11. Permutations and combinations
- Syllabus
- 0606–2028–2029
- Topic
- 11
- Level
- —
Learning objectives
11.1Distinguish permutations and combinations• Recognise the difference between permutations and combinations and know when each should be used.11.2Use n! and nPr/nCr expressions• Know and use the notation n! and the expressions for permutations and combinations of n items taken r at a time.• Includes 0! = 1.11.3Solve arrangement and selection problems• Solve arrangement and selection problems using permutations or combinations.• Problems will be in everyday contexts or based on algebraic problems.• Problems involving repeated objects, circular arrangements, or both permutations and combinations are not included.
Decide whether order matters
A permutation counts arrangements, so changing the order or assigned roles creates a different outcome. A combination counts selections, so the chosen group is the same however its members are listed.
| Ask | If yes | If no |
|---|---|---|
| Would swapping two chosen items create a new result? | permutation | combination |
| Are positions, ranks, seats, roles, or code order specified? | permutation | check whether it is only a group |
| Is the outcome just a team, committee, or subset? | check for assigned roles | combination |
From 8 people, choosing a 3-person committee is a combination: the group {A,B,C} is one outcome. Filling president, secretary and treasurer is a permutation: A,B,C in those roles differs from B,A,C. These are separate counting situations even though both choose 3 of 8.
Words such as “arrange”, “code”, “first and last” or named offices usually signal order. Words such as “choose”, “select”, “team” or “committee” usually signal no order. The wording is a clue, but the swap test is the decisive check.
Do not choose a formula only because the question uses the word “choose”: a chosen group with assigned roles is ordered. Conversely, listing a committee in a different written order does not make a new committee.
Build permutation and combination formulas from factorials
For a non-negative integer n, n!=n(n−1)(n−2)⋯2⋅1. Also 0!=1: there is exactly one way to arrange or select nothing, and this value keeps factorial formulas valid when r=n.
{}^nP_r=\frac{n!}{(n-r)!},\qquad {}^nC_r=\frac{n!}{r!(n-r)!},\qquad 0\le r\le n
The permutation formula keeps the r descending choices n(n−1)⋯(n−r+1). A selected set of r objects appears in r! different orders, so dividing nPr by r! gives nCr. Therefore nPr=r!nCr.
For 8 objects taken 3 at a time, 8P3=8imes7imes6=336, while 8C3=336/3!=56. In an algebraic equation, expand only the factors that do not cancel: if nP5=6n−1P4, both sides contain (n−1)(n−2)(n−3)(n−4), leaving n=6.
The upper and lower values must be whole numbers with 0≤r≤n. Cancel factorials before expanding them into large numbers, and distinguish the lowercase variable n from a fixed numerical value.
Count restricted arrangements and selections
Restrictions change which outcomes are allowed, not the meaning of permutation or combination. First decide whether the outcome is ordered; then organise the restriction with positions, cases, a block, or a complement.
| Restriction | Reliable counting move |
|---|---|
| fixed first/last position, parity, or leading digit | fill the restricted position first, then multiply remaining choices |
| several mutually exclusive possibilities | count each complete case, then add |
| specified objects must stay together in a line | treat them as one block, then multiply by their internal orders |
| at least one selected object has a property | total selections minus selections with none |
If 3 distinct men must stand together with 3 distinct women in a line, treat the men as one block. The block and 3 women form 4 units, arranged in 4! ways, while the men can be ordered internally in 3! ways. The count is 4!imes3!=144.
To select 7 letters from 4 vowels and 9 consonants with at least 2 vowels, the possible vowel counts are 2, 3 or 4. The total is 4C29C5+4C39C4+4C49C3=1344. The cases are added because no selection belongs to two different vowel-count cases.
Stay inside the syllabus boundary: questions with repeated objects, circular arrangements, or a single problem requiring both permutations and combinations are excluded. For permitted problems, make cases disjoint and check that a leading zero or forbidden position has not been counted.