For n identical candies shared among k distinct children, with zero allowed and no caps, the number of distributions is binomial(n + k − 1, k − 1). This stars-and-bars formula counts allocation patterns directly. If every child must get candy or some children have minimums or maximums, adjust the model before calculating; those conditions change what counts as valid.
Define what counts as a distribution
Let xi be the number of candies received by child i. If all candies must be distributed, the total is described by x1 + x2 + ··· + xk = n.
The basic formula applies when candies are identical, children are distinct, each child may receive zero, and there are no capacity limits. For example, “Alice gets 3, Ben gets 4, and Cy gets 3” is a different distribution from “Alice gets 4, Ben gets 3, and Cy gets 3,” because the recipients are named.
- Identical candies: only the number each child receives matters; individual candy identities do not.
- Distinct children: changing which child gets a given amount creates a different allocation.
- All candies distributed: the amounts sum to n.
- Zero allowed: a child may receive nothing unless the question sets a minimum.
- No caps: there is no upper limit on any child’s amount.
If candies are individually distinguishable, children are interchangeable, or some candies may remain undistributed, this is a different counting model; the formulas below do not apply unchanged.
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Count unrestricted distributions with stars and bars
For nonnegative integer solutions to x1 + ··· + xk = n, the count is:
C(n + k − 1, k − 1)
Imagine writing one star for each candy and inserting k − 1 dividers to separate the children’s shares. For three children, for instance, a pattern such as **|***|* represents 2 candies, 3 candies, and 1 candy. Adjacent dividers or a divider at an end represent an empty share. There are n stars and k − 1 dividers, so each arrangement occupies n + k − 1 positions; choosing which positions contain dividers gives the formula. Each pattern corresponds to exactly one allocation vector, so the method counts distributions without listing them one by one. Richard Hammack presents the same representation in Book of Proof.
Example: 10 identical candies for 3 children, with zero allowed
Here n = 10 and k = 3, so the count is C(10 + 3 − 1, 3 − 1) = C(12, 2) = 66. This is the worked result in Xiaohui Xie’s stars-and-bars notes.
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Example: 10 identical candies for 4 children, with zero allowed
The count is C(10 + 4 − 1, 4 − 1) = C(13, 3) = 286, the result shown in Fall 2025 combinatorics course notes.
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For a positive share by every child, first give each child one candy. That uses k candies and leaves n − k to distribute freely. If n ≥ k, the number of allocations is:
C(n − 1, k − 1)
If n < k, no valid distribution exists because there are not enough candies to give one to each child.
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Example: 10 identical candies for 3 children, at least one each
Reserve 1 candy for each child, leaving 7. The remaining nonnegative allocation count is C(7 + 3 − 1, 3 − 1) = C(9, 2) = 36, matching Xie’s worked example.
Handle different minimums by shifting the variables
If child i must receive at least ai candies, write xi = ai + yi, where each yi is nonnegative. The amount left to allocate is n − Σai. When that remainder is nonnegative, count the allocations as:
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When n is less than the sum of the minimums, there are no valid allocations. For example, if two children must receive at least 1 and 2 candies, respectively, from a total of 5, reserve 3 candies first. The remaining 2 can be split nonnegatively between the two, giving C(2 + 2 − 1, 2 − 1) = 3 possibilities. This shifting approach is also illustrated in the Fall 2025 course notes.
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Apply inclusion-exclusion when children have caps
The unrestricted formula includes allocations that exceed a child’s capacity. To impose upper bounds, start with the unrestricted count and remove allocations that violate one or more caps using inclusion-exclusion. For a child with cap m, a violation means xi ≥ m + 1. In a set of violations, subtract m + 1 from that child’s amount, and count the remaining nonnegative solutions. For different caps, use each child’s own threshold.
For a set S of children assumed to exceed their caps, the shifted total is n − Σi∈S(mi + 1). If that total is nonnegative, the number of allocations in that intersection is C(n − Σi∈S(mi + 1) + k − 1, k − 1); otherwise it is zero. Subtract the counts for single violations, add back pairwise intersections, subtract triple intersections, and continue with alternating signs. This corrects for allocations counted in more than one violation set.
Xie’s notes illustrate the method with ordered triples totaling 15 under caps of 5, 6, and 7, obtaining 10 after inclusion-exclusion. That is a bounded-composition example, not a result for a candy setup unless those exact totals and caps are used.
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Choose the right model before calculating
| Question to settle | Model or consequence |
|---|---|
| Are the candies identical or individually distinguishable? | Stars and bars counts only identical items by recipient totals. Distinguishable candies require a different count. |
| Are children distinct or interchangeable? | The formulas here treat children as distinct categories; swapping their amounts can produce a different allocation. |
| May a child receive zero? | Use nonnegative solutions if yes; use the positive-allocation formula if every child needs at least one. |
| Are there minimums? | Reserve each required minimum, then distribute what remains. |
| Are there caps? | Correct the unrestricted count for over-cap allocations, for example with inclusion-exclusion. |
| Must every candy be distributed? | The equation assumes the shares sum to the full candy total. |
Keep the wording of the question attached to the result: “10 identical candies to 3 distinct children, zero allowed” has 66 distributions, while requiring at least one per child gives 36. Without a specified candy total, number of children, and conditions on valid shares, there is no single numerical answer.
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