What's the difference between Permutations with Repetition and regular Permutations?

Regular Permutations arranges n distinct items (or selects and arranges r from n distinct items); Permutations with Repetition arranges all n items where many are identical. Use this formula when items repeat in groups and you must account for all those identical items in the denominator.

Can I use this formula if I have different numbers of repeats for each item type?

Yes, this is exactly what the formula is designed for. Some groups can have 1 item, others 2, 5, or more—it doesn't matter. You simply put each group's factorial in the denominator, regardless of the group sizes.

What if I forget to include one of the repeated groups in the denominator?

You will overcount significantly. For example, if you arrange BOOK using only \frac{4!}{1! \cdot 1!} = 24 (forgetting the 2! for the two O's), you get 24 instead of the correct 12—you've counted each arrangement twice because swapping the O's produces the same word.

Is Permutations with Repetition the same as Combinations?

No. Combinations chooses a subset of distinct items and does not care about order; Permutations with Repetition arranges all n items where some are identical and order definitely matters. The formula for Combinations is \binom{n}{r} = \frac{n!}{r!(n-r)!}, which has a different structure.

Permutations with Repetition

Counts the distinct arrangements of n total objects when some are identical; use this when items repeat in groups and order matters.

\frac{n!}{n_1!\,n_2!\cdots n_k!}

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What each symbol means

What Permutations with Repetition takes

Symbol	Meaning
$n$	The total count of all objects being arranged, equal to the sum of all group sizes. Confusing this with the count of unique items will give you the wrong answer.
$n_i$	The number of identical items in the i-th group, such as how many times a letter appears in a word. Missing or miscounting any group's repetitions skews the entire denominator.
$k$	The number of distinct groups or types of items; undercounting groups means you omit divisors in the denominator and overcount arrangements.

When to use it

Use this when you need to count ordered arrangements of objects where some items are identical.

Level

Usually taught in: Algebra II

Worked examples

1. Arrange letters in a four-letter word with one repeated letter

Problem

How many distinct arrangements of the letters in the word BOOK can you make?

$n = 4, \text{ with groups: } B \text{ (1 time)}, O \text{ (2 times)}, K \text{ (1 time)}$
The word BOOK has 4 letters total, but O is repeated twice, so we have three groups.
$\frac{4!}{1! \cdot 2! \cdot 1!}$
We apply the formula with $n = 4$ and a factorial in the denominator for each of the three groups.
$\frac{24}{1 \cdot 2 \cdot 1} = \frac{24}{2} = 12$
We calculate $4! = 24$ and the denominator as $1 \cdot 2 \cdot 1 = 2$ , giving 12 distinct arrangements.

Answer: $12$

BOOK has 4 letters, but the two O's are identical. If we used just $4!$ , we'd count 24 arrangements, but many are duplicates because swapping the two O's gives the same word. Dividing by $2!$ corrects for this overcounting.

2. Arrange letters in a long word with multiple repeated letters

Problem

How many distinct arrangements of the letters in MISSISSIPPI can you make?

$n = 11, \text{ with groups: } M \text{ (1)}, I \text{ (4)}, S \text{ (4)}, P \text{ (2)}$
MISSISSIPPI has 11 letters total: 1 M, 4 I's, 4 S's, and 2 P's, giving us four groups.
$\frac{11!}{1! \cdot 4! \cdot 4! \cdot 2!}$
We apply the formula with each group's size factored into the denominator.
$11! = 39916800, \quad 1! \cdot 4! \cdot 4! \cdot 2! = 1 \cdot 24 \cdot 24 \cdot 2 = 1152$
We calculate $11! = 39916800$ and the denominator as $1 \cdot 24 \cdot 24 \cdot 2 = 1152$ .
$\frac{39916800}{1152} = 34650$
Dividing gives 34650 distinct arrangements.

Answer: $34650$

MISSISSIPPI has 11 letters with four types of repetitions: 1 M, 4 I's, 4 S's, and 2 P's. The formula divides by each group's factorial to eliminate overcounting from swapping identical items within each group. This single formula handles all repetitions at once.

