Permutations and Combinations

CIE A-Level Mathematics· 35 min read

1. Permutations of Distinct Items★★☆☆☆⏱ 10 min

📘 Definition

Permutation

$^{n} P_{r} = P (n, r)$

An ordered arrangement or selection of $r$ items from $n$ distinct items, where order of selection matters.

Example:

Arranging 2 out of 3 letters A,B/C gives 6 unique permutations.

When there is no repetition of items, the number of permutations is calculated with the formula:

^nP_r = \frac{n!}{(n-r)!} = n(n-1)(n-2)...(n-r+1)

📐 Worked Example

How many permutations are there of 4 letters selected from the word HEPTAGON (all letters are distinct)?

1
1. Count the total number of distinct letters: $n = 8$ , we select $r = 4$ .
2
1. Substitute into the permutation formula:
3
$^8P_4 = \frac{8!}{(8-4)!} = \frac{8!}{4!}$
4
1. Simplify to get: $8 \times 7 \times 6 \times 5 = 1680$

Exam tip:

Always confirm that all items are distinct before using this formula.

2. Permutations with Repetition & Identical Items★★★☆☆⏱ 15 min

There are two common special cases for permutations involving repeated items:

Repetition allowed: When items can be reused (e.g. 4-digit PIN codes), the number of permutations of $r$ items from $n$ is $n^{r}$ .
Identical items in arrangement: When arranging all $n$ items with repeated identical items, divide by the factorial of the count of each repeated item to avoid overcounting.

\text{Distinct arrangements} = \frac{n!}{k_1! \cdot k_2! \cdot ... \cdot k_m!}

Where $k_{i}$ is the number of identical items of type $i$ .

📐 Worked Example

Find the number of distinct arrangements of the letters in the word STATISTICS.

1
1. Count total letters: $n = 10$ .
2
1. Identify repeated letters: S = 3, T = 3, I = 2, A = 1, C = 1.
3
1. Substitute into the formula:
4
$\frac{10!}{3! \cdot 3! \cdot 2!} = \frac{3628800}{6 \times 6 \times 2}$
5
1. Simplify to get 50400 distinct arrangements.

3. Combinations of Distinct Items★★☆☆☆⏱ 12 min

📘 Definition

Combination

$^{n} C_{r} = (r n)$

An unordered selection of $r$ items from $n$ distinct items, where order does not change the outcome.

Example:

Selecting 2 out of 3 letters A,B/C gives only 3 unique combinations.

Combinations are calculated by adjusting permutations to remove duplicate orderings of the same selection. The formula is:

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

A useful identity to simplify calculations is $(r n) = (n - r n)$ .

📐 Worked Example

A committee of 3 people is selected from 8 candidates. How many different committees are possible?

1
1. Order of committee members does not matter, so use combinations: $n = 8$ , $r = 3$ .
2
1. Substitute into the formula:
3
$\binom{8}{3} = \frac{8!}{3! \cdot 5!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1}$
4
1. Simplify to get 56 different committees.

✓ Quick check

Is arranging people in a queue for a photo a permutation or combination?
- Permutation
- Combination
Reveal answer
Permutation —
Order of position matters in a queue, so it is a permutation.

4. Combined Problems with Restrictions★★★★☆⏱ 15 min

Most CIE exam questions combine permutations and combinations with restrictions, such as 'must include at least X' or 'certain items must be together'. A common efficient approach is to calculate the total number of outcomes, then subtract the number of invalid outcomes that do not meet the restriction.

📐 Worked Example

How many 5-person teams can be selected from 6 men and 5 women, if the team must have at least 2 women?

1
1. Calculate total number of ways to select 5 people from 11:
2
$\binom{11}{5} = 462$
3
1. Calculate invalid outcomes (fewer than 2 women = 0 women or 1 woman):
4
0 women (all men): $(5 6) = 6$
5
1 woman + 4 men: $(1 5) \times (4 6) = 5 \times 15 = 75$
6
1. Total invalid = $6 + 75 = 81$
7
1. Valid outcomes = total - invalid = $462 - 81 = 381$

Exam tip:

For 'at least' restrictions, the total minus invalid method is almost always faster than counting every valid case separately.

5. Common Pitfalls

Wrong move:

Treating unordered selections as permutations

Why:

You overcount by a factor of $r!$ because you count every ordering of the same selection as unique

Correct move:

Always check if order matters: if swapping two items does not create a new outcome, use combinations.

Wrong move:

Forgetting to divide by factorials of identical items

Why:

Swapping identical items does not create a new arrangement, so you overcount

Correct move:

List all repeated items explicitly before calculating, then divide by the product of their factorials.

Wrong move:

Adding instead of multiplying for independent selections

Why:

Confusing the addition principle for mutually exclusive cases with multiplication for independent choices

Correct move:

Use multiplication for 'and' choices, addition for 'or' mutually exclusive cases.

Wrong move:

Forgetting to permute items that must be kept together

Why:

When grouping items into a block, you miss the internal arrangements of the block

Correct move:

Count arrangements of the blocks, then multiply by the number of permutations of items within the block.

Wrong move:

Double counting outcomes that satisfy multiple conditions

Why:

When adding cases for overlapping restrictions, you count outcomes that meet multiple conditions twice

Correct move:

Use the principle of inclusion-exclusion or use the total minus invalid method to avoid double counting.

6. Quick Reference Cheatsheet

Scenario	Formula/Method
Permutations, $r$ distinct from $n$ , no repetition	$P (n, r) = \frac{n !}{( n - r )!}$
Permutations, $r$ from $n$ , repetition allowed	$n^{r}$
Arrange $n$ items with repeats $k_{1}, k_{2} ...$	$\frac{n !}{k _{1} ! k _{2} ! ...}$
Combinations, $r$ unordered from $n$	$(r n) = \frac{n !}{r ! ( n - r )!}$
At least X restriction	Total outcomes $-$ invalid outcomes
Items must be kept together	Treat as 1 block, multiply by internal permutations

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 5
Committee selection problem
2023 · 5
Letter arrangement problem
2021 · 5
Restricted team selection

Going deeper

syllabusCIE 9709 Syllabus Page

What's Next

Permutations and combinations are the foundational counting skill for almost all probability questions in CIE A-Level Statistics 1. You will directly use combinations to derive and calculate probabilities for the binomial distribution, the next core topic in this unit. These counting methods also appear regularly in problems involving conditional probability, discrete random variables, and further statistics topics in A2 level. Mastery of this subtopic is critical to scoring well on probability sections of the exam.