I have three natural numbers $a, b, c$ such that $a + b + c = n$ and I'm looking for $\sum abc$.
So far I've figured out that the generating function for $p(n,3)$ might be $\frac{x^3}{(1-x)(1-x^2)(1-x^3)}$ but I'm not sure if I'm correct because none of $a,b,c$ can be zero and I don't know how to approach this problem.
A partition of $n$ items into groups of $a$, $b$ and $c$ items corresponds to selecting $2$ of $n+2$ objects to be partitions walls. Then $abc$ counts the number of ways of selecting one item from each group. Equivalently, we can select $5$ of $n+2$ objects and regard the $2^{nd}$ and $4^{th}$ as partition walls and the $1^{st}$, $3^{rd}$ and $5^{th}$ as selected items. Thus the desired sum is $\binom{n+2}5$.