How many ways can you choose k items from n types when repetition is allowed and order does not matter? Get the result using the formula C(n+k-1, k) with a side-by-side comparison to standard combinations.
Combinations with repetition from 10 elements
How many ways can you choose k items from n types when repetition is allowed and order does not matter? Get the result using the formula C(n+k-1, k) with a side-by-side comparison to standard combinations.
The 10 setting works from different assumptions than the ones next to it. Stars and bars turns the question into arranging dividers, which is why the answer is a plain binomial coefficient in disguise. This page opens the calculator with 10 already selected, so only the remaining fields are left to fill in. Swap in your own figures and the arithmetic is rebuilt from them.
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Five ice cream flavors, three scoops, repeats allowed - how many options?
You walk into an ice cream shop with 5 flavors and order 3 scoops. You can pick the same flavor more than once, and nobody cares about the order on the cone. Chocolate-chocolate-vanilla is the same selection as vanilla-chocolate-chocolate. How many distinct combinations exist? The answer is 35, and you get it from the formula C(n+k-1, k). This calculator does the arithmetic for you and compares the result with standard combinations (no repeats) so you can see exactly how much repetition expands the count.
Result: 35 combinations with repetition vs. only 10 without repetition.
Allowing repeats multiplied the options by 3.5.
When does this formula apply?
Use combinations with repetition whenever you pick k items from n types, repetition is allowed, and order does not matter. The technical term is a multiset of size k from a set of n types. The formula is C(n+k-1, k) = (n+k-1)! / (k! (n-1)!). It is sometimes written with double parentheses as ((n choose k)).
| Scenario | n | k | C(n+k-1, k) | C(n, k) no rep. |
|---|---|---|---|---|
| 3 scoops from 5 flavors | 5 | 3 | 35 | 10 |
| 2 dice (unordered result) | 6 | 2 | 21 | 15 |
| 10 candies among 4 kids | 4 | 10 | 286 | 0 (k>n) |
| 5 coins from 4 denominations | 4 | 5 | 56 | 0 (k>n) |
| Quadratic in 3 variables | 3 | 2 | 6 | 3 |
The four counting methods side by side
Combinatorics has four fundamental counting formulas depending on two questions: does order matter, and is repetition allowed?
| Method | Formula | Order? | Repetition? |
|---|---|---|---|
| Combination | C(n, k) = n! / (k!(n-k)!) | No | No |
| Combination with rep. | C(n+k-1, k) | No | Yes |
| Permutation | P(n, k) = n! / (n-k)! | Yes | No |
| Variation with rep. | n^k | Yes | Yes |
Stars and bars: the intuition behind the formula
Picture k identical stars and (n-1) dividers arranged in a row. Each arrangement maps to exactly one multiset. With 3 types and 4 items:
**|*|*= 2 of type A, 1 of type B, 1 of type C****||= 4 of type A, 0 of type B, 0 of type C|**|**= 0 of type A, 2 of type B, 2 of type C
Total symbols: k + (n-1). Choose k positions for stars out of k + n - 1 total positions: that gives C(n+k-1, k). This bijection between multisets and star-bar strings is why the formula works.
Worked examples
Hand out 10 identical candies to 4 children (each can get zero or more). Here n = 4, k = 10. C(13, 10) = C(13, 3) = 286 ways. Without repetition: impossible (k > n), which makes sense because you cannot give each child at most one candy when there are 10 candies and only 4 kids.
Two dice land. You only care about the set of values, not which die shows which. n = 6 faces, k = 2 dice. C(7, 2) = 21 distinct unordered outcomes. For comparison, there are 36 ordered outcomes (6 x 6), but many are duplicates when order is ignored.
A polynomial of degree d in n variables has C(n+d-1, d) distinct terms. A quadratic (d=2) in 3 variables (x, y, z) has C(4, 2) = 6 terms: x2, y2, z2, xy, xz, yz. A cubic in 4 variables has C(6, 3) = 20 terms.
A store sells 8 types of fruit. You want to buy exactly 4 pieces, and you may buy multiple of the same type. C(11, 4) = 330 possible baskets. Without repetition you could only make C(8, 4) = 70 baskets.
Choose 20 items from 50 types with repetition: C(69, 20) = 7,886,597,962,249,166 (nearly 8 quadrillion). Without repetition: C(50, 20) = 47,129,212,243,960, about 167 times fewer. Repetition has a massive multiplying effect when k is large relative to n.
Common questions
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Reviewed by: Patryk Matyjasik