3. Count training route sequences with uphill and downhill segments

Problem

During a 10-segment training ride, a cyclist plans exactly 4 uphill segments and 6 downhill segments in some order. How many different sequences of ups and downs could make up the route?

$n = 10, \text{ uphill: } n_1 = 4, \text{ downhill: } n_2 = 6$
The route consists of 10 segments total: exactly 4 uphill and 6 downhill, giving two groups.
$\frac{10!}{4! \cdot 6!}$
We apply the formula with two groups of different sizes.
$10! = 3628800, \quad 4! = 24, \quad 6! = 720$
We calculate $10! = 3628800$ , $4! = 24$ , and $6! = 720$ .
$\frac{3628800}{24 \cdot 720} = \frac{3628800}{17280} = 210$
The denominator is $24 \cdot 720 = 17280$ , and dividing gives 210 distinct sequences.

Answer: $210$

The cyclist can arrange 4 identical uphill segments and 6 identical downhill segments in $\frac{10!}{4! \cdot 6!} = 210$ different orders. Each arrangement is a distinct training pattern even though segments of the same type are indistinguishable—only their position in the 10-segment route matters.

Common mistakes

Where Permutations with Repetition usually goes wrong

The mistake	Why it is wrong	The fix
Writing $n!$ without dividing by any factorials in the denominator.	This counts every arrangement as if all items were distinct, but identical items produce duplicate arrangements that should be counted only once.	Always identify groups of identical items and divide $n!$ by the factorial of each group's size: $\frac{n!}{n_1! \cdot n_2! \cdots n_k!}$ .
Forgetting to include one of the groups of identical items in the denominator.	If you divide by only some of the factorials, you still overcount because you have not accounted for all the swaps that produce identical results.	Write out every group of repeated items and include each one in the denominator; every distinct type must contribute exactly one factorial.
Confusing this formula with regular Permutations and using $\frac{n!}{(n-r)!}$ instead.	Regular Permutations counts ways to select and arrange $r$ distinct items from $n$ distinct items; this formula arranges all $n$ items where many are identical.	Use Permutations with Repetition when you have all $n$ items and some are repeated; use regular Permutations when selecting a subset of distinct items.

Tips and when to use something else

Identify every group of identical items—even one missed group ruins the denominator.
Verify your answer on small cases (like 3-letter words) by listing all arrangements by hand.
When you're selecting a subset of objects (not arranging all $n$ ), use Combinations or Permutations instead—this formula requires all $n$ items.
If there are only two distinct types, the formula simplifies to the binomial coefficient $\binom{n}{n_1}$ .

Related formulas

PermutationsPermutations count the number of ways to arrange a set of objects in a specific order, where changing the order creates a different arrangement.CombinationsCombinations count the ways to select items from a group when order doesn't matter; they are used for choosing unordered subsets.Fundamental Counting PrincipleMultiply the number of choices at each step to find the total number of different possible outcomes when making sequential selections or decisions.

Frequently asked questions

What's the difference between Permutations with Repetition and regular Permutations?: Regular Permutations arranges $n$ distinct items (or selects and arranges $r$ from $n$ distinct items); Permutations with Repetition arranges all $n$ items where many are identical. Use this formula when items repeat in groups and you must account for all those identical items in the denominator.
Can I use this formula if I have different numbers of repeats for each item type?: Yes, this is exactly what the formula is designed for. Some groups can have 1 item, others 2, 5, or more—it doesn't matter. You simply put each group's factorial in the denominator, regardless of the group sizes.
What if I forget to include one of the repeated groups in the denominator?: You will overcount significantly. For example, if you arrange BOOK using only $\frac{4!}{1! \cdot 1!} = 24$ (forgetting the $2!$ for the two O's), you get 24 instead of the correct 12—you've counted each arrangement twice because swapping the O's produces the same word.
Is Permutations with Repetition the same as Combinations?: No. Combinations chooses a subset of distinct items and does not care about order; Permutations with Repetition arranges all $n$ items where some are identical and order definitely matters. The formula for Combinations is $\binom{n}{r} = \frac{n!}{r!(n-r)!}$ , which has a different structure.

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Reviewed 2026-09-18