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Consider the problem of counting the number of ways one can distribute $k$ identical objects between $g$ distinct groups, such that no group has more than $c$ objects. The solution of this counting problem is given in many answers (e.g this). The answer is:

$$\sum_{r \geq 0} (-1)^r {g \choose r} {g+k-r (c+1)-1 \choose g-1}$$ Now according to this website, the solution can also be approximated by a normal distribution in the limit of large $g$. I'm not really satisfied with the argument on the website about the reason for this approximation. Is there perhaps an explicit way of showing it? Preferably without even needing to derive the exact solution in the first place.

Sahand Tabatabaei
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  • Can you be more explicit about what you found unsatisfactory? E.g., in the special case of $c=1$, the approximation is equivalent to saying that binomial ${g \choose k} \to$ normal, which I think is very commonly accepted. The case of $c>1$ (and all bin capacities being equal, as in your situation) is really not much different. – antkam Sep 20 '19 at 12:41

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We can restate the "rescaled normal" statement as a probabilistic one.

Alternate Version: Consider an assignment of objects and bins chosen randomly from all assignments satisfying the capacity assumption. Then the total number of objects in this assignment approximately follows a normal distribution.

This way of restating things has a couple nice properties: For one, there's a sum hiding in the background: We're adding up the number of objects in each bin. Furthermore, the commands are independent: The number of objects in bin $i$ is uniform on $\{0,1,\dots c_i\}$ regardless of what happens in the other bins. So as long as the $c_i$ are close enough so that the sum isn't dominated by just a few terms, the Central Limit Theorem will kick in for the sum.

Kevin P. Costello
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