How to show that these two summations are equal

Question

$$\sum_{k=1}^{n-2} (n!-\frac{n!}{(n-k)!})\qquad (1)$$ $$\sum_{k=2}^{n-1} \frac{n!}{k!}(n-k)(k-1)\qquad (2)$$ I've got these summations as a solutions of a combinatorial problem. But I have no idea how to prove equality of them.

Answer 1

Swap the order of summation in the first sum, so that it becomes $$ n! \sum_{k=2}^{n-1} \frac{k!-1}{k!} $$ Now cancel $n!$, so that you need to show $$ \sum_{k=2}^{n-1} \frac{k!-1}{k!} = \sum_{k=2}^{n-1} \frac{(n-k)(k-1)}{k!}.$$ Observe that $$ \frac{ (n-k)(k-1)}{k!} + ((k-1)!-1) \frac{n-k}{(k-1)!} = \frac{k!-1}{k!} + (k!-1)\frac{n-(k+1)}{k!}.$$ Using this to substitute for $(n-k)(k-1)/k!$ in the second sum gives you the result, because the new terms telescope.