Quick answer
P(n, r) counts ordered selections; C(n, r) counts unordered selections (binomial coefficients).
Formula
- P(n,r) = n!/(n−r)!
- C(n,r) = n!/(r!(n−r)!)
- P(n,r) = r! × C(n,r)
Introduction
Read the problem aloud and ask whether two outcomes that use the same items in different orders should be counted separately. If yes, use permutations. If no, use combinations.
Combination notation is developed fully in what is a binomial coefficient; this article contrasts that count with ordered arrangements.
Once you label a problem as combination, compute with calculation methods for C(n, k) or the home calculator for verification.
Side-by-side mental models
Permutation picture: medals on a podium, PIN codes with distinct positions, race finishes with labeled places.
Combination picture: pizza toppings chosen from a list, lottery numbers printed in ascending machine order (order not part of the bet), committees without roles.
Some prompts add roles after selection ("choose a president among the committee"), which is permutation or multiplication principle on top of C(n, k), not a reason to skip the combination step.
Formulas and bridge
- P(n,r) = n × (n−1) × … × (n−r+1)
- C(n,r) = P(n,r) / r!
Memorize the bridge: dividing ordered counts by r! collapses each block of orderings into one combination.
Step-by-step guide
- Underline clue words. Arrange, rank, sequence → permutation. Choose, select, committee → combination.
- Test with a toy set. List outcomes for n = 3, r = 2 if unsure; counts 6 vs 3 settle the question quickly.
- Pick formula and compute. Use symmetry on C(n,r) when helpful; use the product form for P(n,r).
Same club, two questions
Choose president and treasurer from 8 members with roles distinct: P(8, 2) = 56 ordered assignments.
Choose a two-person committee with no roles: C(8, 2) = 28 subsets.
56 = 2! × 28 shows how role labeling reintroduces ordering after an unordered pair is chosen.
