Learn9709Paper 5 · Probability & Statistics 1Permutations and combinationsSelections (combinations)

Permutations and combinations · Subtopic 3 of 3

Selections: Combinations

When order does not matter, count with $^nC_r$ . Split selections into cases, use "at least one $=$ total $-$ none", and turn counts into probabilities by dividing by the total number of selections.

Cambridge 9709 Paper 5≈ 35 minRevision 5 · updated 2026-10-27

Section 1 of 14

Learning objectives

Know when to use $^nC_r$ (selection) rather than $^nP_r$ (arrangement).
Count selections from several groups by multiplying, and add over separate cases.
Handle "at least", "must include" and "refuse to serve together".
Find probabilities as (favourable selections) $\div$ (all selections).

Pause and think

From $4$ friends you choose $2$ to come to a concert. Is choosing "Ali then Bilal" different from choosing "Bilal then Ali"? How does that change the count compared with arranging $2$ of $4$ ?

Section 2 of 14

Core idea

Start herePicking a team, not a line-up

Choosing a team of 2 from Ali, Bea, Cal, Dev: is "Ali and Bea" different from "Bea and Ali"? No. When order doesn't matter, every arrangement is overcounted, and dividing it out gives the number of combinations.

The big idea

The number of ways to choose $r$ from $n$ (order irrelevant) is $\binom nr=\frac{n!}{r!(n-r)!}$ . For "at least" problems, add cases or use the complement.

1. From permutations to combinations

Choosing 2 from 4: twelve ordered picks, six pairs

Each pair shows up twice in the ordered list, so divide by

2!

\dbinom42=6

\binom nr

${}^nP_r=\frac{n!}{(n-r)!}$
Ordered selections.
$\text{each group of }r\text{ appears }r!\text{ times}$
Its internal orderings.
$\binom nr=\frac{{}^nP_r}{r!}=\frac{n!}{r!(n-r)!}$
Divide.

How many ways to choose r from 10?

Symmetric: choosing

3

to take is the same as choosing

7

to leave, so

\binom{10}3=\binom{10}7=120

. The most choices are in the middle:

\binom{10}5=252

2. Committees

Worked exampleAt least two women

A committee of 5 from 6 men and 4 women must contain at least 2 women. How many ways?

Goal: Count.
Strategy: Cases (2, 3, 4 women), checked by complement.

$\binom42\binom63=6\times20=120$
2W 3M.
$\binom43\binom62=4\times15=60;\quad\binom44\binom61=6$
3W 2M; 4W 1M.
$120+60+6=186$
Add cases (they don't overlap).
$\binom{10}5-\binom65-\binom41\binom64=252-6-60=186$
Complement check.

Answer

186

✓ CheckBoth methods agree ✓.

Worked exampleProbability by counting

Choose 3 from 5 red and 4 blue counters. P(exactly 2 red)?

Goal: Probability.
Strategy: Favourable selections ÷ all selections.

$\binom52\binom41=40$
Favourable.
$\binom93=84$
Total.

Answer

\frac{40}{84}=\frac{10}{21}

✓ CheckTree method:

3\times\frac59\cdot\frac48\cdot\frac47=\frac{240}{504}=\frac{10}{21}

✓.

Common misconceptionAdding vs multiplying Within one case (choose women AND men) multiply; between cases (OR) add.

Pause and think

Why is $\binom{10}3=\binom{10}7$ ?

Summary

Order matters: $P$ ; not: $C$ .
Cases or complement for "at least".

Combination: A selection, order irrelevant.

HistoryPascal's triangle is not Pascal's. The triangle of numbers

^nC_r

is named after Blaise Pascal, who wrote about it in 1654. But it appears in the work of the Persian mathematician al-Karajī around 1000, in a 1261 book by the Chinese mathematician Yang Hui (in China it is "Yang Hui's triangle"), and in Indian work on poetry metres even earlier.

Fun factLottery odds. The UK National Lottery chooses

6

numbers from

59

. There are

^{59}C_6=45\,057\,474

possible tickets, so one ticket has about a

1

45

million chance of the jackpot. You are far more likely to be struck by lightning in your lifetime.

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