In how many ways can 40 identical carrots be distributed among 8 different rabbits, while every rabbit needs to get a carrot, and no rabbit get more then 16 carrots.
Thank you for the help!
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You are asking for the number of partitions of $40$ where every term in the partition is less than or equal to $16$. I think there's no known closed-form formula for the number of partitions, so you'll have to simply compute this by systematically trying all possibilities – Feb 01 '16 at 11:14
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1@vrugtehagel This is a combination with repetition rather than a partition. – N. F. Taussig Feb 01 '16 at 12:09
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Suppose rabbit $j$ get $x_j$ carrots. $$ \sum_{j=1}^8 x_j=40 $$ Number of positive solutions less than $17$ of the last equation is coefficient of $z^{40}$ in $$ (z^1+z^2+\cdots+z^{16})^8={z^8(1-z^{16})^8\over(1-z)^8} $$ Or equivalently coefficient of $z^{40}$ in $$ z^8(1-8z^{16}+28z^{32})\sum_{k=0}^{32}\binom{8+k-1}{k}z^{k}\\ =\binom{8+32-1}{32}-8\binom{8+16-1}{16}+28\\ =13419709$$
Jack's wasted life
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This is only a part of a solution.
Give 1 carrot to each rabbit. You have 24 remaining, since the first condition forces you give each rabbit at least one. Next, the problem is: In how many ways can 24 be partitioned 24 so none of the parts is greater then 16.
I have no clue what the answer above is calculating. I only understood that $x_i$ are carrots.
I have opened this Problem here: In how many ways to partition(ordered partition) an natural number n so none of the parts is greater then k?
