Two counting rules do almost everything
Factorial — n! = n × (n−1) × (n−2) × … × 2 × 1, the number of ways of arranging n distinct objects in a row. By convention 0! = 1, which is what makes the formulae below work at their edges.
Before any formula, two principles. The multiplication rule: if one task can be done in m ways and a second, independent task in n ways, then the pair can be done in m × n ways. The addition rule: if you must choose either one option from m or one from n, and the two sets do not overlap, there are m + n choices.
In practice: "and" means multiply, "or" means add. A meal of one starter from 4 and one main from 6 offers 24 meals; a single dish chosen from 4 starters or 6 mains offers 10 choices.
Permutation or combination? Ask whether order matters
This one question decides which formula to use, and almost every lost mark in the chapter comes from answering it wrongly.
A permutation is an arrangement: order matters. Choosing a president and a secretary from a committee, or arranging books on a shelf, or forming a three-digit number. A combination is a selection: order does not matter. Choosing a team of three, or a hand of cards, or which five questions to answer.
| The question says | Order matters? | Use |
|---|---|---|
| arrange, order, line up, seat | yes | ⁿPᵣ |
| form a number, form a word | yes | ⁿPᵣ |
| president and secretary (distinct roles) | yes | ⁿPᵣ |
| choose, select, a team, a committee | no | ⁿCᵣ |
| a hand of cards, a group of friends | no | ⁿCᵣ |
Why ⁿCᵣ = ⁿCₙ₋ᵣ
Choosing which 3 people to take from 10 is exactly the same act as choosing which 7 to leave behind. So ¹⁰C₃ = ¹⁰C₇ = 120. Use it to save work: computing ²⁰C₁₈ directly is unpleasant, but it equals ²⁰C₂ = 190, which is two lines.
Arrangements with repeated objects, and in a circle
Two variations appear regularly and both are corrections to the plain n! count.
When some objects are identical, arrangements that merely swap identical items are not different arrangements, so the total is divided by the factorial of each repeat count. The word MISSISSIPPI has 11 letters with four S, four I and two P, giving 11!/(4!4!2!) distinct arrangements.
When objects are arranged in a circle, there is no first position — rotating the whole ring gives the same arrangement. Fixing one object and arranging the rest around it gives (n − 1)!.
How many arrangements of the letters of the word COMMITTEE are there? In how many of them do the two Ts stay together?
- COMMITTEE has 9 letters, with M twice, T twice and E twice.Count the repeats before anything else; missing one is the usual error.
- Total arrangements
= 9!/(2!2!2!) = 362880/8 = 45 360.Divide by 2! once for each pair of identical letters. - For the Ts together, glue them into one block. There are now 8 items: [TT], C, O, M, M, I, E, E.Treating a required group as a single object is the standard technique.
- These arrange in
8!/(2!2!) = 40320/4 = 10 080ways.M and E are still repeated; the two Ts inside the block are identical, so the block has only one internal order.
45 360 arrangements in all; 10 080 with the two Ts together
The strategies that unlock the hard questions
Three techniques cover almost every awkward counting question on the paper.
- Glue it together. If certain items must be adjacent, treat them as one block, arrange everything, then multiply by the internal arrangements of the block (if the items inside are distinct).
- Count the opposite. "At least one" and "not all together" are far quicker as
total − unwanted. Items never adjacent = all arrangements minus the ones where they are adjacent. - Place the fussy ones first. If a restriction affects certain positions — a number must be even, a particular seat must be filled — satisfy that restriction before counting anything else.
Do not multiply when you should add
A committee of 3 from 5 men and 4 women containing at least one woman is not one calculation. Split into cases — exactly 1, exactly 2, exactly 3 women — compute each with ⁿCᵣ and add them. Or take the complement: all committees minus the all-male ones, ⁹C₃ − ⁵C₃ = 84 − 10 = 74, which is one line.
Before you leave this chapter
- "And" multiplies, "or" adds. 0! = 1.
- Order matters → ⁿPᵣ. Order does not → ⁿCᵣ. Decide this before writing a formula.
- ⁿPᵣ = r! × ⁿCᵣ, and ⁿCᵣ = ⁿCₙ₋ᵣ.
- Identical objects: divide n! by the factorial of each repeat count.
- Circular arrangements: (n − 1)!, because rotating the ring changes nothing.
When repetition is allowed
The ⁿPᵣ formula assumes each object is used at most once. Some situations do not work that way: a four-digit PIN may repeat digits, a coin may land heads twice, a number plate may reuse a letter.
When repetition is permitted the counting is simpler, not harder. Each of the r positions has the full n choices available, independently of the others, so the multiplication rule gives nʳ straight away.
How many 3-letter codes can be made from the 26 letters (a) if letters may repeat, (b) if they may not?
- (a) Each of the three positions has all 26 letters available:
26 × 26 × 26 = 26³.The choices are independent, so the multiplication rule applies directly. = 17 576codes.- (b) Now each letter used removes one option:
26 × 25 × 24.This is ²⁶P₃ — order matters, since a code is an arrangement. = 15 600codes.Fewer, as expected: the repeated-letter codes such as AAB have been excluded.
(a) 17 576 (b) 15 600
Row n of Pascal's triangle lists ⁿC₀ through ⁿCₙ — every possible selection size from a set of n. Notice the symmetry: ⁿCᵣ equals ⁿCₙ₋ᵣ, because choosing what to take is the same as choosing what to leave.