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Topic 11

Permutations and combinations

Syllabus 11.1–11.3 · Papers 1 and 2

🎯What you need to be able to do

  • Tell whether a problem is about arrangement (order matters) or selection (order does not).
  • Use \( n! \), \( {}^{n}P_{r} \) and \( {}^{n}C_{r} \), including \( 0! = 1 \).
  • Count arrangements with restrictions: items together, apart, or in fixed positions; numbers with digit conditions.
  • Count selections with conditions such as “at least one”, “exactly two” or “not both”.
  • Solve algebraic problems involving \( {}^{n}P_{r} \) and \( {}^{n}C_{r} \).

📚The mathematics

Order matters, or it does not

\( n! = n(n - 1)(n - 2) \cdots 2 \times 1 \), \( 0! = 1 \)
permutations \( {}^{n}P_{r} = \dfrac{n!}{(n - r)!} \)
combinations \( {}^{n}C_{r} = \dbinom{n}{r} = \dfrac{n!}{(n - r)!\,r!} \)

\( {}^{n}P_{r} \) counts ordered arrangements of \(r\) items chosen from \(n\) different ones; \( {}^{n}C_{r} \) counts unordered selections. Each selection of \(r\) items can be arranged in \( r! \) ways, which is why \( {}^{n}P_{r} = r! \times {}^{n}C_{r} \). A team, committee, group or hand is a selection; a queue, number, code, word or seating is an arrangement.

The syllabus leaves out three things: repeated objects (such as the letters of BANANA), arrangements in a circle, and problems that need both a selection and an arrangement. You will only meet distinct objects in a line, or selections.

Arrangements with restrictions

  • Fixed positions first. Fill the restricted places, then arrange the rest. A “slot diagram” of boxes makes this visible.
  • Together. Glue the items into one block, arrange the blocks, then multiply by the arrangements inside the block.
  • Not together (all of them): total minus “together”.
  • Numbers. Deal with the digit that has a condition (first digit not 0, last digit even, first digit at least 5) before the others. If two conditions compete for the same digit, split into cases.
Two slot diagrams for arranging the seven letters of COMPUTE. Top: the first and last boxes are reserved for vowels, with 3 and 2 choices, and the five middle boxes have 5, 4, 3, 2 and 1 choices, giving 720. Bottom: the vowels O, U and E are glued into one block, which is arranged with the four consonants in 5 factorial ways and internally in 3 factorial ways, also giving 720.
Restricted places first; “together” means one block. Both give 720 for COMPUTE — a coincidence of these numbers, not a rule.

Selections with conditions

Split the selection by type and multiply within each case, then add the cases: choosing 2 women from 5 and 2 men from 7 is \( {}^{5}C_{2} \times {}^{7}C_{2} \). “At least one” is usually quickest as total minus none. “Two particular people not both chosen” is total minus the selections that contain both.

✏️Worked example

(a) Find the number of different arrangements of the 7 letters of the word COMPUTE in which (i) the first and last letters are both vowels, [2] (ii) the three vowels are together, [2] (iii) the three vowels are not all together. [1] (b) A team of 5 is chosen from 6 boys and 4 girls. Find the number of teams that contain at least 2 girls. [3]

(a)(i) The vowels are O, U, E. Choose the first letter (3 ways) and the last (2 ways), then arrange the other five letters: \( 3 \times 2 \times 5! = 720 \).

(ii) One vowel block plus C, M, P, T is 5 items: \( 5! \) ways. The vowels inside the block: \( 3! \) ways. Total \( 5! \times 3! = 720 \).

(iii) All arrangements minus those with the vowels together: \( 7! - 720 = 5040 - 720 = 4320 \).

(b) Total teams \( {}^{10}C_{5} = 252 \). Remove those with fewer than 2 girls: no girls, \( {}^{6}C_{5} = 6 \); one girl, \( {}^{4}C_{1} \times {}^{6}C_{4} = 4 \times 15 = 60 \). So \( 252 - 6 - 60 = 186 \).

Check it. Count (b) directly: 2 girls \( 6 \times 20 = 120 \), 3 girls \( 4 \times 15 = 60 \), 4 girls \( 1 \times 6 = 6 \); total 186 ✓.
“Not all together” is not “all apart”. \( 7! - 5!\,3! \) still counts arrangements like COUMPET, where two vowels are together. If a question wants no two vowels adjacent, that is a different count.

📝Practise

Counting questions appear on both papers; none of these needs a calculator.

1. Evaluate \( {}^{8}P_{3} \) and \( {}^{8}C_{3} \). [2]
\( {}^{8}P_{3} = 8 \times 7 \times 6 = 336 \); \( {}^{8}C_{3} = \dfrac{336}{3!} = 56 \).
2. Find \(n\) given that \( {}^{n}C_{2} = 45 \). [3]
\( \dfrac{n(n - 1)}{2} = 45 \Rightarrow n^{2} - n - 90 = 0 \Rightarrow (n - 10)(n + 9) = 0 \). Since \( n > 0 \), \( n = 10 \).
3. Four-digit numbers are formed from the digits 1, 2, 3, 4, 5, 6, 7; no digit is used more than once. Find how many (a) can be formed, (b) are even, (c) are greater than 5000 and even. [6]
(a) \( {}^{7}P_{4} = 840 \). (b) Last digit 2, 4 or 6, then \( {}^{6}P_{3} \): \( 3 \times 120 = 360 \). (c) Cases on the first digit. First digit 6: last digit 2 or 4 (2 ways), middle \( {}^{5}P_{2} = 20 \): 40. First digit 5 or 7: last digit 2, 4 or 6 (3 ways): \( 2 \times 3 \times 20 = 120 \). Total 160. (The 6 can be first or last but not both, which is why the cases are split.)
4. Three different maths books and three different science books are placed on a shelf. Find the number of arrangements in which (a) the maths books are together, (b) the subjects alternate. [4]
(a) Maths block + 3 science books = 4 items: \( 4! \times 3! = 144 \). (b) Pattern MSMSMS or SMSMSM (2 ways), with \( 3! \) orders for each subject: \( 2 \times 3! \times 3! = 72 \).
5. A committee of 4 is chosen from 7 men and 5 women. Find the number of committees with (a) exactly 2 women, (b) at least one woman. [4]
(a) \( {}^{5}C_{2} \times {}^{7}C_{2} = 10 \times 21 = 210 \). (b) \( {}^{12}C_{4} - {}^{7}C_{4} = 495 - 35 = 460 \).
6. Five people are chosen from a group of 9, which includes Ana and Budi. Find the number of groups that do not contain both Ana and Budi. [3]
Total \( {}^{9}C_{5} = 126 \). Groups containing both: choose the other 3 from 7, \( {}^{7}C_{3} = 35 \). So \( 126 - 35 = 91 \).
7. Find the value of \(n\) for which \( {}^{n}P_{3} = 6 \times {}^{n}C_{2} \). [3]
\( n(n - 1)(n - 2) = 6 \times \dfrac{n(n - 1)}{2} = 3n(n - 1) \). Since \( n \ge 3 \), divide by \( n(n - 1) \ne 0 \): \( n - 2 = 3 \), \( n = 5 \). (Dividing is safe here because \( n(n-1) \) cannot be zero for a valid \(n\).)

🔗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.

  • Your calculator’s nPr and nCr keys — use them to check Paper 1 answers during revision
  • Khan Academy — permutations and combinations, with many short practice sets