Permutations and Combinations

Additional Mathematics· Syllabus sections 11.1, 11.2, 11.3· 25 min read

1. 1. Factorial Notation and Key Rules★★☆☆☆⏱ 5 min

📘 Definition

Factorial

n! = n \times (n-1) \times (n-2) \times \dots \times 1

For a positive integer n, n! is the product of all positive integers from 1 to n. The special case $0! = 1$ is a standard rule to avoid undefined values in permutation calculations.

Example:

5! = 5 × 4 × 3 × 2 × 1 = 120

Factorials are the foundation of permutation and combination calculations. For non-calculator papers, always cancel common factorial terms before multiplying to avoid large, error-prone calculations.

📐 Worked Example

Simplify $\frac{7 !}{4 !}$ without using a calculator.

1
Step 1: Expand the larger factorial to include the smaller factorial as a term:
$7! = 7 \times 6 \times 5 \times 4!$
2
Step 2: Cancel the common 4! term in numerator and denominator:
$\frac{7!}{4!} = \frac{7 \times 6 \times 5 \times 4!}{4!} = 7 \times 6 \times 5$
3
Step 3: Calculate the final value:
$7 \times 6 \times 5 = 210$

2. 2. Permutations (Order Matters)★★★☆☆⏱ 7 min

📘 Definition

Permutation

^nP_r = \frac{n!}{(n-r)!}

An ordered arrangement of r distinct objects selected from n objects, with no repetition allowed. Use permutations when swapping two selected items creates a different outcome.

Example:

Number of ways to arrange 2 letters from {A,B,C}: $^{3} P_{2} = 6$ (AB, BA, AC, CA, BC, CB)

Common permutation use cases include arranging people in a line, awarding ranked prizes, or arranging distinct letters in a word.

📐 Worked Example

How many ways can 4 distinct books be arranged on a shelf?

1
Step 1: Identify n=4 and r=4 (we are arranging all 4 books):
$^4P_4 = \frac{4!}{(4-4)!} = \frac{4!}{0!}$
2
Step 2: Substitute $0! = 1$ and calculate:
$\frac{4!}{1} = 4 \times 3 \times 2 \times 1 = 24$

3. 3. Combinations (Order Does Not Matter)★★★☆☆⏱ 7 min

📘 Definition

Combination

^nC_r = \frac{n!}{r!(n-r)!}

An unordered selection of r distinct objects selected from n objects, with no repetition allowed. Use combinations when swapping two selected items creates the same outcome.

Example:

Number of ways to select 2 letters from {A,B,C}: $^{3} C_{2} = 3$ (AB, AC, BC)

Common combination use cases include selecting teams, choosing items for a trip, or picking questions to answer on an exam paper.

📐 Worked Example

How many ways can a 3-person committee be selected from 8 eligible students?

1
Step 1: Identify n=8 and r=3 (committee positions have no hierarchy, so order does not matter):
$^8C_3 = \frac{8!}{3!(8-3)!} = \frac{8!}{3!5!}$
2
Step 2: Cancel common 5! terms:
$\frac{8 \times 7 \times 6 \times 5!}{3 \times 2 \times 1 \times 5!} = \frac{8 \times 7 \times 6}{6}$
3
Step 3: Simplify to get the final answer:
$8 \times 7 = 56$

4. 4. Solving Constrained Counting Problems★★★★☆⏱ 8 min

Most exam problems include simple constraints: either specific objects must be included in the selection/arrangement, or specific objects must be excluded. Always handle the fixed constraint first, then calculate the remaining free choices.

📐 Worked Example

A group of 10 students includes 2 siblings. How many 4-person teams can be selected that include both siblings?

