Combinations & Permutations — Practice Problems

A permutation counts arrangements where order matters: $P(n,r) = \frac{n!}{(n-r)!}$. A combination counts selections where order doesn't matter: $C(n,r) = \binom{n}{r} = \frac{n!}{r!(n-r)!}$. Ask: does switching the order create a different outcome? If yes → permutation. If no → combination.