Counting outcomes with permutations and combinations

By the end of this lesson

By the end you can decide whether the outcomes in a probability question are counted with permutations or combinations, and use that count to find a probability when the sample space is far too large to write out.

When listing is hopeless

The counting formula for equally likely outcomes,

$P(E) = \dfrac{n(E)}{n(S)}$

still works when $S$ is enormous. Here $n(E)$ means "the number of outcomes in event $E$ " and $n(S)$ means "the number of outcomes in the sample space". What changes is how we get those numbers.

You cannot list the ways to choose 3 students from 10 — there are $120$ of them, and real exam questions have more. So we count with the permutation and combination formulae from topic 5.2.

The one rule that prevents almost every error: count the top and the bottom of the fraction the same way. If the denominator counts ordered selections, the numerator must too.

Definition

Permutations and combinations

First, factorial: $n!$ (read " $n$ factorial") means $n \times (n-1) \times \cdots \times 2 \times 1$ . For example $4! = 4 \times 3 \times 2 \times 1 = 24$ , and $0!$ is defined to be $1$ .

A permutation is a selection in which the order matters: choosing A then B is different from choosing B then A. The number of ways to choose $r$ objects from $n$ distinct objects in order is $^{n}P_{r} = \dfrac{n!}{(n-r)!}$ .

A combination is a selection in which the order does not matter, so AB and BA are the same selection. The number of ways to choose $r$ objects from $n$ is $^{n}C_{r} = \dfrac{n!}{r!\,(n-r)!}$ .

Since $^{n}P_{r} = r! \times {}^{n}C_{r}$ , a permutation count is a combination count with the chosen objects also arranged among themselves.

How to tell them apart

Ask yourself: if I swapped two chosen items, would that be a different outcome of the experiment?

Selecting two counters from a bag and looking at their colours: swapping them gives the same pair, so use combinations.
Picking a president then a treasurer: swapping the two people gives different job holders, so use permutations.
Dealing the first two cards of a hand and caring which came first: permutations.

In most probability questions about choosing a group, the answer is combinations, because "which items ended up chosen" is all we observe.

Worked example: Example 1: two counters from a bag

A bag contains $5$ red and $3$ blue counters. Two counters are taken out at random, without replacement. Find the probability that both are red.

Have a go first?

Check 1 — counting a pair

CalculateQuestion 1 · Apply

Multiple choiceQuestion 2 · Recall

Answer the 2 questions above to continue.