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There are $n$ balls to be distributed between $k$ bins. Balls of the same color are considered the same. How many distributions are there is there are $r$ red balls and $n-r$ blue balls?

I believe the problem should reduce to the number of ways of distributing $r$ red balls among $k$ bins, which is $r+k-1 \choose k-1$, multiplied by the number of ways of distributing $n-r$ blue balls among $k$ bins, which is $n-r+k-1 \choose k-1$.

Is this reduction justified?

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Yes, for each distribution of red balls you can have every distribution of blue balls. – Ross Millikan Apr 7 '12 at 14:34

To get this off the Unanswered list:

Yes, your calculation is entirely correct. The distribution of the $n-r$ blue balls is independent of the distribution of the $r$ red balls.

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