Permutations and Combinations — A-Level Mathematics Stats Study Guide

For: A-Level Mathematics candidates sitting Paper 5 (Probability & Statistics 1).

Covers: Multiplication and addition counting principles, permutations of distinct objects, unordered combinations, arrangement restrictions, and rules for repeated/identical objects.

You should already know: Basic probability, summation, integration (Pure 1 calculus).

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the A-Level Mathematics style for educational use. They are not reproductions of past Cambridge International examination papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official Cambridge mark schemes for grading conventions.

1. What Is Permutations and Combinations?

Permutations and combinations are the core set of combinatorial counting rules used to calculate the number of possible arrangements or selections of objects from a set, without manually listing every possible outcome. They form the foundation of almost all probability calculations in Paper 5, and are tested in 3-6 mark questions that often pair with discrete probability or binomial distribution topics. Common synonyms you will see in exam questions include "counting problems", "arrangements", and "selections".

2. Multiplication and addition principles

These two foundational rules apply to all counting problems, before you use permutation or combination formulas. The addition principle (rule of sum) applies when you are choosing between mutually exclusive options: if you have two tasks that cannot be completed at the same time, with $n_1$ ways to complete the first and $n_2$ ways to complete the second, the total number of ways to complete either task is $n_1+n_2$. For example, if you can choose a drink from 3 sodas or 4 juices, you have $3+4=7$ total drink options. The multiplication principle (rule of product) applies when you are completing sequential, independent tasks: if you have two tasks that do not interfere with each other, with $n_1$ ways to complete the first and $n_2$ ways to complete the second, the total number of ways to complete both tasks is $n_1\times n_2$. For example, if you choose 1 drink from 7 options and 1 snack from 5 options, you have $7\times 5=35$ total combinations.

Worked example

A café sells 4 breakfast items, 6 lunch items, and 3 dessert items. (a) How many ways can you choose one meal (either breakfast or lunch)? (b) How many ways can you choose a 2-course meal of lunch and dessert?

(a) Mutually exclusive choice: $4+6=10$ ways. (b) Independent choices: $6\times 3=18$ ways.

Exam tip: Always confirm events are mutually exclusive before adding, and independent before multiplying. If events overlap, use inclusion-exclusion or split into non-overlapping cases.

3. Permutations of distinct objects, $n!$ and ${}^n{P}_r$

A permutation is an ordered arrangement of a subset of objects from a set of distinct items. First, the factorial function $n!$ calculates the number of ways to arrange all $n$ distinct objects in a sequence: $$n!=n\times (n-1)\times (n-2)\times ...\times 1$$ By definition, $0!=1$, to simplify boundary cases in formulas. For example, arranging 5 students in a line has $5!=120$ total permutations. If you only want to arrange $r$ objects from the set of $n$ distinct objects, use the permutation formula ${}^n{P}_r$, which counts ordered selections with no repetition: $${}^n{P}_r=\frac{n!}{(n-r)!}$$ This formula is derived from the multiplication principle: you have $n$ choices for the first position, $n-1$ for the second, down to $n-r+1$ for the $r$-th position, and the product of these terms simplifies to the ratio above.

Worked example

How many 4-character unique access codes can be created from the 26 letters of the alphabet, with no repeated letters?

We are selecting 4 ordered letters from 26, no repetition: ${}^{26}{P}_4=\frac{26!}{22!}=26\times 25\times 24\times 23=358,800$ total codes.

4. Combinations ${}^n{C}_r$ — order doesn't matter

A combination is an unordered selection of $r$ objects from a set of $n$ distinct items, where the order of selection does not affect the outcome. You will use combinations for questions that mention "selecting", "choosing", or forming groups with no hierarchy. The combination formula is derived from the permutation formula: each unordered group of $r$ objects has $r!$ possible ordered permutations, so we divide the permutation count by $r!$ to eliminate duplicate groups: $${}^n{C}_r=\left(\genfrac{}{}{0ex}{}{n}{r}\right)=\frac{{}^n{P}_r}{r!}=\frac{n!}{r!(n-r)!}$$

Worked example

How many ways can you select a 3-person committee from 12 club members?

