GREENCALC JOURNAL

Combinations with Repetition: Types, Totals and Stars and Bars

Count quantities of each type, not draw order Suppose you choose three scoops from five available flavors, may repeat a flavor and ignore scoop order. An.

Small bowls of colored beads separated by simple wooden dividers.
Combinations with Repetition: Types, Totals and Stars and Bars — conceptual editorial illustration.

Small bowls of colored beads separated by simple wooden dividers. Original AI-generated artwork produced for GreenCalc; not a measured diagram or empirical data display.

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Count quantities of each type, not draw order

Suppose you choose three scoops from five available flavors, may repeat a flavor and ignore scoop order. An outcome is a vector of five nonnegative flavor counts summing to three. Stars and bars represents three selected items as stars and separates the five flavor counts using four bars. Choosing star positions among seven positions gives C(7,3) = 35 multisets.

Apply the general expression

For n distinct types and r selected items, combinations with repetition equal C(n+r−1,r). The Combination Calculator uses exact integer arithmetic for this count. Setting n = 5, r = 3 and repetition to yes gives 35. Without repetition, selecting three distinct flavors would instead give C(5,3) = 10. If order matters, neither unordered result describes the sequence count.

Check capacity constraints

The standard expression permits any type to supply all r items. It does not handle a limit of two scoops per flavor, unavailable inventory or a requirement to choose at least one from particular categories. Those conditions change the allowed vectors. Write the constraint explicitly instead of treating a multiset count as a universal inventory formula.

Do not assume the multisets are equally likely

If three independent uniform draws are made from five flavors, ordered draw sequences are equally likely, but flavor-count multisets are not. Three different flavors can appear in six orders; three identical flavors appear in only one order. Thus a count of 35 does not mean every unordered outcome has probability 1/35 under independent drawing. The mechanism matters as much as the count. Read subset versus sequence counting for the role distinction. For a collision event under independent repeated draws, the Birthday Paradox Calculator answers a different probability question rather than counting all possible multisets.

Check the model behind your result

Use the Combination Calculator for the supported calculation and review its stated assumptions before interpreting the output. Find related tools in the Statistics category, or browse the GreenCalc homepage for other calculation topics.

By GreenCalc Editorial Team · Updated 2026-10-26

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