Permutations with replacement count ordered arrangements where the same item can be reused, like PIN codes or license plates, a fundamentally different (and usually much larger) count than permutations without repetition.

How to use this calculator

1. Enter n (the number of available items or choices per position).
2. Enter r (the number of positions or selections).
3. Read the calculated number of permutations with replacement.

What this calculator does

Permutation with replacement counts the number of ordered arrangements of r selections from n items, where each selection is made independently and repetition is allowed, unlike a standard permutation, an item chosen once can be chosen again in a later position.

Why this formula is just exponentiation

Since each of the r positions independently has n possible choices (repetition being allowed means earlier choices don’t reduce the options for later positions), the total count is simply n multiplied by itself r times, n^r, a much simpler formula than the factorial-based permutation-without-replacement case, precisely because nothing gets “used up” as selections are made.

Where this shows up in practice

A 4-digit PIN where digits can repeat (like 1111 or 1234) is a classic permutation-with-replacement problem, 10 possible digits for each of 4 positions gives 10^4 = 10,000 possible PINs. The same logic applies to license plates, product codes, and any scenario where positions are filled independently from a fixed set of options with repetition allowed.

Frequently asked questions