Count unordered selections of k items from n types when repeats are allowed. The page traces how the inputs become repeated combinations before rounding.
It models scoops by flavor, nonnegative integer solutions, inventory bundles, and allocations of identical units among labeled groups.
Items of the same type are treated as indistinguishable and order is ignored. Upper limits on individual types invalidate the unrestricted stars-and-bars formula.
The Combinations with Repetition case starts with Number of item types and Selection size. Recalculate Repeated combinations from those entries. A nearby Selection size can challenge the Combinations with Repetition relationship, but its Repeated combinations belongs to a separate Combinations with Repetition record.
Represent the k selected units as stars and place n−1 separators to divide them among the labeled types. Count all positions for stars or separators.
Stars and bars converts a repeated unordered selection into k stars separated among n categories, giving C(n+k−1,k). Combinations with Repetition can also be compared with ordered reuse.
Start the Combinations with Repetition review with Number of item types. Compare Number of item types with its source, then test Selection size in a second Combinations with Repetition run without changing the first Combinations with Repetition case.
Link Number of item types to its Combinations with Repetition role. Link Selection size to its Combinations with Repetition role. The retained Combinations with Repetition formula identifies the Combinations with Repetition model.
For a small Combinations with Repetition example using Number of item types, Selection size, solve once with the displayed formula and once through direct enumeration or a probability tree. Two methods that organize outcomes differently are less likely to share the same hidden assumption error.
For Combinations with Repetition, when the exact list becomes too large, retain a simpler identity as a check: complementary probabilities sum to one, a Pascal row is symmetric, and adjacent factorial results have a known ratio. State which identity was used beside the reported answer.
Selecting seven items from five types with reuse allowed gives C(11,7)=330 possible count profiles. This Combinations with Repetition example can be compared with stars-and-bars coefficient.
Yes.
No.
This unrestricted formula no longer suffices.
It counts k stars and n−1 separators.