Hard

Permutations with Repeated Objects

Simple Explanation

When some of the n objects being arranged are actually indistinguishable copies of each other (like repeated letters in a word), the count of DISTINCT arrangements is smaller than n! — divide by the factorial of each repeated group's count: n! / (n₁!×n₂!×...×nₖ!).

Why Do We Need It?

Plain n! overcounts whenever some objects are identical, since swapping two identical objects does not actually create a new, visibly different arrangement — this formula corrects for that overcounting.

Formula

Permutations with Repeated Objects

n! / (n₁! × n₂! × ... × nₖ!)

The number of distinct arrangements of n objects, when some of them are indistinguishable copies of each other, grouped into k types.

n: — the total number of objects (including repeats)
n₁, n₂, ..., nₖ: — the number of repeated copies within each distinct type

When to use it: Whenever arranging a set that includes repeated (indistinguishable) items, like letters in a word with repeated letters.

Worked Example

Count distinct arrangements with repeated letters

How many distinct arrangements are there of the letters in the word "LEVEL"?

Why Does This Work?

Treating all 5 letters as if they were distinct would give 5!=120 arrangements — but every genuine arrangement is counted 2!×2!=4 times in that total, once for each way the two identical L's (2! ways) and the two identical E's (2! ways) could be swapped without visibly changing anything. Dividing by that overcounting factor gives the true count of distinct arrangements.

Real-Life Example

Counting distinct sequences of colored beads

A craftsperson arranges a string of beads that includes several beads of the same color, and wants to know how many visually distinct patterns are possible.

Since beads of the same color are indistinguishable from each other, this formula gives the correct count of genuinely different-looking patterns, avoiding the overcounting of plain n!.

Practice

How many distinct arrangements are there of the letters in the word "BANANA"?

Hard

Common mistake

Dividing by the factorial of only ONE repeated group when there are multiple repeated groups — every distinct repeated group needs its own factorial in the denominator.

Quick Review

n! / (n₁!×n₂!×...×nₖ!), for k groups of repeated objects.
Corrects the overcounting from treating identical objects as if they were distinguishable.
Every repeated group contributes its own factorial to the denominator.