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I want to put n indistinguishable balls in 12 distinguishable boxes, we number the boxes from 1-12. How many ways are there to distribute the balls if boxes except 1-4 get at least 1 ball and boxes 1-4 get at least 2 balls?

I understand that we can use stars and bars method to get the ways to distribute n ball in 12 boxes, which would be $\binom{n+12-1}{12-1} = \binom{n+11}{11}$. But how should I approach the conditions when there should be at least 1 balls and 4 boxes should have at least 2 balls?

Jon Shaw
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    I suggest that you study this answer, by first skimming it, and paying particular attention to Addendum-2, at the end of the answer. The answer was originally intended to deal with upper bounds on the variables. I added Addendum-2 to also show how lower bounds on the variables should be handled. – user2661923 Feb 23 '23 at 01:32

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So you know how $16$ of your indistinguishable balls have to be distributed. You care about the distribution of the "extras."

$y_1=x_1-2, \ldots, y_4=x_4-2, y_5=x_5-1, \ldots, y_{12}=x_{12}-1, m=n-16$.

Solve for $\underset{i=1}{\overset {12}{\sum}}y_i=m$ using stars-and-bars. There are $\binom{n-16+11}{11}=\binom{n-5}{11}$ solutions.

Robert Shore
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