What is the difference between permutations and combinations, and when does each apply?
Permutations count ordered arrangements: P(n,k) = n!/(n-k)!. Combinations count unordered selections: C(n,k) = n!/[k!(n-k)!]. The rule is simple — if the order of selection matters, permute; if it doesn't, combine. Combinations are always smaller by a factor of k!.
How to think about it
Combinatorics is the backbone of discrete probability: every classical probability calculation requires correctly counting the sample space and the favorable outcomes.
Permutations — order matters
Arranging k items chosen from n distinct items, where position matters:
P(n, k) = n × (n-1) × ··· × (n-k+1) = n! / (n-k)!Example: the number of ways to assign Gold, Silver, Bronze medals among 10 athletes:
P(10, 3) = 10 × 9 × 8 = 720Combinations — order does not matter
Choosing k items from n without caring about their order:
C(n, k) = P(n,k) / k! = n! / [k! × (n-k)!]Example: choosing 3 people from 10 to form a committee (no roles):
C(10, 3) = 720 / 6 = 120The relationship: C(n,k) = P(n,k) / k! — divide out the k! orderings we don’t want to distinguish.
With vs without replacement
| Order matters | Order doesn’t matter | |
|---|---|---|
| Without replacement | n!/(n-k)! | C(n,k) |
| With replacement | nᵏ | C(n+k-1, k) |
The “with replacement, unordered” case (stars-and-bars) appears in Poisson models and text-as-bag-of-words counting.
Applied example: card hand probability
Five-card hand from a 52-card deck:
Total hands: C(52,5) = 2 598 960
Hands with exactly 2 aces: C(4,2) × C(48,3) = 6 × 17 296 = 103 776
P(exactly 2 aces) = 103 776 / 2 598 960 ≈ 3.99 %