S1 2 Permutations and combinations | A Level Maths 9709

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🎯What you need to be able to do

Decide whether order matters, and so whether to use a permutation or a combination.
Count arrangements of \(n\) distinct objects, and of \(r\) chosen from \(n\).
Count arrangements when some objects are identical.
Handle restrictions: objects together, objects apart, fixed positions, at least/at most.
Count selections where order does not matter, including from separate groups.
Use counting results to calculate probabilities.

📚The mathematics

The central question

Before any formula, ask: would swapping two of the chosen items give a different outcome?

Yes — order matters — it is a permutation (arrangement). Words like arrange, order, line up, code, first/second/third, or distinct roles such as chair and secretary.
No — order does not matter — it is a combination (selection). Words like choose, select, committee, team, sample.

Getting this wrong is not a small error: the two answers differ by a factor of \( r! \).

The formulae

\( n! = n(n-1)(n-2)\cdots 1 \)

\( ^{n}P_r = \dfrac{n!}{(n-r)!} \)

\( ^{n}C_r = \dbinom{n}{r} = \dfrac{n!}{r!\,(n-r)!} \)

Note \( 0! = 1 \) by convention, and \( ^{n}C_r = {}^{n}C_{n-r} \) — choosing which 3 of 10 to take is the same as choosing which 7 to leave, which often saves arithmetic.

The relationship \( ^{n}P_r = {}^{n}C_r \times r! \) says exactly what the distinction is: first choose the \(r\) objects, then arrange them.

Repeated objects

The number of distinct arrangements of \(n\) objects of which \(p\) are alike of one kind, \(q\) alike of another, and so on, is

\[ \frac{n!}{p!\,q!\,\cdots} \]

So the letters of BANANA give \( \dfrac{6!}{3!\,2!} = 60 \) arrangements — three A's and two N's. Divide by a factorial for each repeated letter; missing one inflates the answer by exactly that factor.

Restrictions

Three standard techniques, and choosing the right one is most of the difficulty:

Objects must be together — treat the block as a single item, arrange everything, then multiply by the arrangements within the block. Two people who must sit together among 6: \( 5! \times 2! = 240 \).
Objects must be apart — usually easiest as total minus together. For the same 6 people, \( 6! - 240 = 480 \). Alternatively, arrange the others first and slot the restricted items into the gaps between them.
Fixed positions — deal with the restricted positions first, then fill the rest. If a number must be even, choose the last digit first.

“At least” usually means take the complement. Counting “at least one woman” directly means adding the cases for exactly one, exactly two, and so on; subtracting the “no women” case from the total is one calculation instead of several. Conversely, be careful not to use the complement when the phrase is “exactly” — and never add cases that overlap, since that double-counts.

Selections from separate groups

When choosing from distinct groups, multiply the independent choices. To pick 3 from 7 men and 2 from 5 women: \( ^{7}C_3 \times {}^{5}C_2 = 35 \times 10 = 350 \).

When several cases satisfy the condition, count each and add. “A committee of 4 with at least 2 women” means exactly 2, exactly 3, and exactly 4 women, each computed separately and summed. Multiply within a case, add between cases — that one rule organises the whole topic.

Using counting for probability

With equally likely outcomes,

\[ \mathrm{P}(A) = \frac{\text{number of favourable outcomes}}{\text{total number of outcomes}} \]

Count numerator and denominator the same way — both as permutations or both as combinations. Mixing the two is the classic error here, and it produces a probability that may not even lie between 0 and 1, which is itself a useful check.

✏️Worked example

A committee of 5 is to be chosen from 6 men and 4 women. (a) In how many ways can the committee be chosen? (b) In how many ways can it be chosen if it must contain at least 3 women? (c) Find the probability that a randomly chosen committee contains exactly 2 women. (d) One particular committee consists of 2 women and 3 men. They sit in a row for a photograph. In how many ways can they be arranged if the two women sit together?

(a) Order does not matter for a committee, so this is a combination:

\[ ^{10}C_5 = \frac{10!}{5!\,5!} = 252 \]

(b) With only 4 women available, “at least 3” means exactly 3 or exactly 4. Count each case and add:

3 women, 2 men: \( ^{4}C_3 \times {}^{6}C_2 = 4 \times 15 = 60 \)

4 women, 1 man: \( ^{4}C_4 \times {}^{6}C_1 = 1 \times 6 = 6 \)

Total \( = 66 \) ways.

