Enter the number of objects (n) and sample size (r) to calculate combinations with repetition. The calculator applies the formula C'(n,r) = (r+n−1)! / [r!(n−1)!] and returns the total number of combinations where items can be chosen more than once. You also get a full list of all possible combinations in the results table below.
Combinations with Repetition C'(n,r)
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n + r − 1
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Numerator (n+r−1)!
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Denominator r!(n−1)!
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Combinations with repetition (also called multiset combinations or combinations with replacement) count the number of ways to choose r items from n types when each type can be chosen more than once and the order does not matter. For example, picking 2 scoops from 3 ice-cream flavors where you can pick the same flavor twice.
The formula is C'(n,r) = (r + n − 1)! / [r! × (n − 1)!]. Here n is the number of distinct object types and r is how many you are selecting. It is essentially the standard combinations formula C(n+r−1, r) applied to the expanded pool.
In regular combinations C(n,r), each object can be chosen at most once. In combinations with repetition C'(n,r), the same object type can appear multiple times in a selection. Regular combinations use n! / [r!(n−r)!], while combinations with repetition use (n+r−1)! / [r!(n−1)!].
It depends on how many you are choosing. If you choose r = 2 from n = 5 numbers with repetition, C'(5,2) = 6! / (2! × 4!) = 15. If you choose r = 3, C'(5,3) = 7! / (3! × 4!) = 35. Enter your specific values above to get the exact count.
1) Identify n (number of object types) and r (selection size). 2) Compute n + r − 1. 3) Calculate (n+r−1)! for the numerator. 4) Calculate r! × (n−1)! for the denominator. 5) Divide numerator by denominator. The result is C'(n,r).
Suppose you have 3 fruit types (apple, banana, cherry) and want to pick 2 with repetition. C'(3,2) = (2+3−1)! / (2! × 2!) = 4! / (2 × 2) = 24 / 4 = 6. The six selections are: {apple,apple}, {apple,banana}, {apple,cherry}, {banana,banana}, {banana,cherry}, {cherry,cherry}.
This calculator supports n and r up to 20 each. Beyond that, the factorial values become extremely large and the number of individual combinations in the list grows rapidly. For theoretical work with very large values, you can apply the formula directly using logarithms.
No. In all combination problems — with or without repetition — the order of selection does not matter. {apple, banana} is the same combination as {banana, apple}. If order matters, you would instead use permutations with repetition, which gives n^r possible arrangements.