Permutations and Combinations · 0606 Topic 11

Permutations (nPr)

Rig, founder of IGCSE Add Math Malaysia

Written by Rig, our founder

8 years teaching IGCSE & SPM maths · Updated 12 June 2026

nPr=n!(nr)!{}^nP_r = \dfrac{n!}{(n-r)!} counts the ways to choose rr items from nn and arrange them, selection with order. It’s the package for “how many ways can rr of these nn things fill rr distinct positions”.

8 runners; how many ways can gold, silver and bronze be awarded? 8P3=8!5!=8×7×6=336{}^8P_3 = \dfrac{8!}{5!} = 8 \times 7 \times 6 = \mathbf{336}

Notice the cancellation: 8P3{}^8P_3 is the slot product 8×7×68 \times 7 \times 6. The formula and the slots are the same object, which means if you ever doubt the formula, draw the slots; if the slots are tedious, use the formula. Both earn the marks.

The order test

Use nPr{}^nP_r when the positions are distinguishable: medals, ranked prefects (chairman/secretary/treasurer), letters arranged into “words”, digits into numbers, books onto a shelf choosing 4 of 9. If swapping two chosen items produces a different outcome, order matters → nPr{}^nP_r (or slots). If a swap changes nothing, a committee, a team, a handful of cards, order doesn’t matter and you want (nr)\binom{n}{r}. The relationship: nPr=(nr)×r!{}^nP_r = \binom{n}{r} \times r!, choose the set, then arrange it; many harder questions are exactly that two-step in disguise.

Restrictions inside permutations

Same discipline as everywhere in the topic, constrained positions first:

From 9 different books, arrange 4 on a shelf with a particular book at the left end. Anchor it: 1 way. Fill the other three positions from the remaining 8: 8×7×6=3368 \times 7 \times 6 = \mathbf{336}

“Two particular items both selected and adjacent” combines the block method with selection, glue, count units, multiply by internal arrangements. Write each factor with a one-word label (“block”, “internal”, “rest”); labelled factors are followable, followable is markable.

Common mistakes

  • nPr{}^nP_r used for committees (order doesn’t matter there)
  • nn and rr swapped in the formula (rPn{}^rP_n is meaningless, the big number leads)
  • Restrictions applied after free counting
  • nPr=(nr)×r!{}^nP_r = \binom{n}{r} \times r! forgotten, making “choose then arrange” questions look unfamiliar
  • Calculator nPr{}^nP_r button trusted blind where the slot logic would have caught a misread

Full topic context: P&C notes · the unordered twin: combinations (nr)\binom{n}{r}.

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