Home / IGCSE Add Maths 0606 / Permutations and combinations / Arrangement and selection problems
Cambridge IGCSE · Topic 11.3
IGCSE Add Maths 0606 / Topic 11 / IGCSE

Does the order matter?

Every counting question in this topic turns on one question: are the objects being arranged, or only chosen? Read the wording, pick \({}^{n}P_{r}\) or \({}^{n}C_{r}\), split the problem into cases joined by 'or' and 'and', and use the complement when 'at least' would need too many cases.

Topic 11All permutations and combinations lessons →
The key idea

Order matters (arranged, in order, positions, codes, numbers, first/second/third): use \({}^{n}P_{r} = \dfrac{n!}{(n - r)!}\). Order does not matter (chosen, selected, committee, team, group): use \({}^{n}C_{r} = \dfrac{n!}{r!\,(n - r)!}\). Separate cases joined by 'or' are ADDED; stages that happen together ('and') are MULTIPLIED. For 'at least one', the complement is often quickest: total minus the number with none.

Read the question. Does the ORDER matter? Yes No ARRANGE: permutation arranged, in a row, in order, positions, 1st/2nd/3rd prizes, codes, numbers, roles n! ÷ (n − r)! CHOOSE: combination chosen, selected, picked, committee, team, group, hand of cards, sample n! ÷ (r! (n − r)!) Then split into cases case 1 OR case 2: ADD the counts this stage AND that stage: MULTIPLY at least one: total − none
Fig. 1: deciding between a permutation and a combination from the wording, then combining the cases
Section 01

Arranged or chosen?

A permutation counts ordered arrangements; a combination counts unordered selections. The same 3 people from 8 give \({}^{8}C_{3} = 56\) committees but \({}^{8}P_{3} = 336\) ways to hand out gold, silver and bronze, because each committee of 3 can be put in order in \(3! = 6\) ways. The verbs in the question decide which you need (Fig. 1).

Check with a small listing. From the letters A, B, C, take two at a time.

Chosen (combination)Arranged (permutation)
AB, AC, BC: \({}^{3}C_{2} = 3\)AB, BA, AC, CA, BC, CB: \({}^{3}P_{2} = 6\)

If your formula and the listing disagree on a tiny case, the formula is the wrong one.

Worked example Prizes or a committee
Eight students enter a competition. In how many ways can (a) first, second and third prizes be awarded, (b) three of them be chosen to represent the school?
  1. (a) The prizes are different, so the order matters: \({}^{8}P_{3} = 8 \times 7 \times 6\).
  2. \({}^{8}P_{3} = 336\).
  3. (b) The three representatives form a group with no order: \({}^{8}C_{3} = \dfrac{8 \times 7 \times 6}{3!}\).
  4. \({}^{8}C_{3} = \dfrac{336}{6} = 56\).
Answer: (a) \(336\)   (b) \(56\)
Worked example Codes or a selection of letters
The six letters of the word PLANET are written on separate cards. Find (a) the number of 4-letter codes that can be made with four of the cards placed in a row, (b) the number of ways of choosing 4 of the cards.
  1. All six letters are different, so there is no repetition to worry about.
  2. (a) A code is read in order: \({}^{6}P_{4} = 6 \times 5 \times 4 \times 3 = 360\).
  3. (b) A set of cards has no order: \({}^{6}C_{4} = \dfrac{360}{4!} = \dfrac{360}{24} = 15\).
Answer: (a) \(360\)   (b) \(15\)
Your turn
A club has 10 members. In how many ways can a captain and a vice-captain be chosen?
Examiner trap

The word 'choose' alone does not mean combination. 'Choose a captain and a vice-captain' gives two different roles, so it is a permutation. Ask whether swapping two of the objects gives a different outcome.

Section 02

Cases: add for or, multiply for and.

Restrictions split a problem into cases. Within one case the stages happen together, so their counts MULTIPLY; the separate cases cannot happen together, so their totals ADD. 'At least' and 'at most' list the allowed cases; when 'at least one' would need many cases, count the opposite and subtract from the total.