(c) Exactly 2 women means 2 women and 3 men:

\[ ^{4}C_2 \times {}^{6}C_3 = 6 \times 20 = 120 \]

so the probability is \( \dfrac{120}{252} = \dfrac{10}{21} = 0.476 \).

(d) This part is about arrangement, so it is a permutation — the switch from (a) to (d) is exactly the distinction this topic turns on. Treat the two women as a single block. That leaves 4 items to arrange (the block plus 3 men), and the women can be ordered two ways inside the block:

\[ 4! \times 2! = 24 \times 2 = 48 \]

Check it. In (c), the probability 0.476 is comfortably between 0 and 1 and close to a half, which is plausible — 2 women out of 5 is near the proportion you would expect from a pool that is 40% women. In (b), note that \( 66 \) is well under \( 252 \), as any restricted count must be. And in (d), \( 48 \) is less than \( 5! = 120 \), as forcing people together must reduce the arrangements.

Multiply within a case, add between cases. In (b) the two cases are combined by addition because a committee cannot have both exactly 3 and exactly 4 women; within each case the choices of women and men are combined by multiplication because both happen. Writing \( ^{4}C_3 \times {}^{6}C_2 \times {}^{4}C_4 \times {}^{6}C_1 \) — multiplying across the cases — is a common and badly wrong answer.

📝Practise

Work through these, then reveal the answer.

1. How many different arrangements are there of the letters of the word STATISTICS?

There are 10 letters: S appears 3 times, T appears 3 times, I appears 2 times, and A and C once each. So the number of arrangements is \( \dfrac{10!}{3!\,3!\,2!} = \dfrac{3\,628\,800}{6 \times 6 \times 2} = \dfrac{3\,628\,800}{72} = 50\,400 \).

2. In how many ways can 8 people be seated in a row if two particular people must not sit next to each other?

Total arrangements: \( 8! = 40\,320 \). Arrangements with the pair together: treat them as a block, giving \( 7! \times 2! = 5040 \times 2 = 10\,080 \). So the number with them apart is \( 40\,320 - 10\,080 = 30\,240 \). Counting "apart" directly is far harder — the complement is the route.

3. How many 4-digit numbers greater than 3000 can be formed from the digits 1, 2, 3, 4, 5 without repetition?

Deal with the restricted position first. The leading digit must be 3, 4 or 5, giving 3 choices. The remaining three positions are filled from the 4 unused digits in \( ^{4}P_3 = 4 \times 3 \times 2 = 24 \) ways. Total \( = 3 \times 24 = 72 \). Order matters throughout, since these are numbers.

4. A team of 4 is chosen from 5 boys and 6 girls. Find the number of teams containing at least one boy.

Total teams: \( ^{11}C_4 = 330 \). Teams with no boys (all girls): \( ^{6}C_4 = 15 \). So teams with at least one boy \( = 330 - 15 = 315 \). Adding the cases for exactly 1, 2, 3 and 4 boys would give the same answer with four times the work.

5. Five cards are drawn from a set of ten cards numbered 1 to 10. Find the probability that all five are odd.

There are 5 odd numbers among the ten. Total selections: \( ^{10}C_5 = 252 \). Selections that are all odd: \( ^{5}C_5 = 1 \). So the probability is \( \dfrac{1}{252} = 0.00397 \). Both numerator and denominator are counted as combinations, which is what keeps the answer valid.

6. In how many ways can 6 people be arranged in a row if two particular people, A and B, must sit together and a third person C must sit at one of the ends?

Handle C first: 2 choices of end. That leaves 5 people to arrange in the remaining 5 seats, with A and B together: treat A and B as a block, giving 4 items, so \( 4! \times 2! = 48 \). Total \( = 2 \times 48 = 96 \). Dealing with the most restrictive condition first — C's position — keeps the count manageable.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

Khan Academy — permutations, combinations and counting principles
Brilliant — combinatorics problems with worked solutions
Cambridge examiner reports — the permutation/combination confusion is flagged nearly every session