11.1 Permutations vs Combinations


2026 📋 Syllabus Objectives

By the end of these notes, you should be able to:

  • Recognise the difference between a permutation and a combination.
  • Know when to use each one when solving a problem.

What Are We Counting?

In many real-life situations, we need to count the number of ways something can be done. For example:

  • How many ways can 5 students sit in a row?
  • How many ways can you choose 3 books from a shelf of 10?

These two questions look similar, but they are actually very different. The key question you must always ask is:

Does the order matter?

This single question tells you whether you need a permutation or a combination.


🔑 The Core Difference

PermutationCombination
What it countsArrangementsSelections
Does order matter?✅ YES❌ NO
Keyword cluesarrange, order, rank, sequence, code, queuechoose, select, pick, group, team, committee

Understanding Permutations

A permutation is a way of arranging items where the order matters.

Think of it this way: if you arrange the letters A, B, C, then ABC and BAC are counted as two different arrangements because the order is different.

Real-life example: Imagine a race with 3 runners — Alice, Bob, and Charlie. The results 1st: Alice, 2nd: Bob, 3rd: Charlie is a completely different outcome from 1st: Bob, 2nd: Alice, 3rd: Charlie. The order in which they finish matters. This is a permutation situation.

Other examples where order matters:

  • Creating a 4-digit PIN code (1234 ≠ 4321)
  • Arranging books on a shelf
  • Assigning 1st, 2nd, and 3rd place in a competition
  • Forming a password using letters and numbers

The permutation formula:

When you choose r items from a group of n items and arrange them (order matters), the number of ways is:

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

where n! (called n factorial) means: n×(n1)×(n2)××2×1n \times (n-1) \times (n-2) \times \dots \times 2 \times 1

Quick example: How many ways can you arrange 3 letters chosen from A, B, C, D, E?

Here, n=5n = 5 and r=3r = 3.

5P3=5!(53)!=5!2!=5×4×3×2×12×1=1202=60^5P_3 = \frac{5!}{(5-3)!} = \frac{5!}{2!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1} = \frac{120}{2} = 60

So there are 60 different arrangements.

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