1
Step 1: Account for the 'must include' constraint: both siblings are already selected, so we only need to choose 2 more members from the remaining 8 students.
2
Step 2: Calculate the combination for the remaining positions:
$^8C_2 = \frac{8!}{2!6!} = \frac{8 \times 7}{2 \times 1} = 28$

✓ Quick check

Is the number of ways to award 1st, 2nd and 3rd place prizes to 10 runners a permutation or combination problem?
- Permutation
- Combination
Reveal answer
Permutation —
Order matters for ranked prizes: 1st place is different from 2nd place, so this is a permutation problem.

5. 5. Algebraic Problems with $^nP_r$ and $^nC_r$★★★★☆⏱ 8 min

A very common exam question gives you an equation involving $^{n} P_{r}$ or $^{n} C_{r}$ and asks you to find the value of n. The method is always the same: replace each term with its factorial definition, cancel the common factorial factors, and you are left with a simple polynomial equation (usually a quadratic) to solve. Remember to reject any solution that is negative or too small for the expression to be defined (you need $n \geq r$ ).

📐 Worked Example

Given that $^{n} C_{4} = 6^{n} C_{2}$ , find the value of n.

1
Step 1: Write each combination using $^{n} C_{r} = \frac{n !}{r ! ( n - r )!}$ :
$\frac{n!}{4!(n-4)!} = 6 \times \frac{n!}{2!(n-2)!}$
2
Step 2: Divide both sides by $n!$ (which is non-zero), then cross-multiply:
$2!\,(n-2)! = 6 \times 4!\,(n-4)!$
3
Step 3: Substitute $2! = 2$ and $4! = 24$ , expand $(n - 2)! = (n - 2) (n - 3) (n - 4)!$ , then cancel $(n - 4)!$ :
$2(n-2)(n-3) = 144$
4
Step 4: Divide by 2 and expand into a quadratic:
$(n-2)(n-3) = 72 \;\Rightarrow\; n^2 - 5n - 66 = 0$
5
Step 5: Factorise and solve, rejecting the negative root:
$(n-11)(n+6) = 0 \;\Rightarrow\; n = 11$

📐 Worked Example

Show that $^{n} C_{3} +^{n} C_{2} = \frac{1}{6} (n^{3} - n)$ , and hence solve $^{n} C_{3} +^{n} C_{2} = 35$ .

1
Step 1: Write each term as a product using the factorial definition:
$^nC_3 = \frac{n(n-1)(n-2)}{6}, \quad {}^nC_2 = \frac{n(n-1)}{2}$
2
Step 2: Put both terms over a common denominator of 6:
$^nC_3 + {}^nC_2 = \frac{n(n-1)(n-2)}{6} + \frac{3n(n-1)}{6}$
3
Step 3: Factor out $n (n - 1)$ from the numerator:
$= \frac{n(n-1)[(n-2)+3]}{6} = \frac{n(n-1)(n+1)}{6}$
4
Step 4: Since $n (n - 1) (n + 1) = n (n^{2} - 1) = n^{3} - n$ , the identity is shown:
$^nC_3 + {}^nC_2 = \frac{1}{6}(n^3 - n)$
5
Step 5: Hence set the expression equal to 35 and solve the resulting cubic by inspection (n must be a positive integer with $n \geq 3$ ):
$\frac{1}{6}(n^3 - n) = 35 \;\Rightarrow\; n^3 - n - 210 = 0 \;\Rightarrow\; n = 6$

6. 6. Together, Not Together, and At Least / At Most★★★★☆⏱ 8 min

Two constraint types appear again and again in 0606 counting questions. The first is when certain items must stay together (adjacent) or must be kept apart (not together) in a row. The second is when a selection must contain at least, or at most, a given number of a certain kind of item. Each has a reliable standard method.

📘 Definition

Block (bundling) method for 'together'

\text{arrangements} = (\text{units})! \times k!

When k specific items must be adjacent in a row, tie them together into a single block. Arrange that block alongside the other separate items as units, giving (number of units)! arrangements, then multiply by k! for the internal order of the items inside the block. For 'not together', use the complement: (total arrangements) minus (arrangements with the items together).

Example:

2 people must be adjacent among 5 in a row: tie them into 1 block, giving 4 units, so $4! \times 2! = 48$ .

