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I found the following closed form solution for the abovementioned problem:

$${1\over k^n}\cdot{k!\over (k-m)!}\cdot{\{{n\over m}\}}$$ with ${\{{n\over m}\}}$ being the Stirling Number of the second kind.

Although it seems to have some intuition and seems to work for a sample problem for which I have the solution this closed form is not from a trusted source. Unfortunately I can't find any other source.

My question: Could anyone acknowledge this closed form solution and/or give me a hint where to find a citable source.

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There are $k^n$ possible dice configurations, presumably all equally likely. $\frac{k!}{(k-m)!}$ is the number of distinct combinations of $m$ (out of $k$ total) faces. The Stirling number of the second kind is the number of ways to separate n elements into m sets. –  Steve Huntsman Feb 21 '10 at 19:28
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In other words, this is not far from the definition of the Stirling number of the second kind. See equation 11 in mathworld.wolfram.com/StirlingNumberoftheSecondKind.html. Set x=k and divide both sides by k^n. –  Douglas Zare Feb 21 '10 at 19:35
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1 Answer

up vote 5 down vote accepted

Applied probability by Kenneth Lange deals with this problem on page 74. It is on Google books, here is the URL.

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