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So I've been able to find answers on here as to how to distribute n identical objects between r parties by applying the following formula:

C( n+r-1, r-1)

However, this is only for the case where each of the parties can have x objects, where 0 ≤ x ≤ n.

How does this formula change when there is a minimum requirement for each party?


For example, I have 10 identical balls to distribute to 4 unique students. Each student must receive at least 1 ball. How many ways can I distribute the balls?

How would I solve that?

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    $\begingroup$ This is equivalent (since you can't distinguish the balls) to randomly distributing $6$ to the students after giving each one ball. $\endgroup$ – AlexR Feb 2 '15 at 21:34
  • $\begingroup$ @AlexR Oh that makes so much sense :) $\endgroup$ – David Feb 2 '15 at 21:35
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The standard way to handle this kind of problem is to think in terms of "stars and bars" for deriving binomial coefficient answers to these counting problems. The standard derivation is that you have $r$ numbers that must sum to $n$ and they are all positive. Then if you think about having $n$ stars in a row, and you can place $r-1$ bars in between them in distinct locations to get a solution, then there are ${{n-1} \choose {r-1}}$ ways. If you have different requirements on the numbers, e.g. each one has a distinct minimum, then you subtract 1 less than each minimum from $n$ and that gives you a bijection between the original problem with positive numbers and your modified problem with minimums.

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  • $\begingroup$ Would you mind walking me through the example problem then? If I have 10 identical balls to distribute between 4 unique students? So I would be subtracting just 1 from n in this case, making the answer C(n-1, r-1) = C(9, 3), but that would increase to, say 5, if there was a 2 ball minimum? Making the answer then be C(n-5, r-1) = C(5, 3)? $\endgroup$ – David Feb 2 '15 at 21:35
  • $\begingroup$ Ah yes that's correct! Hooray! Thanks for your help $\endgroup$ – David Feb 2 '15 at 21:43

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