📐 Worked Example

Five different people stand in a row for a photograph. (a) In how many arrangements do two particular people, A and B, stand next to each other? (b) In how many arrangements are A and B NOT next to each other?

1
Part (a), Step 1: Tie A and B together into a single block. This block, together with the other 3 people, gives 4 units to arrange in a row:
$\text{units} = 4 \;\Rightarrow\; 4! = 24$
2
Part (a), Step 2: Inside the block, A and B can be ordered as AB or BA, giving $2!$ internal arrangements. Multiply the two stages:
$4! \times 2! = 24 \times 2 = 48$
3
Part (b), Step 1: Find the total number of unrestricted arrangements of all 5 people:
$5! = 120$
4
Part (b), Step 2: 'Not together' is the complement of 'together', so subtract the part (a) result:
$120 - 48 = 72$

📐 Worked Example

A team of 8 players is chosen from a squad of 15. Four of the players are close friends who will only play if all four are chosen together, or none of them is chosen. In how many ways can the team be selected?

1
Step 1: Selection is unordered, so use combinations. Unlike the row block method there is no internal $\times k!$ here, because the team has no order. The condition 'all four together or none' splits into two separate cases: all four friends in, or all four out.
2
Step 2: Case 1 — all four friends included. They fill 4 of the 8 places, so choose the remaining 4 players from the other 11:
${}^{11}C_{4} = 330$
3
Step 3: Case 2 — all four friends excluded. Choose all 8 players from the other 11:
${}^{11}C_{8} = {}^{11}C_{3} = 165$
4
Step 4: The two cases are mutually exclusive, so add them:
$330 + 165 = 495$

📐 Worked Example

A team of 4 people is chosen from 6 men and 4 women. In how many ways can the team be chosen if it must contain at least 2 women? Selection is unordered.

1
Step 1: 'At least 2 women' means exactly 2, exactly 3, or exactly 4 women. Split into cases and count each with combinations, since order does not matter.
2
Step 2: Exactly 2 women (and so 2 men): choose 2 of the 4 women and 2 of the 6 men:
${}^{4}C_{2} \times {}^{6}C_{2} = 6 \times 15 = 90$
3
Step 3: Exactly 3 women (and 1 man):
${}^{4}C_{3} \times {}^{6}C_{1} = 4 \times 6 = 24$
4
Step 4: Exactly 4 women (and 0 men):
${}^{4}C_{4} \times {}^{6}C_{0} = 1 \times 1 = 1$
5
Step 5: Add the cases for the final total:
$90 + 24 + 1 = 115$
6
Check with the complement method (total teams minus those with 0 or 1 woman); both methods agree:
${}^{10}C_{4} - ({}^{6}C_{4} + {}^{4}C_{1}\,{}^{6}C_{3}) = 210 - (15 + 80) = 115$

7. 7. Forming Numbers and Passwords★★★★☆⏱ 9 min

A very common 0606 question gives you a set of digits and asks how many k-digit numbers you can make from them, usually with an extra condition: the number must be even, odd, divisible by 5, or greater than some value. Treat each digit position as a slot to be filled and use the multiplication principle, but always deal with the restricted positions first.

📘 Definition

Position (slot) method for forming numbers

\text{total} = (\text{choices for restricted slots}) \times (\text{choices for the remaining slots})

Fill the most restricted position first, then work through the rest. The last (units) digit is fixed by any even/odd/divisibility condition: even numbers end in 0, 2, 4, 6 or 8; odd numbers end in 1, 3, 5, 7 or 9; multiples of 5 end in 0 or 5. The first digit can never be 0, or the number would have fewer digits. When 0 is an allowed last digit AND is banned from the first slot, the two restrictions clash over the same digit, so split the count into cases on the last digit.

Example:

Three-digit even numbers from {1,2,3,4,5}, no repeats: the units digit must be 2 or 4 (2 ways), then the first two slots take 2 of the other 4 digits, $^{4} P_{2} = 12$ , giving $2 \times 12 = 24$ .

