Find the Sum using bijection Find the sum of $S=\displaystyle\sum_{i,j,k \ge 0, i+j+k=17} ijk$.
I am looking for a solution that uses some bijection.
I couldn't find any bijection.
I am able to do the problem by other method by observing that,
$S$ is the coefficient of $x^{17}$ in,
$(x+2x^2+3x^3+4x^4........)^3$ and then sum the last formal power series and then find the coefficient of  $x^{17}$ in that sum.
But the hint says find a bijection.
so, please help.
 A: Suppose I need to pick $5$ people from a group $19$ people conveniently numbered $1, \ldots, 19$. 
One method would be to first pick the $(i+1)$th person and the $(i+j+2)$th person. 
Then, I will select one of the $i$ people numbered $1, \ldots i$, and one of the $j$ people numbered $i+2, \ldots, i+j+1$, and one of the $k = 17-i-j$ people numbered $i+j+3,\ldots,19$. 
For each triple $(i,j,k)$ of non-negative integers such that $i+j+k = 17$, there is one way to pick the $(i+1)$th person and the $(i+j+2)$th person, and $ijk$ ways to choose the other three people. This gives us a total of $\displaystyle\sum_{i,j,k \ge 0, i+j+k = 17}ijk$ ways to pick $5$ people. 
Can you find a bijection between this method of picking people and a simpler method of picking people? If so, how many ways are there to pick $5$ people from $19$ in the simpler method?
A: For this you'll require the sums up to fourth powers:
$\begin{align}
\sum_{0 \le k \le n} k
  &= \frac{n (n + 1)}{2} \\
\sum_{0 \le k \le n} k^2
  &= \frac{n (n + 1) (2 n + 1)}{6} \\
\sum_{0 \le k \le n} k^3
  &= \frac{n^2 (n + 1)^2}{4} \\
\sum_{0 \le k \le n} n^4
  &= \frac{n^2 (6 n^3 + 15 n^2 + 10 n - 1)}{30}
\end{align}$
So you want:
$\begin{align}
\sum_{\substack{i + j + k = n \\
       i, j, k \ge 0}}
         i j k
  &= \sum_{0 \le i \le n}
       i \sum_{0 \le j \le n - i}
           j  (n - i - j) \\
  &= \sum_{0 \le i \le n}
       i \sum_{0 \le j \le n - i} (n j - i j - j^2) \\
  &= \sum_{0 \le i \le n}
       i \left(
           n \frac{(n - i) (n - i + 1)}{2}
             - i \frac{(n - i) (n - i + 1)}{2}
             - \frac{(n - i) (n - i + 1) (2 n - 2 i + 1}{6}
         \right) \\
  &= \sum_{0 \le i \le n}
       \frac{i (n^3 - n) - i^2 (3 n^2 - 1) + 3 i^3 n - i^4}{6} \\
  &= \frac{(n - 1) n (3 n^3 + 3 n^2 - 12 n - 10)}{360}
\end{align}$
Phew!
Thanks to maxima for routine algebra help. Any transcription errors are mine alone.
