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Permutation Calculator

Count ordered arrangements nPr, with or without repetition

Updated · Free, no signup

On: each position can reuse any item (n^r).

Number of permutations

720

Number of permutations (numeric)

720

Shown as 0 if the count is too large for a standard number (over about 10^308).

Without repetition nPr

720

With repetition nʳ

1,000

Combinations nCr (order ignored)

120

Digits in the answer

3

  • Each of the 120 groups can be ordered 3! ways, giving 720 arrangements.

Arrangements for each r (no repetition)

rnPrnCr
011
11010
29045
3720120
45,040210
530,240252
6151,200210
7604,800120
81,814,40045
93,628,80010
103,628,8001

About the Permutation Calculator

This permutation calculator counts how many ways you can arrange r items chosen from n when order matters. Enter n and r to get nPr, and switch on repetition when the same item can be used more than once — the way a PIN code can repeat digits.

Use it for race finishing orders (gold, silver, bronze), seating plans, passwords and lock codes, scheduling, or probability homework. The combination count nCr is shown alongside so you can see how much larger the count becomes when order matters: every group of r items can be ordered r! ways.

Every count is computed exactly with whole-number arithmetic. Answers up to 21 digits are shown in full; longer ones (such as 52! for a shuffled deck) are shown in scientific notation with the exact digit count. The table lists nPk for small k so you can see how quickly arrangements grow.

How to use the permutation calculator

  1. 1Enter the total number of distinct items n.
  2. 2Enter how many positions r you are filling.
  3. 3Turn on repetition if the same item can appear more than once.
  4. 4Read the number of permutations and compare with the combination count.

Formula and method

P(n, r) = n! ÷ (n − r)! · with repetition: nʳ

Without repetition, the first position can be filled n ways, the second n − 1 ways, and so on for r positions, so nPr = n × (n − 1) × … × (n − r + 1) = n! ÷ (n − r)!. The calculator multiplies these r factors directly with exact big-integer arithmetic rather than computing huge factorials, so even 1000P500 is exact.

With repetition, every one of the r positions can be any of the n items, giving n × n × … × n = nʳ. Permutations and combinations are linked by nPr = nCr × r!, because each unordered group of r items can be arranged r! ways.

n
Number of distinct items available
r
Number of positions to fill (items arranged)
!
Factorial: n! = n × (n − 1) × … × 1, with 0! = 1

Worked examples

Gold, silver and bronze among 10 runners

Gold can go to any of 10 runners, silver to any of the remaining 9 and bronze to any of 8: 10 × 9 × 8 = 720 podiums. Ignoring order there are only 720 ÷ 3! = 120 groups of medallists.

4-letter codes using A–Z with repeats

Each of the 4 positions can be any of 26 letters, so there are 26⁴ = 456,976 possible codes.

Ordered 5-card sequences from a deck

52 × 51 × 50 × 49 × 48 = 311,875,200 ordered deals. Dividing by 5! = 120 gives the familiar 2,598,960 unordered poker hands.

Seating 8 guests in 8 chairs

Arranging all 8 guests is 8! = 40,320 different seating plans.

Shuffling a full deck: 52P52

There are 52! ≈ 8.07 × 10⁶⁷ ways to order a deck of cards — a 68-digit number, far more than could ever be dealt.

Frequently asked questions

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

A permutation counts arrangements where order matters (first, second, third place), while a combination counts groups where it does not (a committee). nPr is always r! times nCr.

How do you calculate permutations with repetition?+

Raise the number of choices to the power of the number of positions: nʳ. A 4-digit PIN using 0–9 has 10⁴ = 10,000 possibilities.

What is 0! and why does nPn equal n!?+

0! is defined as 1, the number of ways to arrange nothing. When r = n, nPn = n! ÷ 0! = n!, the number of ways to order all the items.

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

If all letters are different, use n! (for example, 5 letters give 120). If letters repeat, divide by the factorial of each repeat count: “LETTER” has 6! ÷ (2! × 2!) = 180 arrangements.

Can r be larger than n?+

Only with repetition. Without repetition you cannot fill more positions than you have distinct items, so nPr = 0 when r > n.

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