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Given that there are $x$ ways to distribute $r$ distinct objects to $n$ distinct boxes (subject to any contraint, for example each box must have at least 1 object), and then we change the objects to be from $k$ classes instead (in which there are $r_1$ identical objects in class 1, and $r_2$ identical objects in class 2,...), then is it true that the way to distribute the $r_1+r_2+...+r_k=r$ items, subject to the same constraints, is simply $\frac{x}{r_1!r_2!...r_k!}$?

The application of this is to solve the problem of distributing, let's say 20 apples, 30 bananas and 40 oranges to 8 children so that each child gets at least one fruit. I know of the inclusion-exclusion method on every child, but I think it would be tedious if the number of children becomes larger than 5.

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  • $\begingroup$ For three or more, just determine the formulas, and let a computer calculate them. $\endgroup$ – Pieter21 Dec 1 '15 at 18:49
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Your formula isn't true. For a counterexample put $r=2$ distinct objects in $n=1$ box. There is only $x=1$ way to do this. If you put both objects in the same class, your formula gives $\frac{1}{2!}=\frac{1}{2}$. The issue is that permutations of objects in the same box don't give different outcomes when counting $x$.

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