# Permutations and Combinations

> CIE A-Level Mathematics · Probability & Statistics 1
> Source: https://www.owlsprep.com/study/cie-9709-u4-permutations-and-combinations/

This subtopic covers core counting methods for arrangements (ordered permutations) and selections (unordered combinations) of items. It is a foundational skill for probability, binomial distributions and all further statistics topics in CIE A-Level.

**Prerequisites:** Factorials and the fundamental counting principle

## Learning objectives

- Distinguish between ordered permutations and unordered combinations
- Calculate permutations with and without repetition/identical items
- Calculate combinations for unordered selections
- Solve restricted counting problems for CIE exam questions

## How often is this tested?

Based on OwlsPrep's analysis of official CIE 9709 S1 past papers (2016–2025): **Permutations and Combinations** appears **86 times** in the last 10 years — **19.6%** of all S1 questions (86 of 438).

Most-tested forms: Arrangements with restrictions (64), Selections (51), Permutations (nPr) (29), Combinations (nCr) (23).

## Permutations of Distinct Items

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

*Notation:* $^nP_r = P(n,r)$

*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. 2. Substitute into the permutation formula:
3. $$^8P_4 = \frac{8!}{(8-4)!} = \frac{8!}{4!}$$
4. 3. Simplify to get: $8 \times 7 \times 6 \times 5 = 1680$

> **Exam tip:** Always confirm that all items are distinct before using this formula.

## Arrangements with Identical Items

This section covers the syllabus skill of arranging identical items. First we separate it from a related prerequisite (reusing items) that is often confused with it:

1. **Reusing items — prerequisite, multiplication principle (not the syllabus 'repetition')**: When items may be reused (e.g. each position of a 4-digit PIN code), the number of arrangements of $r$ items from $n$ is $n^r$. This follows from the basic counting principle and is *different* from the identical-items case below.
2. **Identical items — the syllabus 'repetition' (e.g. the letters of NEEDLESS)**: When arranging all $n$ items where some are identical, 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. 2. Identify repeated letters: S = 3, T = 3, I = 2, A = 1, C = 1.
3. 3. Substitute into the formula:
4. $$\frac{10!}{3! \cdot 3! \cdot 2!} = \frac{3628800}{6 \times 6 \times 2}$$
5. 4. Simplify to get 50400 distinct arrangements.

## Combinations of Distinct Items

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

*Notation:* $^nC_r = \binom{n}{r}$

*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 $\binom{n}{r} = \binom{n}{n-r}$.

**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. 2. 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. 3. Simplify to get 56 different committees.

**Check your understanding**

1. Is arranging people in a queue for a photo a permutation or combination?

   - Permutation
   - Combination

   *Why:* Order of position matters in a queue, so it is a permutation.

## Combined Problems with Restrictions

Most CIE exam questions combine permutations and combinations with restrictions, such as 'must include at least X', 'certain items must be together', or 'certain items must not 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. 2. Calculate invalid outcomes (fewer than 2 women = 0 women or 1 woman):
4. 0 women (all men): $\binom{6}{5} = 6$
5. 1 woman + 4 men: $\binom{5}{1} \times \binom{6}{4} = 5 \times 15 = 75$
6. 3. Total invalid = $6 + 75 = 81$
7. 4. Valid outcomes = total - invalid = $462 - 81 = 381$

**Worked example:** In how many ways can 8 people stand in a line if two particular people, A and B, must **not** stand next to each other?

1. 1. Count all arrangements with no restriction: $8! = 40320$.
2. 2. Count the arrangements where A and B **are** together by gluing them into one block (7 items to arrange) and multiplying by 2 for the internal order AB or BA: $2 \times 7! = 10080$.
3. 3. The 'not together' arrangements are the total minus the 'together' arrangements: $40320 - 10080 = 30240$.

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

## Common pitfalls

- **Wrong:** Treating unordered selections as permutations
  - Why it fails: You overcount by a factor of $r!$ because you count every ordering of the same selection as unique
  - Correct: Always check if order matters: if swapping two items does not create a new outcome, use combinations.
- **Wrong:** Forgetting to divide by factorials of identical items
  - Why it fails: Swapping identical items does not create a new arrangement, so you overcount
  - Correct: List all repeated items explicitly before calculating, then divide by the product of their factorials.
- **Wrong:** Adding instead of multiplying for independent selections
  - Why it fails: Confusing the addition principle for mutually exclusive cases with multiplication for independent choices
  - Correct: Use multiplication for 'and' choices, addition for 'or' mutually exclusive cases.
- **Wrong:** Forgetting to permute items that must be kept together
  - Why it fails: When grouping items into a block, you miss the internal arrangements of the block
  - Correct: Count arrangements of the blocks, then multiply by the number of permutations of items within the block.
- **Wrong:** Double counting outcomes that satisfy multiple conditions
  - Why it fails: When adding cases for overlapping restrictions, you count outcomes that meet multiple conditions twice
  - Correct: Use the principle of inclusion-exclusion or use the total minus invalid method to avoid double counting.

## 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 $n$ | $\binom{n}{r} = \frac{n!}{r!(n-r)!}$ |
| At least X restriction | Total outcomes $-$ invalid outcomes |
| Items must be kept together | Treat as 1 block, multiply by internal permutations |

## 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.

- [Discrete Random Variables](https://www.owlsprep.com/study/cie-9709-u4-discrete-random-variables/)
- [Probability](https://www.owlsprep.com/study/cie-9709-u4-probability/)
- [Normal distribution](https://www.owlsprep.com/study/cie-9709-u4-normal-distribution/)

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