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I would like to distribute balls into bins. The balls are indistinguishable in their own color group and the bins are distinguishable and have the same minimum capacity. Here I read that we can take each color as a separate setup and multiply the number of combinations afterwards (please correct me if I'm wrong).

So we should be left with the problem of distributing $n$ indistinguishable balls into $m$ distinguishable bins where each bin has the same minimum capacity of $c$. Another constraint is that every ball has to be distributed to the bins.

I found this question which should be the exact same problem. But when I try out the answer on a simple example I don't get the expected result.

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oezguensi
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    Please give an example of how the algorithm in the other question gives an unexpected result. – almagest Jan 03 '20 at 08:01
  • Let's have 2 bins and 4 balls. Each bin has minimum of 2 balls. So the result should be (4 - 2 * 2 + 2 - 1) choose (2 - 1) which equals to one. I would have anticipated an ordering of the balls such as 1 2 | 3 4; 1 3 | 2 4; 1 4 | 2 3; 2 3 | 1 4; 3 4 | 1 2. I may have missed some, but those are way over 1 possible solution. – oezguensi Jan 03 '20 at 11:00
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    The solution to the referenced question is correct. For 2 bins and 4 balls with a min of 2 balls for each bin, there is obviously only one solution, namely 2 balls in each bin. You have treated the balls as distinguishable. All the solutions in your comment above are the same single solution. – almagest Jan 03 '20 at 16:14
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    On a separate point. If you have several colours, you can only treat each colour separately if you have a min for each colour. If you simply require say 2 balls min per bin, irrespective of colour, then you cannot just multiply the answers for the separate colours. – almagest Jan 03 '20 at 16:15
  • @almagest Thank you, the balls should be distinguishable. That was the problem. Also, sorry for not clarifying. For each color the minimum capacity is one. So in that case I should be able to separate the colors in distinct problems, right? – oezguensi Jan 04 '20 at 02:27

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