Statistics
Permutation Calculator
Count the ordered arrangements of r items chosen from a set of n. Switch between selections without repetition (nPr = n! / (n − r)!) and with repetition (nʳ), see every step, and compare against the matching combination.
Set up the arrangement
order mattersRepetition
Permutations · nPr
5P3 =60
60 ordered arrangements of 3 from 5.
5C3 (order ignored)
10
with repetition · nʳ
125
Updates as you type
Try an example
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The basics
What is a permutation?
A permutation is an ordered arrangement. The moment the sequence matters — first place versus second, the order of letters in a code — you are counting permutations. The whole idea rests on two pieces: the factorial, and the rule that order makes two otherwise identical selections count as different.
nPr = n! / (n − r)!
Order matters
You have n choices for the first slot, one fewer for the second, and so on for r slots. Multiplying those falling factors gives the count — the tail of the factorial cancels cleanly.
5P3 = 5 × 4 × 3 = 60
n! = n × (n − 1) × … × 1
The factorial
The factorial of n is the product of every whole number from 1 up to it. It counts the orderings of a whole set, and every permutation formula is built from it.
5! = 5 × 4 × 3 × 2 × 1 = 120
Method
How to calculate nPr by hand
You never need the giant factorials in full — the bottom cancels the tail of the top, leaving a short product.
- 1
Write the formula
Place n on top and n − r in the denominator.
5P3 = 5! / (5 − 3)! = 5! / 2! - 2
Cancel the tail
The 2! at the bottom cancels the bottom of 5!, so only the top r factors survive.
5! / 2! = 5 × 4 × 3 - 3
Multiply the surviving factors
Multiply the r descending numbers together to get the number of arrangements.
5 × 4 × 3 = 60
Worked examples
See it across the common cases
The same counting logic covers ordinary selections, full arrangements, repeated slots, and words with repeated letters.
Pick 3 runners from 5 for gold, silver, bronze.
Given 5P3 = 5! / (5 − 3)! = 5! / 2!
Keep the top 3 factors: 5 × 4 × 3
Multiply: 5 × 4 × 3 = 60
5P3 = 60
Compare
Permutation vs combination
Both pick r from n. The only question is whether reordering creates a new outcome — which is exactly the factor of r! between them.
| Permutation (nPr) | Combination (nCr) | |
|---|---|---|
| Order | Matters — ABC ≠ ACB | Ignored — ABC = ACB |
| Formula | n! / (n − r)! | n! / (r! (n − r)!) |
| Relationship | nPr = nCr × r! | nCr = nPr / r! |
| Count for 5, 3 | 60 | 10 |
| Typical use | Rankings, codes, seating | Teams, hands, subsets |
Where it shows up
Why permutations matter
Anywhere a sequence carries meaning, the count of arrangements is a permutation.
races & rankings
Standings
Assigning gold, silver, and bronze from a field of runners is an ordered selection — swapping two finishers is a different result.
PINs & passwords
Security
Counting how many codes a lock or PIN allows is a permutation with repetition, which is what makes brute-force attacks scale so fast.
schedules & seating
Logistics
Ordering tasks on a timeline or guests around a table are arrangements where position changes the outcome.
Worth knowing
Rules and edge cases
A handful of identities make permutation problems quick to sanity-check.
nP0 = 1
Choosing nothing
There is exactly one way to arrange an empty selection — do nothing. The formula returns 1, never 0.
nP1 = n
Choosing one
Picking a single item from n just gives n choices, with nothing to order.
nPn = n!
Arranging everything
When r equals n, the denominator is 0! = 1, so the count collapses to the full factorial.
nPr = nCr × r!
Bridge to combinations
Every combination fans out into r! ordered permutations, which is the exact gap between the two counts.
r > n → 0
Too many chosen
You cannot arrange more distinct items than exist, so an ordinary permutation with r larger than n is impossible.
repeats → divide
Repeated items
When items repeat, divide by the factorial of each repeat count so identical rearrangements are not double-counted.
Quick chart
nPr for every r at a glance
Pick a value of n and read off the number of ordered arrangements for every r from 0 up to n.
Total items n =
Every nPr from r = 0 up to r = n (exact integers up to n = 18).8P0 = 1
1
8P1 = n
8
8P2
56
8P3
336
8P4
1,680
8P5
6,720
8P6
20,160
8P7
40,320
8P8 = n!
40,320
FAQ
Common questions
A permutation is an arrangement of items in which the order matters. Choosing r items from a set of n and lining them up in a specific sequence is a permutation — so ABC and CBA use the same letters but count as two different permutations.
The number of permutations of r items chosen from n is nPr = n! / (n − r)!, where n! (n factorial) is the product of every whole number from 1 up to n. In practice you just multiply the top r descending factors: n × (n − 1) × … × (n − r + 1).
In a permutation order matters, so ABC and ACB are counted separately. In a combination order does not matter, so ABC and ACB are the same selection. There are always at least as many permutations as combinations: nPr = nCr × r!.
When each position can reuse the same items (like digits in a PIN), every one of the r slots has all n choices, so the count is n^r instead of n! / (n − r)!. A 4-digit PIN from 10 digits gives 10^4 = 10,000 codes.
nP0 = 1, because there is exactly one way to arrange nothing — the empty arrangement. nP1 = n, since you simply pick one of the n items. nPn = n!, because arranging all n items in order is just the number of orderings of the whole set.
When some items repeat, divide the total factorial by the factorial of each repeat count: total! / (n1! · n2! · …). LEVEL has 5 letters with L×2 and E×2, so it has 5! / (2! · 2!) = 30 distinct arrangements rather than 120.
Not for ordinary permutations: you cannot place more distinct items than you have, so nPr = 0 (or undefined) when r > n. It is only possible when repetition is allowed, where each slot independently has n choices and the count n^r keeps growing.
Permutation counts grow explosively. Factorials stay finite in standard 64-bit arithmetic only up to 170!, and exact whole-number results hold to about 18!. Beyond that the tool shows an approximation in scientific notation instead of an exact integer.
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Ugo CharlesBuilt and fact-checked against our editorial standards · Last updated June 2026
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