Permutations Calculator
Calculate the number of ordered arrangements of r items chosen from n.
Last updated · View algorithm, constants and default inputs (JSON)
About this calculator
Permutations count arrangements where order matters, so picking A then B is different from B then A. That is why there are always more permutations than combinations for the same n and r.
Worked example
calculated with the default values below
Total items (n) 10 · Items arranged (r) 3 Permutations 720 · Combinations 120 · Permutations with repetition 1000 · n factorial 3628800
General background references
Disclaimer
These calculators give general mathematical results. They suit study, planning and estimation — verify anything that feeds a graded submission, a published figure or a financial commitment.
Implementation source
This record describes the implementation used by this page. It is not independent validation. External references, when available, support only the stated formula or definition; estimates and custom rules retain their own assumptions.
Formula and variables
P(N,R) = N! ÷ (N − R)!; C(N,R) = N! ÷ (R! × (N − R)!); P_rep = N^R
- N
- Total items
- R
- Items chosen
- P
- Permutations
- C
- Combinations
- P_rep
- Permutations with repetition
Inputs must be whole numbers from 0 to 100. Without repetition, permutations and combinations are 0 when R > N; permutations with repetition and N! are still calculated. Counts are returned as exact integer text.
Frequently asked questions
Why are permutations always larger than combinations for the same n and r?
Permutations count each ordering separately, while combinations group all orderings of the same r items together. The permutation count is therefore at least as large; the two are equal when r is 0 or 1. For n=10 and r=3 the calculator shows 720 permutations and 120 combinations, a ratio of 3! = 6. With repetition, items may be reused, so the count can remain positive even when r is greater than n.