Permutations & combinations
A permutation is an ordered arrangement, counted by P(n,r) = n!/(n−r)!; a combination is an unordered selection, counted by C(n,r) = n!/(r!(n−r)!). The only difference is the r! dividing out the orderings, so the single question to answer first is always: does order matter? Committees do not care about order; podium finishes do.
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Method: how to approach it
The order below is what examiners expect to see, and each step carries its own marks.
- Ask whether order mattersRankings, sequences, passwords and podiums are ordered. Committees, hands and subsets are not.
- Use P(n,r) for ordered selectionsn choices, then n−1, and so on for r factors.
- Use C(n,r) for unordered onesCount the ordered version and divide by r! to remove the duplicate orderings.
- Watch for restrictionsHandle "must include X" by fixing X and counting the rest; handle "must not be adjacent" by counting the complement.
Worked example
From 10 people, how many ways to choose a committee of 3? And how many ways to choose a president, secretary and treasurer?
- Committee: order does not matter, so C(10,3) = 10!/(3!·7!).
- = (10 × 9 × 8)/(3 × 2 × 1) = 720/6 = 120.
- Officers: the roles are distinct, so order matters — P(10,3) = 10 × 9 × 8.
- = 720.
Answer. 120 committees and 720 sets of officers — exactly 3! = 6 times as many, one per role assignment.
Where marks get dropped
These are the specific errors that cost credit on permutations & combinations questions — QED's rubric penalises each of them separately.
- Using C(n,r) when roles are distinct. Naming the positions makes the selection ordered.
- Dividing by r! twice. C(n,r) already removes the orderings; dividing again undercounts.
- Forgetting that C(n,r) = C(n, n−r), which often makes the arithmetic much lighter: C(20,18) = C(20,2) = 190.
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Permutations & combinations — frequently asked questions
What is the quick way to compute C(n,r)?
Write r factors descending from n over r!. C(10,3) = (10·9·8)/(3·2·1) — never expand the full factorials.
How do circular arrangements differ?
Fix one person to break the rotational symmetry, giving (n−1)! arrangements of n people around a table.
What about repeated items?
Arranging letters with repeats divides by the factorial of each repeat count: MISSISSIPPI gives 11!/(4!4!2!1!).
The rest of Combinatorics
Counting principles, permutations, combinations. Each subtopic below has its own method, worked example and mark-losing traps.
- 1Sum & product rules
- 2Permutations & combinations
- 3Binomial theorem & Pascal’s triangle
- 4The pigeonhole principle
- 5Inclusion–exclusion
- 6Counting with repetition (stars & bars)
- 7Derangements & counting surjections
- 8Double counting & bijective proofs
- 9Setting up & solving counting recurrences
- 10Generating functions — an introduction
- 11Multinomial coefficients & repeated items
- 12Hockey stick & Vandermonde identities
- 13Catalan numbers & lattice-path counting
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