MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
In other words, this is not far from the definition of the Stirling number of the second kind. See equation 11 in Set x=k and divide both sides by k^n. – Douglas Zare Feb 21 '10 at 19:35
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.