Order does not matter for committee membership, so use combinations: ${}^{12}{C}_3=\frac{12!}{3!9!}=220$ total committees.

Exam tip: Always circle keywords in the question to decide between permutations and combinations: if positions, ranks, or order are specified, use permutations; if unordered groups are specified, use combinations.

5. Restrictions — fixing positions, requiring or excluding objects

Most exam questions add restrictions to counting problems, and these are the most common source of mark loss. We cover the three core restriction types below:

1. Fixed positions: If an object must be placed in a specific position, calculate that position first, then count permutations/combinations for the remaining positions. For example, arranging 5 letters A,B,C,D,E with A in the first position: fix A first, then arrange the remaining 4 letters for $4!=24$ total arrangements.
2. Required objects: If specific objects must be included in the selection, select those objects first, then select the remaining items from the leftover pool. For example, choosing 4 people from 10, with Alice and Bob required: select Alice and Bob first, then choose 2 more from the remaining 8 for ${}^8{C}_2=28$ total groups.
3. Excluded objects: If specific objects cannot be included, remove them from the pool before counting. For example, arranging 5 letters A,B,C,D,E without C: arrange the remaining 4 letters for $4!=24$ total arrangements. Two advanced, commonly tested restrictions are adjacent and separated objects:

• Adjacent objects: Treat the group of objects that must be together as a single unit, count permutations of the units, then multiply by the internal permutations of the group. For example, arranging A,B,C,D,E with A and B adjacent: treat [AB/BA] as one unit, so 4 total units with $4!=24$ permutations, multiplied by $2!$ for internal order of A and B, giving $48$ total arrangements.
• Separated objects: Arrange all unrestricted objects first, then place the restricted objects in the gaps between (and around) the unrestricted objects to avoid adjacency. For example, arranging A,B,C,D,E with A and B separated: arrange C,D,E first for $3!=6$ permutations, there are 4 gaps to place A and B, so ${}^4{P}_2=12$ ordered placements, giving $6\times 12=72$ total arrangements.

6. Repetition allowed vs not

All formulas above assume no repetition of objects, but many exam questions include identical objects or allow repeated selections:

1. Permutations with repetition allowed: If you can reuse objects for ordered selections, the total count is $n^r$, where $n$ is the number of available objects and $r$ is the number of positions. For example, 3-digit passcodes from 10 digits with repetition allowed: $10^3=1000$ total codes.
2. Permutations of identical objects: If you are arranging a set of objects with $k_i$ copies of each identical item, divide the total factorial by the product of the factorials of the duplicate counts: $$\text{Total permutations}=\frac{n!}{k_1!k_2!...k_m!}$$ For example, arranging the letters of "STATISTICS" (10 letters, 3 S, 3 T, 2 I): $\frac{10!}{3!3!2!}=50400$ total arrangements.
3. Combinations with repetition allowed: For unordered selections where objects can be reused, the formula is ${}^{n+r-1}{C}_r$, though this is only tested in basic scenarios (e.g. distributing identical candies to children) in A-Level Mathematics.

Common Pitfalls (and how to avoid them)

