 geography books
and 1 French book.
a. Find the number of different arrangements of the books if there are no restrictions.
b. Find the number of different arrangements if the mathematics books have to be
kept together.
c. Find the number of different arrangements if the mathematics books have to be
kept together and the geography books have to be kept together.
[7]
Answer:
Mr. Karim
32
Let's solve each part step by step.
Given Data:
We have 8 books consisting of:
•
•
•
4 Mathematics books
3 Geography books
1 French book
(a) No Restrictions
Since all books are distinct, the total number of ways to arrange them is:
8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 40,320
Thus, the number of different arrangements is 40,320.
(b) Mathematics books kept together
If the 4 Mathematics books must stay together, we treat them as a single unit (block). Now,
instead of 8 books, we have:
•
•
•
1 unit of Mathematics books
3 Geography books
1 French book
So, we have 5 units to arrange, which can be done in:
5! = 5 × 4 × 3 × 2 × 1 = 120.
Within the Mathematics book block, the 4 books can be arranged among themselves in:
4! = 4 × 3 × 2 × 1 = 24.
Total arrangements:
5! × 4! = 120 × 24 = 2,880.
Thus, the number of different arrangements is 2,880.
Mr. Karim
33
(c) Mathematics and Geography books kept together
Now, both the 4 Mathematics books and the 3 Geography books must be kept together. This
means we treat them as two separate units (blocks):
•
•
•
1 unit of Mathematics books
1 unit of Geography books
1 French book
So, we have 3 units to arrange:
3! = 3 × 2 × 1 = 6.
Within the Mathematics book block, the 4 books can be arranged in:
4! = 24.
Within the Geography book block, the 3 books can be arranged in:
3! = 6.
Total arrangements:
3! × 4! × 3! = 6 × 24 × 6 = 864.
Thus, the number of different arrangements is 864.
Final Answers
(a) 40,320
(b) 2,880
(c) 864
Question 15:
Jack has won 7 trophies for sport and wants to arrange them on a shelf. He has 2 trophies for
cricket, 4 trophies for football and 1 trophy for swimming. Find the number of different
arrangements if
a. there are no restrictions,
b. the football trophies are to be kept together,
Mr. Karim
34
c. the football trophies are to be kept together and the cricket trophies are to be kept
together.
[7]
Answer:
Let's go step by step for each part.
Given Data
Jack has 7 trophies, categorized as:
•
•
•
2 Cricket trophies
4 Football trophies
1 Swimming trophy
(a) No Restrictions
Since all trophies are distinct, the total number of ways to arrange them is:
7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5,040.
Thus, the number of different arrangements is 5,040.
(b) Football Trophies Kept Together
If the 4 Football trophies must stay together, we treat them as a single unit (block). Now,
instead of 7 trophies, we have:
•
•
•
1 unit of Football trophies
2 Cricket trophies
1 Swimming trophy
So, we have 4 units to arrange, which can be done in:
4! = 4 × 3 × 2 × 1 = 24.
Within the Football block, the 4 trophies can be arranged among themselves in:
4! = 24.
Total arrangements:
4! × 4! = 24 × 24 = 576.
Mr. Karim
35
Thus, the number of different arrangements is 576.
(c) Football and Cricket Trophies Kept Together
Now, both the 4 Football trophies and the 2 Cricket trophies must be kept together. This
means we treat them as two separate units (blocks):
•
•
•
1 unit of Football trophies
1 unit of Cricket trophies
1 Swimming trophy
So, we have 3 units to arrange:
3! = 3 × 2 × 1 = 6.
Within the Football block, the 4 trophies can be arranged in:
4! = 24.
Within the Cricket block, the 2 trophies can be arranged in:
2! = 2.
Total arrangements:
3! × 4! × 2! = 6 × 24 × 2 = 288.
Thus, the number of different arrangements is 288.
Final Answers
(a) 5,040
(b) 576
(c) 288
Mr. Karim