How to simplify the sum

Question

Given the sum: $$ \sum_{i = t}^{k} (i - 1)! (k - i)! \binom{k - t}{i - t} $$

Using WolframAlpha I got result $\frac{k!}{t}$. But I cannot obtain such result myself.

How to simplify the sum?

Answer 1

First note that $\binom{k-t}{i-t}=\binom{k-t}{k-i}$, so

$$(i-1)!(k-i)!\binom{k-t}{i-t}=(i-1)!(k-i)!\binom{k-t}{k-i}=\frac{(i-1)!(k-t)!}{(i-t)!}\;,$$

and we can rewrite your sum as

$$\begin{align*} (k-t)!\sum_{i=t}^k\frac{(i-1)!}{(i-t)!}&=(k-t)!\sum_{\ell=0}^{k-t}\frac{(t-1+\ell)!}{\ell!}\\ &=(k-t)!(t-1)!\sum_{\ell=0}^{k-t}\binom{t-1+\ell}{t-1}\\ &=(k-t)!(t-1)!\binom{k}t\\ &=\frac{k!(k-t)!(t-1)!}{t!(k-t)!}\\ &=\frac{k!}t\;. \end{align*}$$

The step in which the summation disappeared uses the hockey stick identity.