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.
On
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.
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?