# How many permutations are there?

Counts permutations exactly: nPr ordered arrangements of r items from n, arrangements with repetition (n^r), or the distinct arrangements of the letters of a word.

- Page: https://www.acalculator.org/statistics/permutation-calculator
- JSON spec: https://www.acalculator.org/statistics/permutation-calculator.json
- Version: e520b4bb840d

## Default answer

Example with the default inputs (What are you arranging? r of n items, Items to choose from (n) 10, Items arranged (r) 3): There are 720 permutations (10! ÷ (10 − 3)! = 10! ÷ 7!).

## Inputs

| Key | Label | Description |
| --- | --- | --- |
| mode | What are you arranging? | r of n different items, r picks where an item can repeat, or all the letters of a word. |
| n | Items to choose from (n) | How many different items there are. |
| r | Items arranged (r) | How many positions are filled, in order. |
| w | Word or letters | The letters to arrange, such as MISSISSIPPI. Spaces are ignored; capital and small letters count as the same. |

## Outputs

| Key | Label | Description |
| --- | --- | --- |
| permutations | Permutations | The number of different ordered arrangements. |
| approx | In scientific notation | The same count to 6 significant digits, shown when it has more than 15 digits. |
| formula | Formula | The formula with your numbers. |

## Method

nPr = n! ÷ (n − r)! = n (n − 1) … (n − r + 1); with repetition, n^r; letters of a word, n! ÷ (n₁! n₂! … nₖ!) where nⱼ counts each repeated letter.

## Assumptions

- n and r are whole numbers from 0 to 1,000. Without repetition, r is at most n.
- Letters: spaces are ignored and capital and small letters count as the same letter; any other character counts as a letter too.
- Counts are exact whole numbers; the scientific form is rounded half up to 6 significant digits.

## Worked examples

1. mode = npr, n = 12, r = 9 gives permutations = 79,833,600. Source: OpenStax, Algebra and Trigonometry 2e, §13.5 Counting Principles, P(12, 9) = 79,833,600, https://openstax.org/books/algebra-and-trigonometry-2e/pages/13-5-counting-principles.
2. mode = letters, w = DISTINCT gives permutations = 10,080, formula = 8! ÷ (2! for I, 2! for T). Source: OpenStax, Algebra and Trigonometry 2e, §13.5 Counting Principles, DISTINCT: 8! ÷ (2! 2!) = 10,080, https://openstax.org/books/algebra-and-trigonometry-2e/pages/13-5-counting-principles.
3. mode = letters, w = Mississippi gives permutations = 34,650. Source: NIST DLMF §26.4 Multinomial Coefficients, equation 26.4.2, https://dlmf.nist.gov/26.4.
4. mode = repeat, n = 10, r = 4 gives permutations = 10,000, formula = 10^4. Source: OpenStax, Algebra and Trigonometry 2e, §13.5, https://openstax.org/books/algebra-and-trigonometry-2e/pages/13-5-counting-principles.
5. mode = npr, n = 52, r = 52 gives approx = 8.06582 × 10⁶⁷. Source: OpenStax, Algebra and Trigonometry 2e, §13.5, https://openstax.org/books/algebra-and-trigonometry-2e/pages/13-5-counting-principles.
6. mode = npr, n = 7, r = 0 gives permutations = 1, formula = 7! ÷ (7 − 0)! = 7! ÷ 7!. Source: OpenStax, Algebra and Trigonometry 2e, §13.5: P(n, 0) = n! ÷ n! = 1, https://openstax.org/books/algebra-and-trigonometry-2e/pages/13-5-counting-principles.

## FAQ

### What is a permutation?

A permutation is an arrangement of items in order. ABC and CBA are different permutations of the same three letters. When order does not matter, count combinations instead.

### What is the nPr formula?

nPr = n! ÷ (n − r)!, the number of ways to fill r ordered places from n different items. It is n × (n − 1) × … × (n − r + 1): 12P9 = 12 × 11 × … × 4 = 79,833,600.

### What is the difference between a permutation and a combination?

A permutation counts orders; a combination does not. Choosing 3 of 10 people for gold, silver and bronze is 10P3 = 720. Choosing 3 of 10 for a team is 10C3 = 120, which is 720 ÷ 3!.

### How do I count permutations with repetition?

When each of the r places can take any of the n items again, there are n^r arrangements. A 4-digit PIN from the digits 0 to 9 has 10⁴ = 10,000 possibilities.

### How many ways can the letters of a word be arranged?

Divide n! by the factorial of the count of each repeated letter. MISSISSIPPI has 11 letters: 4 I, 4 S and 2 P, so 11! ÷ (4! × 4! × 2!) = 34,650 distinct arrangements.

### What is 0!, and what is nP0?

0! = 1 by definition, so nP0 = n! ÷ n! = 1: there is exactly one way to arrange nothing. And nPn = n!, every order of all n items.

### Why does the calculator say r cannot be more than n?

Without repetition you cannot fill more places than you have different items. Choose "With repetition" if an item can be used more than once.

## Sources

- OpenStax, Algebra and Trigonometry 2e, §13.5 Counting Principles (P(n, r) = n! ÷ (n − r)!; permutations of non-distinct objects n! ÷ (r₁! r₂! … rₖ!); the multiplication principle). https://openstax.org/books/algebra-and-trigonometry-2e/pages/13-5-counting-principles (retrieved 2026-10-01)
- NIST Digital Library of Mathematical Functions, §26.4 Lattice Paths: Multinomial Coefficients, equation 26.4.2. https://dlmf.nist.gov/26.4 (retrieved 2026-10-01)