<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] This problem is somewhat specific and nice, since when using symmetry it reduces to the summation of two terms.

We obtain
\begin{align*}
&\color{blue}{\sum_{\substack{a+b+c=5\\a,b,c\geq 1}}(6-a)(6-b)(6-c)}\\
&\qquad=6\sum_{\substack{a+b+c=5\\1\leq a<b<c\leq 3}}(6-a)(6-b)(6-b)\tag{1}\\
&\qquad\quad +3\sum_{\substack{a+b+c=5\\1\leq a=b<c\leq 3}}(6-a)^2(6-c)\tag{2}\\
&\qquad\quad +3\sum_{\substack{a+b+c=5\\1\leq a<b=c\leq 3}}(6-a)(6-c)^2\tag{3}\\
&\qquad\quad +\sum_{\substack{a+b+c=5\\1\leq a=b=c\leq 3}}(6-a)^3\tag{4}\\
&\qquad=0+3\cdot5^2\cdot3+3\cdot5\cdot4^2+0\tag{5}\\
&\qquad=9\cdot25+15\cdot16\tag{6}\\
&\,\,\color{blue}{\qquad=465}
\end{align*}

Comment:

*

*In (1) we observe that the conditions $a+b+c=5$ and $a,b,c\geq 1$ imply that each element $a,b,c$ is less or equal $3$.
We sum up all triples $(a,b,c)$ whereby each two elements are pairwise different. This can be done in $3!=6$ ways. After appropriately renaming the variables we can consider therefore $6$ ordered triples $(a,b,c)$ with $1\leq a<b<c\leq 3$.


*In (2) we sum up all triples $(a,b,c)$ whereby $a$ is equal to $b$ and different to $c$. This can be done in $\frac{3!}{2!1!}=3$ different ways.


*In (3) we have a similiar situation as in (2). Here we consider the constellation $a<b=c$ which can be done again in $\frac{3!}{2!1!}=3$ different ways.


*In (4) we cover the final case $a=b=c$ which can be done in $\frac{3!}{3!}=1$ different way.


*In (5) we observe

*

*there is no admissible triple $(a,b,c)$ with $1\leq a<b<c\leq 3$ and $a+b+c=5$.


*there is just one admissible triple with $1\leq a=b<c\leq 3$ and $a+b+c=5$, namely $\color{blue}{(a,b,c)=(1,1,3)}$.


*there is just one admissible triple with $1\leq a<b\leq c\leq 3$ and $a+b+c=5$, namely $\color{blue}{(a,b,c)=(1,2,2)}$.


*there is no admissible triple $(a,b,c)$ with $1\leq a=b=c\leq 3$ and $a+b+c=5$.
Conclusion: In (6) we note that thanks to symmetry only two admissible triples $\color{blue}{(a,b,c)\in\{(1,1,3),(1,2,2)\}}$ provide a contribution to the result.
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