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Permutation Calculator

Count the ways to arrange r items chosen from n, where order matters, with or without repetition, as an exact whole number.

Level: Grades 11–12

Printable worksheet

Permutation Calculator

Inputs
Repetition

Result

Permutations, P(n, r)

336

There are 336 ordered arrangements of 3 items from 8 without repetition.

Number of digits
3

See the step-by-step solution ↓

Step-by-step solution

  1. Order matters and no item is used twice, so use the permutations formula (! means multiply by every whole number below, down to 1)

    P(n,r)=n!(n−r)!

  2. Put in your numbers: n = 8 items and r = 3 positions

    P(8,3)=8!(8−3)!=8!5!

  3. Cancel 5! from the top and bottom, since it appears in both. That leaves the top 3 factors of 8! multiplied together

    8×7×6=336

Test yourself

20 quick questions built from this calculator's formula: work out results, predict what happens when an input changes, and work backward from an answer. Every answer comes with the working.

What this calculator does

Find out how many different ways you can line things up when the order matters. Each such arrangement is called a permutation. Gold, silver and bronze finishers in a race, the digits of a PIN code and a team's batting order are all permutations. Swap two positions, and you get a different outcome. Counting them tells you how many different orders are possible. That's the starting point for many questions about chances and security.

The calculator counts arrangements of r items taken from n items. It works without repetition, where each item is used at most once, written nPr or P(n, r). It also works with repetition, where items can be reused, as with digits in a code. It does the math with exact whole numbers. So even huge results such as 52! are right to every digit.

Where it's used

Why this math matters outside the classroom:

  • Race podiums. With 8 runners, gold, silver and bronze can go to P(8, 3) = 8 × 7 × 6 = 336 different ordered trios.
  • PIN codes. A 4-digit PIN has 10⁴ = 10,000 possibilities with repeats allowed and 10 × 9 × 8 × 7 = 5,040 without.
  • Club officers. A club of 15 can fill president, vice president and treasurer in 15 × 14 × 13 = 2,730 ways.
  • Batting orders. A coach can arrange 9 players into a batting order in 9! = 362,880 ways.
  • License plates. A plate of 3 letters then 3 digits, repeats allowed, has 26³ × 10³ = 17,576,000 possible arrangements.

Grade level and standards: Grades 11–12; Common Core State Standards for Mathematics HSS.CP.B.9. © Copyright 2010. National Governors Association Center for Best Practices and Council of Chief State School Officers. All rights reserved.

How to use it

  1. Choose Without repetition if each item can fill only one position. Choose With repetition if the same item can appear in several positions.
  2. Enter n, the number of different items available.
  3. Enter r, the number of positions to fill.
  4. Read the count and the working. Large answers appear in scientific notation, with the exact digits listed underneath.

What the inputs mean

Repetition
Without repetition fits awarding places to different people. With repetition fits a code where a digit may be used again.
Number of items to choose from (n)
The number of different items, such as runners in a race or symbols available for a code. Whole numbers from 0 to 10,000.
Number of positions to fill (r)
The number of positions to fill, in order. Without repetition, r can't be more than n.

Formula and method

P(n, r) = n! ÷ (n − r)! = n × (n − 1) × … × (n − r + 1)
With repetition: N = n^r

The ! sign means factorial: multiply a whole number by every whole number below it, down to 1. So 4! = 4 × 3 × 2 × 1. Without repetition, the first position can be filled in n ways. The second can be filled in n − 1 ways, since one item is used up, and so on for r positions. Multiply these choices together, and you get n × (n − 1) × … × (n − r + 1). That's the same as n! divided by (n − r)!, because the leftover factors cancel out.

With repetition, any of the n items can fill every position. So the count is n multiplied by itself r times, written n^r. N is that total count. By agreement, 0! = 1 and 0^0 = 1. Both mean there's exactly one way to arrange nothing.

Worked example

Eight runners start a race. How many different ways can the gold, silver and bronze medals be awarded?

  1. Write the formula: P(8, 3) = 8! ÷ (8 − 3)! = 8! ÷ 5!.
  2. Cancel 5!, leaving the top three factors: 8 × 7 × 6.
  3. Multiply: 336.

Result: There are 336 possible podium finishes. Now say items could be reused, as in a 3-symbol code drawn from 8 symbols. Then there would be 8^3 = 512 possibilities.

Understanding the result

The count is the number of different ordered outcomes. If every arrangement is equally likely, the chance of one particular arrangement is 1 divided by the count. In the podium example, guessing all three medalists in order has a 1 in 336 chance.

Permutations always outnumber combinations (groups where order doesn't matter) of the same size by a factor of r!. That's because each group of r items can be put in r! orders. 336 podiums ÷ 3! = 56 possible groups of three medalists.

Assumptions and limitations

  • The n items are all different. Rearranging a word with repeated letters, such as the anagrams of 'LETTER', needs a different formula that divides by the repeats.
  • Order matters. If it doesn't, use the combination calculator.
  • With repetition, every item is available for every position, with no limit.
  • n and r are limited to 10,000. Results with more than 1,000 digits show how many digits there are, not every digit.
  • Circular arrangements, such as seating around a round table, aren't handled directly. They're the straight-line count divided by the number of rotations.
  • The calculator counts arrangements; it doesn't list them.

Frequently asked questions

What is the difference between nPr and nCr?

nPr counts ordered arrangements, and nCr counts groups where order doesn't matter. They're linked by nPr = nCr × r!. Picking a president, secretary and treasurer from 10 people is a permutation (720 ways). Picking a three-person committee is a combination (120 ways).

How many 4-digit PIN codes are possible?

Digits can repeat, so use permutations with repetition: 10^4 = 10,000 codes, from 0000 to 9999. If no digit could repeat, it would be P(10, 4) = 10 × 9 × 8 × 7 = 5,040.

Why is n! the number of ways to arrange n items?

The first position can take any of the n items. The second can take any of the remaining n − 1, and so on down to 1. Multiplying gives n!. It's the permutation count with r = n.

How big do factorials get?

Very big, very fast. 10! is 3,628,800, and 20! has 19 digits. 52!, the number of ways to shuffle a deck of cards, has 68 digits: about 8.065817517 × 10^67. The calculator keeps every digit exact.

How this calculator is checked

Method: Standard counting rules: falling factorial n!/(n−r)! and n^r for arrangements with repetition. The calculation is covered by automated tests with known reference values and edge cases. Last updated 2026-10-18. Found a problem? Tell us.

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