Combinations with Repetition Calculator
Count combinations with repetition (multisets): ways to choose r items from n types when repeats are allowed and order does not matter.
Count the ways to choose r items from n types when you may pick the same type more than once and the order does not matter — scoops from a list of flavours, coins from a set of denominations, donuts from a bakery tray. This count is also called the number of multisets, or “n multichoose r”. For small cases the calculator lists every selection.
How to use this tool
- Enter n, the number of different types available.
- Enter r, how many items you choose in total. It can be larger than n.
- Read the number of combinations with repetition, with the matching counts without repetition and with order mattering for comparison.
Formula
C(n + r − 1, r) = (n + r − 1)! ÷ (r! × (n − 1)!)
The “stars and bars” argument explains it: line up r stars (the items) and n − 1 bars (dividers between the types). Every arrangement of those n + r − 1 symbols is exactly one selection, and choosing which r positions hold stars gives the formula.
Example
Choosing 3 scoops from 5 flavours, repeats allowed: C(5 + 3 − 1, 3) = C(7, 3) = 7! ÷ (3! × 4!) = 5040 ÷ 144 = 35 combinations. Without repeats there would be only C(5, 3) = 10, and if the order of the scoops mattered there would be 5³ = 125.
Frequently asked questions
How is this different from permutations with repetition?
Permutations with repetition count ordered sequences, nʳ — vanilla-then-mint differs from mint-then-vanilla. Combinations with repetition count only which items you end up with, so those two are the same selection.
Can r be bigger than n?
Yes, because types repeat. Choosing 10 donuts from 3 kinds gives C(12, 10) = 66 possible boxes.
What if every type must appear at least once?
Take one of each type first, then choose the remaining r − n freely: C(r − 1, n − 1). For 10 donuts from 3 kinds with at least one of each kind, that is C(9, 2) = 36.
Last updated: 2026-10-30