📐 Worked Example

Using the digits 1, 2, 3, 4, 5 with no digit repeated, how many three-digit even numbers can be formed?

1
Step 1: A number is even exactly when its units (last) digit is even. Among {1,2,3,4,5} the even digits are 2 and 4, so fill this restricted position first:
$\text{units digit} \in \{2,4\} \Rightarrow 2 \text{ choices}$
2
Step 2: One digit is now used, leaving 4 digits for the hundreds and tens positions. Arrange 2 of these 4 in order (there is no 0 to worry about here):
$^4P_2 = 4 \times 3 = 12$
3
Step 3: Multiply the stages using the multiplication principle:
$2 \times 12 = 24$
4
Step 4 (check): There are $^{5} P_{3} = 60$ three-digit numbers in total. The odd ones have units digit in {1,3,5} (3 choices) with the other two slots giving $^{4} P_{2} = 12$ , i.e. $3 \times 12 = 36$ . So the even count is $60 - 36 = 24$ , which agrees.
$60 - 36 = 24$

📐 Worked Example

Using the digits 0, 1, 2, 3, 4 with no digit repeated: (a) how many three-digit numbers can be formed, and (b) how many of them are even?

1
Part (a), Step 1: The hundreds (first) digit cannot be 0, otherwise the result is not a three-digit number. Choose it from {1,2,3,4}:
$\text{hundreds digit} \in \{1,2,3,4\} \Rightarrow 4 \text{ choices}$
2
Part (a), Step 2: The tens and units positions are filled from the 4 digits that remain (this pool now includes 0). Arrange 2 of these 4:
$^4P_2 = 4 \times 3 = 12$
3
Part (a), Step 3: Multiply the stages:
$4 \times 12 = 48$
4
Part (b), Step 4: An even number ends in an even digit; among {0,1,2,3,4} those are 0, 2 and 4. Because 0 is also banned from the hundreds slot, split into cases on the units digit so the two rules do not clash.
5
Part (b), Step 5: Case 1 — units digit = 0. The hundreds digit is then any of {1,2,3,4} (4 choices) and the tens digit is any of the 3 that remain:
$1 \times 4 \times 3 = 12$
6
Part (b), Step 6: Case 2 — units digit = 2 or 4 (2 choices). Now 0 is still unused but banned from the hundreds slot, so the hundreds digit has $4 - 1 = 3$ choices, and the tens digit is any of the 3 that remain:
$2 \times 3 \times 3 = 18$
7
Part (b), Step 7: Add the cases for the total number of even three-digit numbers:
$12 + 18 = 30$
8
Part (b), Step 8 (check): From part (a) there are 48 numbers in total. The odd ones have units digit in {1,3} (2 choices), a non-zero hundreds digit (3 choices) and any tens digit (3 choices), giving $2 \times 3 \times 3 = 18$ odd. So the even count is $48 - 18 = 30$ , which agrees.
$48 - 18 = 30$

The same slot method works for codes and passwords built from a mixed set of characters — letters, digits and symbols together. Fill any restricted positions first, then fill the rest, and use the complement (total minus unwanted) for an 'at least one' condition.

📐 Worked Example

A password is 4 characters long and is formed from 5 different letters, 3 different digits and 2 different symbols (10 distinct characters in total). No character may be repeated. (a) How many passwords are possible? (b) How many begin with a letter and end with a digit? (c) How many contain at least one letter?

