Multiset Permutations

Combinatorics · Axiom Academy

How to count arrangements when some objects are identical 1. The Problem with Identical Objects Consider arranging the letters in the word AAB . If all letters were different, we'd have 3! = 6 arrangements. But the two A's are identical . Let's see what happens when we list all possible arrangements: When we have k identical objects, swapping them among themselves doesn't create a new arrangement. Let's visualize this with colored balls: 2 red balls (identical) and 1 blue ball. 3. The Multiset Permutation Formula If we have n total objects where n₁ are of type 1, n₂ are of type 2, ..., and nₖ are of type k, then: Why this works: We start with n! total arrangements (treating all as distinct), then divide by n₁! for each group of identical objects to remove the overcounting. Let's apply the formula to find how many distinct arrangements exist for the word BANANA . 5. Classic Example: MISSISSIPPI The word MISSISSIPPI is a famous example with multiple repeated letters.

This is the written version of the interactive lesson above. See the full Combinatorics course.