<s>[INST] How many ways $5$ identical green balls and $6$ identical red balls can be arranged into $3$ distinct boxes such that no box is empty? 
How many ways $5$ identical green balls and $6$ identical red balls can be arranged into $3$ distinct boxes such  that no box is empty?

My attempt :
Finding coefficient of $x^{11}$ in the expansion of $$( x + x^2 + x^3 + x^4 + x^5+x^6 )^3 ( x + x^2 + x^3 + x^4 + x^5 )^ 3$$  and arranging them which was wrong when inspected
Please help me out
 [/INST] When we neglect the nonempty condition we can distribute the $5$ green balls in ${5+2\choose2}=21$ ways into the three distinct boxes, and independently of this the $6$ red balls in ${6+2\choose2}=28$ ways. This gives $21\cdot28=588$ possible distributions.
In the same way we can compute the number of distributions where the third box has to remain empty. This gives a total of ${5+1\choose1}\cdot{6+1\choose1}=42$ distributions, and the same number arises when another box has to remain empty.
The total number $N$ of admissible distributions therefore is given by
$$N=588-3\cdot 42+3=465\ .$$
At the end we have added $3$, because the three distributions where two boxes remain empty have been subtracted twice in the $3\cdot42$ term.
</s>