1. Arrangements & Factorials
Why does n × (n−1) × (n−2) × ... work?
Imagine you have 3 letters — A, B, C — and 3 empty seats in a row. Think about filling the seats one at a time:
- Seat 1: any of the 3 letters could sit here → 3 choices
- Seat 2: one letter is already used, so only 2 remain → 2 choices
- Seat 3: only 1 letter is left → 1 choice
Multiply the choices at each stage: 3 × 2 × 1 = 6. That's every possible order:
ABC, ACB, BAC, BCA, CAB, CBA. This "count the options at each position, then multiply" logic is the
engine behind this entire chapter — everything else is a variation on it.
What exactly is a factorial?
"Factorial" is just a mathematical operation — the same category as +, −, ×, ÷ — written with an exclamation mark.
5! is read "5 factorial" and means 5 × 4 × 3 × 2 × 1 = 120.
Two special facts worth knowing cold:
- Negative factorials don't exist — you cannot arrange −3 objects, so it's undefined.
- 0! = 1 — this feels weird at first, but think about it: there are no positive integers below zero, so "arranging zero items" can only be done in exactly one way (do nothing). This isn't just a convention — it's what keeps all the formulas in this chapter working correctly.
Good to Know
Most standard calculators can't handle factorials above about 70! — the number becomes too enormous. Try your own calculator and find the biggest x where x! still works.
Worked Example
Find how many distinct arrangements there are of the letters in the word MATHS.
MATHS has 5 different letters, so there are 5 options for the first space, 4 for the second, 3 for the third, and so on:
5 × 4 × 3 × 2 × 1 = 120 distinct arrangements
Simplifying factorial fractions
Big factorials divided by smaller ones cancel out beautifully. Think of 7!/4! written out in full:
This gives us two handy identities worth memorising:
Worked Example
(i) Show, by writing 8! and 5! in full and cancelling, that 8!/5! = 8 × 7 × 6.
(ii) Hence, simplify n!/(n−3)!
(i) Write both out fully:
5! = 5×4×3×2×1 8! = 8×7×6×5×4×3×2×1
Cancel the shared 5×4×3×2×1 from top and bottom → left with 8 × 7 × 6. ✓
(ii) "Hence" means reuse part (i)'s pattern. If n = 8, then n!/(n−3)! = 8!/5! = 8×7×6.
Notice: 6 = n−2, 7 = n−1, and the leading term is n itself. So generally:
n!/(n−3)! = n(n−1)(n−2)
Practice Question
Simplify: n!/(n−4)!
Practice Question
How many distinct arrangements are there of the letters in the word NUMBER?