<s>[INST] Dividing balls into boxes with capacity limits Problem:
In how many ways can you divide $13$ identical balls into $3$ different boxes
$k_1$, $k_2$, $k_3$, such that $k_1$ contains no more than $5$ balls, $k_2$ contains no more than $6$ balls and $k_3$ contains no more than $4$ balls?
My idea: 
So my idea is to use the following theorem:
"There are $C(n+r-1,r)$ r-combinations from a set with $n$ elements when repetition of elements is allowed." But I'm not sure.
Any great hints would be appreciated.
 [/INST] The answer can simply be given using generating functions. We simply need to calculate the coefficient of $x^{13}$in the expansion of $$(1+x+x^2+x^3+x^4)(1+x+x^2+x^3+x^4+x^5)(1+x+x^2+x^3+x^6)$$
$$=\frac {(1-x^5)}{(1-x)}\cdot  \frac {(1-x^6)}{(1-x)}\cdot  \frac {(1-x^7)}{(1-x)}$$
$$=\frac {1-x^5-x^6-x^7+x^{11}+x^{13}+x^{12}-x^{18}}{(1-x)^3}$$
Hence the answer simply goes as $$\binom {15}{13} -\binom {10}{8}- \binom {9}{7}-\binom {8}{6}+\binom {4}{2}+1+\binom {3}{1}=6$$
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