1
Part (a): There are 10 distinct characters filling 4 ordered positions with no repeats, so this is a straightforward permutation:
$^{10}P_4 = 10 \times 9 \times 8 \times 7 = 5040$
2
Part (b), Step 1: Fill the two restricted positions first. The first character is one of the 5 letters and the last is one of the 3 digits:
$5 \times 3 = 15$
3
Part (b), Step 2: The two middle positions are filled from the 8 characters that remain, arranged in order:
$^8P_2 = 8 \times 7 = 56$
4
Part (b), Step 3: Multiply the stages:
$15 \times 56 = 840$
5
Part (c): 'At least one letter' is quickest by the complement. A password with no letter at all uses only the 5 non-letter characters (3 digits + 2 symbols):
$^5P_4 = 5 \times 4 \times 3 \times 2 = 120$
6
Subtract these from the total in part (a):
$5040 - 120 = 4920$

8. Common Pitfalls

Wrong move:

Using permutations for unordered selections (e.g. $^{5} P_{2}$ for 2-person teams from 5 people)

Why:

Permutations count each unordered pair twice (e.g. Alice+Bob and Bob+Alice as separate) when they are identical for unordered selections.

Correct move:

Use combinations for unordered selections, permutations for ordered arrangements.

Wrong move:

Forgetting $0! = 1$ , leading to division by zero when calculating $^{n} P_{n}$

Why:

The definition of factorial explicitly sets $0! = 1$ to avoid undefined values when arranging all n objects.

Correct move:

Memorize $0! = 1$ as a non-negotiable rule for all factorial calculations.

Wrong move:

Attempting to solve circular arrangement problems with standard permutation formulas

Why:

Circular arrangements have different counting rules and are explicitly excluded from the 0606 syllabus.

Correct move:

All 0606 arrangement problems are linear, so use standard $^{n} P_{r}$ formulas only.

Wrong move:

Mixing permutations and combinations in the same problem

Why:

0606 problems are designed to use either permutations OR combinations, not both; mixing methods leads to overcounting.

Correct move:

First confirm if order matters, then use only the relevant formula for the entire problem.

Wrong move:

Calculating full factorial values instead of cancelling first on non-calculator papers

Why:

Large factorial values lead to error-prone multiplications and waste time.

Correct move:

Cancel common factorial terms in numerator and denominator before multiplying.

9. Quick Reference Cheatsheet

Concept	Formula	Use Case
Factorial	$n! = n \times (n - 1) \times ... \times 1$ , $0! = 1$	Foundational calculation for all counting problems
Permutation	$^{n} P_{r} = \frac{n !}{( n - r )!}$	Ordered arrangements, no repetition of objects
Combination	$^{n} C_{r} = \frac{n !}{r ! ( n - r )!}$	Unordered selections, no repetition of objects
Must include (combination / selection)	Fix the k required items, choose the rest: $^{n - k} C_{r - k}$	Selecting r from n where k specific items must be included (order does not matter)
Must include (permutation / arrangement)	Place the k required items in positions ( $^{r} P_{k}$ ), then arrange the other $r - k$ from the remaining ( $^{n - k} P_{r - k}$ )	Arranging r from n where k specific items must be included (order matters)
Must exclude constraint	Reduce n by number of excluded items, keep r the same	Problems forbidding specific objects from selection

10. Frequently Asked

Do I get the nPr and nCr formulas in the 0606 exam?

No, these formulas are not included in the 0606 formula booklet, so you must memorise them for both calculator and non-calculator papers.

Can I use calculator nPr/nCr functions on Paper 2?

Yes, you may use built-in calculator functions for permutations and combinations on Paper 2, but always show your working to earn method marks.

What's Next

Now that you have mastered permutations and combinations for CIE IGCSE Additional Mathematics 0606, you can apply these counting skills to other syllabus areas, most directly the binomial theorem, where the same $^{n} C_{r}$ coefficients reappear in the expansion of $(a + b)^{n}$ . Permutation and combination questions make up an average of 5-8% of your total exam marks, so practice past paper questions to build speed and accuracy, especially for non-calculator paper problems where you will need to simplify factorial fractions manually. Remember that circular arrangements, repeated objects, and mixed permutation/combination problems are out of scope for 0606, so you do not need to practice these for your exam. Next, move on to the Series unit (binomial theorem and progressions), where these counting skills are reused to write out binomial expansions.