When should I use permutations instead of combinations?

Use permutations when position changes the outcome. If ABC and BAC should count separately, order matters and you need a permutation formula.

What does repetition mean here?

Repetition means the same item can be chosen again. A PIN code allows repetition; a podium of distinct winners does not.

Why is r > n blocked in some modes?

Without repetition, you cannot choose more distinct items than exist. If you need more picks than the pool contains, repetition must be allowed.

What happens when r = 0?

There is exactly one way to choose nothing: the empty arrangement or empty selection. That is why the result is 1 when r = 0.

Why do the answers grow so quickly?

Counting problems multiply choices across positions. Even modest values of n and r can produce very large exact integers.

Permutations & Combinations Calculator

Choose whether order matters and whether repetition is allowed, then get the exact permutations or combinations count with the matching formula.

Does order matter?

Allow repetition?

Total items (n)

Chosen items (r)

Examples

Enter values above to see results

How It Works

Formula

P(n,r) = \frac{n!}{(n-r)!}

P_{\text{rep}}(n,r) = n^r

C(n,r) = \frac{n!}{r!(n-r)!}

C_{\text{rep}}(n,r) = \frac{(n+r-1)!}{r!(n-1)!}

Start with the two yes/no decisions. If order matters, use a permutation formula. If order does not matter, use a combination formula. Then decide whether each item can be used once or reused, and the calculator switches to the matching one of the four standard counting cases.

Use the toggle pair as a 2-by-2 decision table:

| Order matters? | Repetition allowed? | Use | | --- | --- | --- | | Yes | No | $P(n,r)=\frac{n!}{(n-r)!}$ | | Yes | Yes | $P_{\text{rep}}(n,r)=n^r$ | | No | No | $C(n,r)=\frac{n!}{r!(n-r)!}$ | | No | Yes | $C_{\text{rep}}(n,r)=\frac{(n+r-1)!}{r!(n-1)!}$ |