About the Permutation & Combination Calculator
This permutation and combination calculator counts the number of ways to choose or arrange items, the core of combinatorics and probability. All five results appear together; tap one to see the formula, the numbers substituted in and every simplification step.
It uses exact integer arithmetic (BigInt), so even huge values such as 100! — a 158-digit number — are shown with every digit, alongside scientific notation (9.3326 × 10¹⁵⁷).
How to use
- Enter n, the total number of distinct items, and r, how many you choose or arrange.
- Permutations, combinations, both “with repetition” variants and the factorial appear instantly.
- Tap a result to see its exact value, digit count, scientific notation and the step-by-step solution.
- Use “Copy value” to copy every digit of a large result.
Permutation and combination formulas
| Type | Notation | Formula | Example (n=5, r=2) |
|---|---|---|---|
| Permutations | P(n, r), nPr | n! / (n−r)! | 5 × 4 = 20 |
| Combinations | C(n, r), nCr, “n choose r” | n! / (r!(n−r)!) | 20 / 2 = 10 |
| Permutations with repetition | nʳ | nʳ | 5² = 25 |
| Combinations with repetition | C(n+r−1, r) | (n+r−1)! / (r!(n−1)!) | C(6, 2) = 15 |
| Factorial | n! | n × (n−1) × … × 1 | 5! = 120 |
Permutation or combination?
Ask whether order matters. Electing a president and a vice-president from 10 people is a permutation (P(10, 2) = 90) because the roles differ; picking a 2-person committee is a combination (C(10, 2) = 45).
If an item can be used more than once, you need the “with repetition” version: a 4-digit code using only the digits 1–3 has 3⁴ = 81 possibilities, and buying 2 pieces of fruit from apples, pears and oranges (two of the same allowed) gives C(4, 2) = 6 options.
Useful values
- Poker hands (5 cards from 52): C(52, 5) = 2,598,960
- Powerball white balls: C(69, 5) = 11,238,513
- 10! = 3,628,800; 20! = 2,432,902,008,176,640,000
- 0! = 1, C(n, 0) = C(n, n) = 1, C(n, r) = C(n, n − r)
FAQ
What happens if r is bigger than n?
For permutations and combinations without repetition you need r different items, so the answer is 0. With repetition, r can be larger than n.
Why is 0! equal to 1?
There is exactly one way to arrange zero items (do nothing), and the rule n! = n × (n − 1)! only works for n = 1 if 0! = 1. That is also why C(n, 0) = C(n, n) = 1.
How large can the numbers be?
n and r can each go up to 1000. 1000! has 2,568 digits, and the calculator gives every one of them exactly.
Why is the number of combinations with repetition C(n + r − 1, r)?
Line up r stars (the chosen items) and n − 1 bars (dividers between the n types). Every arrangement is one selection, and you only need to choose which r of the n + r − 1 positions are stars.