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My question might be a bit poorly articulated as I am not sure what I'm asking is actually called.

I am faced with an Exclusion/Inclusion problem that goes like this:

You have $25$ identical cakes that are distrubuted to $10$ children, every child must have atleast $1$ cake but no more than $4$.

After a bit of calculating you wind up with something like this:

$x_1 + x_2 +x_3 + x_4... + x_{10} = 15$ (removing the first argument of 1 cake each).

You wind up with your first number which in this case would be $(24,15)$. After that it seems that you remove 4 from this number $((20,11), (16,7)$ and finally $(12,3))$.

What I don't get is how you determine the number of which do exclude from here. My thought process would be that if each child gets 1 cake each, we then exclude the posibilites of children getting 3 more cakes, which would correspond with the argument of a lower bound of 1 and a highest bound of 4 cakes.

I hope this question makes sense and that someone out there can help me.

Thanks!

N. F. Taussig
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user246914
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    See related problem here. You will need to count how many ways no child is predetermined to violate the condition minus the number of ways one child is predetermined to violate the condition plus the number of ways two children are predetermined to violate the condition minus... for a total of $\binom{15+10-1}{10-1} - \binom{10}{1}\binom{11+10-1}{10-1}+\binom{10}{2}\binom{7+10-1}{10-1} -\binom{10}{3}\binom{3+10-1}{10-1}+\binom{10}{4}\binom{-1+10-1}{10-1}$... – JMoravitz Jun 09 '15 at 13:38
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    This answer and the first part of this answer go into some detail about exactly that point. – Brian M. Scott Jun 09 '15 at 16:14

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