<s>[INST] Summation over $a+b+c=5$ Let $a,b,c$ be positive integers. Compute
$$\sum_{a+b+c=5} (6-a)(6-b)(6-c).$$

The first thing I notice is symmetry, so that I can permute $3!=6$ ways, but i'm not really sure how that works with the condition $a+b+c=10.$ The other method is to fix $a$, but that is reall time-consuming and unfeasible if say $a+b+c=20.$ Is there a clever method to evaluate this sum?
I would like to have a generalized method, please.
 [/INST] By using generating functions, we show that for any non-negative integer $n$,
$$\begin{align}\sum_{a+b+c=n} (6-a)(6-b)(6-c)
&=[x^n]\left(\sum_{k=1}^{\infty}(6-k)x^k\right)^3\\
&=[x^n]\frac{(x(5-6x))^3}{(1-x)^6}\\
&=[x^n]\frac{125x^3-450x^4+540x^5-216x^6}{(1-x)^6}\\
&=125\binom{n+2}{5}-450\binom{n+1}{5}+540\binom{n}{5}-216\binom{n-1}{5}\\
&=-\frac{(n-2)(n-1)(n^3-87n^2+2072n-12960)}{120}.
\end{align}$$
Therefore for $n=5,10,20$ we obtain $465$, $-36$, $-4788$ respectively.
</s>