Probability & Combinatorics
Intermediate

Permutations of a Multiset

Arrangements of n items where some are identical.

Formula

n!n1!n2!⋯nk!\frac{n!}{n_1! n_2! \cdots n_k!}

Variables

nTotal items
n_iCount of each identical group

Example

MISSISSIPPI: 11!/(4!4!2!) = 34650

How it works

This formula counts distinct ways to arrange a collection of items when some items are indistinguishable from each other. Start with n! ways to arrange all items as if they were different, then divide by each n_i! because swapping identical items within the same group does not create a new arrangement. In other words, it corrects the overcounting caused by repeated objects.

When to use it

Use it when counting permutations of letters in a word with repeated letters, arranging people or objects where some are identical, or finding the number of distinct orderings in multistep arrangements with repeats. It is especially useful in counting problems where only the pattern matters, not which identical copy is which.

Worked example, step by step

Find the number of distinct arrangements of the letters in BALLOON. Step 1: Count total letters: n = 7 Step 2: Identify repeated groups: L appears 2 times, O appears 2 times; other letters are unique Step 3: Apply the formula: 7! / (2!2!) Step 4: Compute: 7! = 5040 and 2!2! = 4 Step 5: Divide: 5040 / 4 = 1260 Final answer: 1260 distinct arrangements

Common mistakes to avoid

  • Forgetting to divide by every repeated group, not just one of them
  • Using the formula when all items are distinct, which reduces to n! instead
  • Treating identical items as different and counting duplicate arrangements multiple times

Did You Know?

Repeated letters reduce the number of distinct arrangements.

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Reviewed & maintained by the FormulaWon editorial teamLast reviewed September 2026