<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] $$
\begin{array}{l}
 a + b + c = 5\quad \left| {\,1 \le a,b,c\left( { \le 3} \right)} \right.\quad  \Rightarrow  \\ 
  \Rightarrow \quad \left( {6 - a} \right) + \left( {6 - b} \right) + \left( {6 - c} \right)
 = x + y + z = 13\quad \left| {\,3 \le x,y,z, \le 5} \right.\quad  \Rightarrow  \\ 
  \Rightarrow \quad \left( {x,y,z} \right) \in S = \left\{ {\left( {3,5,5} \right) \vee \left( {4,4,5} \right)
 \vee permut.} \right\}\quad  \Rightarrow \quad  \\ 
  \Rightarrow \quad \sum\limits_{\left\{ {\begin{array}{*{20}c}
   {\,1 \le a,b,c\left( { \le 3} \right)}  \\
   {a + b + c = 5}  \\
\end{array}} \right.} {\left( {6 - a} \right)\left( {6 - b} \right)\left( {6 - c} \right)}  =  \\ 
  = \sum\limits_{\left( {x,y,z} \right) \in S} {xyz}  = 3\left( {3 \cdot 5 \cdot 5} \right)
 + 3\left( {4 \cdot 4 \cdot 5} \right) = 225 + 240 = 465 \\ 
 \end{array}
$$
</s>