Permutation
Study guide: Exam P-1
A Permutation P(n, k) counts the number of ways to select and arrange k objects from n distinct objects in a specific order.
- Permutations assume without replacement: you cannot select the same object twice
- Permutations differ from combinations by a factor of
k!: the number of ways to order the selected objects - Use permutations when sequence matters: ranking, scheduling, or assigning distinct roles
P(n, k) = \frac{n!}{(n-k)!}
= n \times (n-1) \times \cdots \times (n-k+1)
Assigning Ranked Prizes to Adjusters From 8 adjusters, an insurer awards a 1st, 2nd, and 3rd place performance bonus. How many distinct award outcomes are possible?
Answer Order matters (1st ≠ 2nd ≠ 3rd place), so this is a permutation:
P(8, 3) = \frac{8!}{(8-3)!} = \frac{8!}{5!} = 8 \times 7 \times 6 = 336There are 336 distinct ways to assign the three ranked prizes, compared to only\binom{8}{3} = 56unordered committees.