Worked example At least two women
A committee of 4 is chosen from 6 men and 5 women. Find the number of committees with at least 2 women.
  1. Allowed cases: 2 women and 2 men, OR 3 women and 1 man, OR 4 women.
  2. 2W 2M: \({}^{5}C_{2} \times {}^{6}C_{2} = 10 \times 15 = 150\).
  3. 3W 1M: \({}^{5}C_{3} \times {}^{6}C_{1} = 10 \times 6 = 60\). 4W: \({}^{5}C_{4} = 5\).
  4. Add the cases: \(150 + 60 + 5 = 215\).
  5. Check by the complement: \({}^{11}C_{4} - {}^{6}C_{4} - {}^{5}C_{1} \times {}^{6}C_{3} = 330 - 15 - 100 = 215\).
Answer: \(215\)
Worked example Two people who will not serve together
A team of 5 is chosen from 10 people, including Ali and Ben, who refuse to be in the team together. How many teams are possible?
  1. Total with no restriction: \({}^{10}C_{5} = 252\).
  2. Teams with both: put Ali and Ben in, then choose 3 more from the other 8: \({}^{8}C_{3} = 56\).
  3. Allowed teams: \(252 - 56 = 196\).
  4. Check by cases: neither \({}^{8}C_{5} = 56\), Ali only \({}^{8}C_{4} = 70\), Ben only \(70\); \(56 + 70 + 70 = 196\).
Answer: \(196\)
Your turn
A group of 3 is chosen from 4 boys and 3 girls. How many groups contain at most one girl?
Examiner trap

Never count 'at least one woman' as 'pick one woman, then any 3 of the rest': \({}^{5}C_{1} \times {}^{10}C_{3} = 600\) counts a committee with two women twice (once for each woman picked first). Use separate cases or the complement.

Stage 1 · Learn

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In how many ways can 3 books be chosen from 7 to take on holiday?

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Which of these needs a permutation rather than a combination?

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How many 4-digit numbers can be made from the digits 1 to 7 if no digit is repeated?

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A committee of 3 is chosen from 5 boys and 4 girls. How many committees include at least one girl?

Section 03

Digits and unknown n.

Number problems are arrangements: a 4-digit number is an ordered row of digits. Fill the RESTRICTED positions first (the last digit for even or divisible by 5, the first digit for size, and never 0 at the front), then arrange the remaining digits in the remaining places. Algebraic problems give an equation in \({}^{n}C_{r}\) or \({}^{n}P_{r}\): write out the formula, cancel, solve, and keep only positive whole-number \(n\) that is large enough.

Worked example Four-digit numbers from 1 to 6
Four-digit numbers are formed from the digits 1, 2, 3, 4, 5, 6, with no digit used more than once. Find how many are (a) even, (b) greater than 5000, (c) even and greater than 5000.
  1. (a) Last digit 2, 4 or 6: 3 ways. The other 3 places from the 5 digits left: \({}^{5}P_{3} = 60\). Total \(3 \times 60 = 180\).
  2. (b) First digit 5 or 6: 2 ways, then \({}^{5}P_{3} = 60\). Total \(2 \times 60 = 120\).
  3. (c) Case first digit 5: last digit 2, 4 or 6 (3 ways), middle two from 4 digits: \({}^{4}P_{2} = 12\), giving \(36\).
  4. Case first digit 6: last digit 2 or 4 only (6 is used), so \(2 \times 12 = 24\).
  5. The cases are separate, so add: \(36 + 24 = 60\).
Answer: (a) \(180\)   (b) \(120\)   (c) \(60\)
Worked example Find n
Find \(n\) given that (a) \({}^{n}C_{2} = 3 \times {}^{n}C_{1}\), (b) \({}^{n+1}P_{2} = 72\).
  1. (a) \(\dfrac{n(n - 1)}{2} = 3n\). Divide by \(n\) (as \(n \ne 0\)): \(\dfrac{n - 1}{2} = 3\), so \(n = 7\).
  2. Check: \({}^{7}C_{2} = 21\) and \(3 \times {}^{7}C_{1} = 21\).
  3. (b) \({}^{n+1}P_{2} = (n + 1)n = 72\), so \(n^{2} + n - 72 = 0\).
  4. \((n + 9)(n - 8) = 0\), so \(n = 8\) or \(n = -9\); reject \(-9\).
  5. Check: \({}^{9}P_{2} = 9 \times 8 = 72\).
Answer: (a) \(n = 7\)   (b) \(n = 8\)
Your turn
Given that \({}^{n}C_{2} = 45\), find \(n\).
Examiner trap

In part (c) above, the restrictions on the first and last digits interact: when 6 is used at the front it cannot also be the even last digit. Split into cases by the first digit instead of multiplying \(2 \times 3\).

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Given that \({}^{n}C_{2} = 21\), find the value of \(n\).

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Four-digit numbers are formed from the digits 1, 2, 3, 4, 5 with no repetition. How many are even?

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A team of 5 is chosen from 7 boys and 6 girls. How many teams contain exactly 3 girls?

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Four-digit numbers are formed from the digits 2, 3, 5, 7, 8 with no repetition. How many are greater than 5000?

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Arranged or chosen: decided from the wording.

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Go deeper · practice
Six original Cambridge-style questions
Prizes against committees, digit problems with 0 and with two restrictions, at least by cases, a pair who will not serve together, finding \(n\), and a multi-part word question.