Another balls and bins problem, but I couldn't find one like this after browsing a while.

Say I have $N$ balls of $R$ different colors (N/R balls of each color) and I need to put them into $R$ different bins so that each bin have an equal number of ball (so $N$ is divisible by $R$). How many ways are there to distribute the balls?

edit: Assume that $N > R$ , for example, if we want to put $36$ balls of $4$ different colors (9 balls of each color) into $4$ bins so that each bin always has $9$ balls, how many ways are there to do it?

  • 2
    $\begingroup$ Surely this will depend on the colour breakdown of the $N$ balls, i.e. how many balls there are of each colour. $\endgroup$ – Sarastro Dec 4 '12 at 7:11
  • $\begingroup$ To follow on the comment by @Sarastro, maybe we should assume there are $\frac NR$ balls of each color? $\endgroup$ – Marc van Leeuwen Dec 4 '12 at 7:28
  • $\begingroup$ Yes you guys are right, I've updated the original question to reflect that. Thank you! $\endgroup$ – Clarence Huang Dec 4 '12 at 7:33

We have $M=N/R$ balls of each color, from $R$ colors. Assuming all the balls of same colours are not distinguishable, but the bins are, each arrangement is specified by the non-negative integers $t_{i,j}=$ "numbers of balls of color $i$ in bin $j$", with the restrictions $\sum_i t_{i,j} = M$ ($M$ balls in each bin) and $\sum_j t_{i,j} = M$ ($M$ balls of each color).

That is, we need to count the number of $R \times R$ matrices (with non-negative integer entries: "contigency tables") with all rows and columns summing $M$. This a well studied problem, and far from trivial. Closed forms are known for small $R$. See eg:


http://arxiv.org/pdf/math/9806076v1.pdf (section 6)



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