Permutations
P(n, r) = n! / (n − r)! n = total items r = items arranged ! = factorial

Read as "n permute r". Order matters. Counts all possible ordered arrangements of r items chosen from n.

n in A1, r in B1
=PERMUT(A1,B1)

When Order Matters: Permutations Explained

A permutation is an ordered arrangement of items. Unlike combinations, the sequence matters: the arrangement A-B-C is different from B-A-C, which is different from C-A-B. P(n,r) = n! / (n-r)! counts all possible ordered arrangements when choosing r items from n, where every different ordering is counted as a distinct permutation.

Real-world permutation questions: how many ways can 3 runners finish first, second, and third in a race of 10? How many unique 4-digit PINs can be formed from the digits 1-9 without repetition? How many ways can a playlist of 8 songs be ordered from a library of 50? Each of these involves ordered selection where different orders represent genuinely different outcomes.

Full vs Partial Permutations

A full permutation arranges all n items (r = n): P(n,n) = n!. The number of ways to arrange 5 books on a shelf is 5! = 120. A partial permutation selects and arranges r items from n: P(n,r) = n! / (n-r)!. The number of ways to award gold, silver, and bronze from 10 athletes is P(10,3) = 10! / 7! = 10 × 9 × 8 = 720.

Notice that partial permutations can be computed without fully calculating the factorials: P(10,3) = 10 × 9 × 8 (just the first 3 terms of 10!). This shortcut is worth remembering for manual calculation and helps build intuition: you have 10 choices for first, then 9 remaining for second, then 8 for third.

Permutations with Repetition

The formula P(n,r) = n!/(n-r)! assumes no item can be chosen twice (sampling without replacement). If repetition is allowed (sampling with replacement), the count is simply n^r. A 4-digit PIN from 10 digits (0-9) with repetition allowed: 10^4 = 10,000 possible PINs. A 4-digit PIN from 10 digits without repetition: P(10,4) = 10×9×8×7 = 5,040.

License plates often illustrate this: a plate with 3 letters followed by 4 digits (repetition allowed) has 26^3 × 10^4 = 175,760,000 possible combinations — enough to serve a moderately sized state. Without repetition, the count would be P(26,3) × P(10,4) = 15,600 × 5,040 = 78,624,000 — still large but noticeably smaller.

Permutations in Computer Science

Generating all permutations of a set is a fundamental algorithm in computer science, used in brute-force problem solving, combinatorial optimization, and cryptography. Algorithms like Heap's algorithm generate all n! permutations with O(n!) time complexity — which becomes prohibitively slow for large n (20! ≈ 2.4 × 10^18). For n ≥ 20, smarter approaches like dynamic programming and heuristic optimization are necessary. Understanding permutation counting helps set realistic expectations for algorithm performance.

Frequently Asked Questions

Order matters for rankings (1st, 2nd, 3rd place), passwords, phone numbers, and lock combinations. Use permutations when different orderings of the same items count as different outcomes.
P(n,r) = C(n,r) × r! — the number of combinations times all the ways to arrange those r items. Permutations are always greater than or equal to combinations.
When r = n, P(n,n) = n! — all possible orderings of the entire set. For example, 3 items can be arranged in 3! = 6 ways.