Permutations and Combinations (AA HL)
Counting problems ask "how many ways", but the right method depends entirely on whether order matters. Line five people up in a queue and swapping two of them creates a new arrangement; pick five people for a committee and swapping two changes nothing. This page separates the two cases and gives the formula for each. It's part of the broader Binomial Theorem topic.
28 questions on this sub-topic.
The two formulas
Covered under IB syllabus reference AHL1.10: counting principles, including permutations and combinations. Permutations where some objects are identical, and circular arrangements, are not required. Both formulas below are given in the formula booklet.
Permutations (order matters)
\({}^nP_r=\dfrac{n!}{(n-r)!}\)
Use when you're arranging or ordering \(r\) items chosen from \(n\) distinct items - queues, rankings, passwords, seating in a row.
Combinations (order doesn't matter)
\(\binom{n}{r}={}^nC_r=\dfrac{n!}{r!(n-r)!}\)
Use when you're selecting a group of \(r\) items from \(n\) with no regard to order - committees, teams, hands of cards.
Need the full syllabus wording and formula-booklet reference table? See Binomial Theorem. GDC methods (nCr and nPr keys) are covered in the parent topic's GDC section.
Worked examples
The letters A, B, C, D, E are arranged in a row.
(a) Find the total number of arrangements.
(b) Find the number of arrangements in which A and B are always adjacent.
Worked solution
(a) \(5! = 120\) A1
(b) Treat AB as a single unit: 4 objects can be arranged in \(4!\) ways M1
, and AB can be arranged internally in \(2!\) ways M1
. \(4! \times 2! = 24 \times 2 = 48\) A1
5 differently coloured beads are threaded onto a straight string.
(a) Find the number of distinct arrangements.
(b) Two specific beads must be placed at the two ends (in either order). Find the number of arrangements.
Worked solution
(a) Permutations of 5 distinct beads in a row: \(5!\) M1
\(= 120.\) A1
(b) The 2 specific beads can occupy the two ends in \(2!\) ways, and the remaining 3 beads fill the middle in \(3!\) ways: M1
\(2! \times 3! = 2 \times 6 = 12.\) A1
Common mistakes
- Using \({}^nC_r\) when order actually matters. If the question is about arranging, ranking, or seating - where swapping two chosen items gives a genuinely different outcome - it's a permutation, not a combination. Read for the word "arrange" or "order" before picking a formula.
- Forgetting to arrange within a "glued" block. When two items must stay adjacent, gluing them together and counting the block as one object is only half the job - you still need to multiply by the number of ways to arrange the glued items inside the block.
- Double counting when items are fixed in place first. If some items are already fixed (e.g. two specific beads pinned to the two ends), count only the arrangements of the remaining items and multiply by the arrangements of the fixed items separately - don't include the fixed items again in a factorial that assumes they're still free to move.
Ready to practise properly?
28 permutations-and-combinations questions, marked instantly like the real exam.
Quick answers
What is the difference between a permutation and a combination?
A permutation \({}^nP_r\) counts arrangements where order matters, while a combination \({}^nC_r\) counts selections where order doesn't matter - the same items chosen in a different order count as one combination but as several permutations.
How do I count arrangements where two items must stay together?
Glue the two items into a single block, arrange that block together with the remaining items as usual, then multiply by the number of ways to arrange the two items inside the block (\(2!\) if it's just two of them).