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I'm trying to figure out how to get all combinations of putting x*n balls into n boxes.

There should be the same number of balls in each box.

If it was about putting n balls into n boxes, the result would be n! and the pseudocode:

for b1 in set:
   for b2 in set-b2:
      for b3 in set-b2-b3:
          results.append(b1,b2,b3,b4)

How can I get this way combinations of x*n instead of n balls so in one box there is x balls exactly?

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  • $\begingroup$ This is a math question and belongs to math.se. The answer is $(xn)!/x!^n$. $\endgroup$
    – Yuval Filmus
    Apr 24, 2016 at 10:35
  • $\begingroup$ @YuvalFilmus Counting the number of combinations is math, but an algorithm to generate them is CS. $\endgroup$
    – Gilles 'SO- stop being evil'
    Apr 26, 2016 at 11:54
  • $\begingroup$ Generating them is even easier using the trivial recursive approach. The difficult part is counting the number of solutions. $\endgroup$
    – Yuval Filmus
    Apr 26, 2016 at 11:56

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