• Wrong move: Using ${}^n{P}_r$ for unordered selections or ${}^n{C}_r$ for ordered arrangements. Why: Students mix up "select" and "arrange" keywords. Correct move: Circle question keywords: if positions, order, or hierarchy are mentioned, use permutations; for unordered groups, use combinations.
• Wrong move: Forgetting to multiply by internal permutations for adjacent object groups. Why: Students only count permutations of the combined units, ignoring order inside the group. Correct move: After counting unit permutations, multiply by the factorial of the group size (adjust for identical objects inside the group if needed).
• Wrong move: Using $n!$ directly for sets with identical objects, no division. Why: Students assume all objects are distinct unless stated otherwise. Correct move: Scan for duplicates (e.g. repeated letters, identical balls) and divide by the factorial of each duplicate count.
• **Wrong move: Adding counts of non-mutually exclusive cases, leading to double counting. Why: Students don't check for overlap between cases. Correct move: Split problems into fully non-overlapping cases, or subtract overlapping counts using the inclusion-exclusion principle.
• Wrong move: Using ${}^n{P}_r$ for permutations with repetition allowed. Why: Students apply the no-repetition formula by default. Correct move: Check if repetition is allowed: if yes, use $n^r$ for ordered selections.

Practice Questions (Paper 5 Style)

Question 1

A school has 6 male and 5 female teachers. (a) How many ways to select a committee of 4 teachers with at least 2 male teachers? (3 marks) (b) How many ways to arrange 8 of these teachers in a line for a staff photo, such that no two female teachers are adjacent? (4 marks)

Solution

(a) Total unrestricted committees: ${}^{11}{C}_4=330$. Subtract invalid committees with <2 male teachers: 0 male = ${}^5{C}_4=5$, 1 male = ${}^6{C}_1{\times }^5{C}_3=60$. Valid committees: $330-5-60=265$. (b) No two female adjacent: first arrange male teachers, place female teachers in gaps. We need 8 total teachers, so $m+f=8$, $f\le m+1$ so $m\ge 4$.

• Case 1: 4 male, 4 female: ${}^6{C}_4{\times }^5{C}_4\times 4!{\times }^5{P}_4=15\times 5\times 24\times 120=216000$
• Case 2: 5 male, 3 female: ${}^6{C}_5{\times }^5{C}_3\times 5!{\times }^6{P}_3=6\times 10\times 120\times 120=864000$
• Case 3: 6 male, 2 female: ${}^6{C}_6{\times }^5{C}_2\times 6!{\times }^7{P}_2=1\times 10\times 720\times 42=302400$ Total: $216000+864000+302400=1,382,400$

Question 2

(a) How many distinct arrangements are there of the letters in the word "ARITHMETIC"? (2 marks) (b) How many of these arrangements have the two T's adjacent and the two I's separated? (3 marks)

Solution

(a) 10 total letters, 2 T's and 2 I's are identical: $\frac{10!}{2!2!}=907200$ (b) Treat the two T's as a single unit, so we have 9 total items, with two identical I's. To keep I's separated: first arrange the 7 non-I items ([TT], A,R,H,M,E,C): $7!=5040$. There are 8 gaps between these items, choose 2 gaps to place the I's: ${}^8{C}_2=28$. Total arrangements: $5040\times 28=141120$

Question 3

A 4-digit PIN is created using digits 0-9, repetition is allowed. The PIN cannot start with 0, and must contain at least one even digit. How many valid PINs are there? (3 marks)

Solution

Total unrestricted PINs (no leading 0): $9\times 10^3=9000$. Subtract PINs with no even digits (all odd: 1,3,5,7,9): $5^4=625$. Valid PINs: $9000-625=8375$

Quick Reference Cheatsheet

What's Next

Permutations and combinations are the foundational building block for all probability topics in A-Level Mathematics Paper 5. You will use these counting rules to calculate event probabilities for discrete random variables, binomial distributions, and sampling scenarios later in the syllabus. A strong grasp of these concepts will also help you tackle cross-topic questions that combine probability with hypothesis testing or data representation, which are common in the final 10-mark paper questions. If you struggle with any of the rules, worked examples, or practice questions in this guide, you can ask Ollie for step-by-step explanations, extra practice problems, or personalized tips to avoid common exam mistakes. Head to the homepage to access AI-powered tutoring, past paper solutions, and more study resources for A-Level Mathematics Paper 5.

Aligned with the Cambridge International AS & A Level Mathematics 9709 syllabus. OwlsAi is not affiliated with Cambridge Assessment International Education.