Simplifying $ {m\choose l}^{-1}\sum_ii^p{n\choose i}{m-n\choose l-i} $ I am trying to simplify the following expression:
$$
{m\choose l}^{-1}\sum_ii^p{n\choose i}{m-n\choose l-i}
$$
where the sum is over all allowed $i$ values, i.e. $\max(0,l-m+n)\le i\le l$ and all variables are non-negative integers. I know that the expression has a value of 1 for $p=0$ from the Chu–Vandermonde identity, but I am wondering if I can simplify the sum away for $p=1$ and $p=2$. Thank you!
 A: For $p=1$ we have
$$\sum_ii\binom{n}i\binom{m-n}{\ell-i}=n\sum_i\binom{n-1}{i-1}\binom{m-n}{\ell-i}=n\binom{m-1}{\ell-1}\;.$$
With a bit more work we can use the ame idea to dispose of the case $p=2$:
$$\begin{align*}
\sum_ii^2\binom{n}i\binom{m-n}{\ell-i}&=n\sum_ii\binom{n-1}{i-1}\binom{m-n}{\ell-i}\\
&=n\sum_i(i-1)\binom{n-1}{i-1}\binom{m-n}{\ell-i}+\\
&\quad\quad+n\sum_i\binom{n-1}{i-1}\binom{m-n}{\ell-i}\\
&=n(n-1)\sum_i\binom{n-2}{i-2}\binom{m-n}{\ell-i}+n\binom{m-1}{\ell-1}\\
&=n(n-1)\binom{m-2}{\ell-2}+n\binom{m-1}{\ell-1}\;.
\end{align*}$$
A: In seeking to simplify
$$\sum_{q=0}^l q^p {n\choose q} {m-n\choose l-q}$$
we use
$$q^p = \sum_{r=1}^p {q\choose r} {p\brace r} r!,$$
(distribute $q$ different possible values into $p$ slots and classify
according to the number $r$ of different values that are present).
We thus obtain
$$\sum_{q=0}^l \sum_{r=1}^p {q\choose r} {p\brace r} r!
{n\choose q} {m-n\choose l-q}
\\ = \sum_{r=1}^p  {p\brace r} r!
\sum_{q=0}^l {q\choose r}
{n\choose q} {m-n\choose l-q}.$$
Now we have as in the first answer
$${n\choose q} {q\choose r} =
\frac{n!}{(n-q)! \times r! \times (q-r)!}
= {n\choose r} {n-r\choose n-q}$$
so that we obtain
$$\sum_{r=1}^p  {p\brace r} r! {n\choose r}
\sum_{q=0}^l {n-r\choose n-q} {m-n\choose l-q}
\\ = \sum_{r=1}^p  {p\brace r} r! {n\choose r}
[z^{l}] (1+z)^{m-n}
\sum_{q=0}^l {n-r\choose n-q} z^q.$$
Here the coefficient extractor enforces the range and we find
for the term covered by the extractor
$$[z^{l}] (1+z)^{m-n}
\sum_{q\ge 0} {n-r\choose n-q} z^q
= [z^{l}] (1+z)^{m-n}
\sum_{q=0}^n {n-r\choose n-q} z^q
\\ = [z^l] (1+z)^{m-n} z^n
\sum_{q=0}^n {n-r\choose q} z^{-q}
= [z^l] (1+z)^{m-n} z^n
\left(1+\frac{1}{z}\right)^{n-r}.$$
Returning to the full sum we find
$$\sum_{r=1}^p  {p\brace r} r! {n\choose r}
[z^l] (1+z)^{m-n} z^r (1+z)^{n-r}.$$
We thus get the closed form involving $p$ terms:
$$\bbox[5px,border:2px solid #00A000]{
\sum_{r=1}^p  {p\brace r} r! {n\choose r}
{m-r\choose l-r}}$$
or alternatively
$$\bbox[5px,border:2px solid #00A000]{
\sum_{r=1}^p  {p\brace r} n^{\underline{r}}
{m-r\choose l-r}.}$$
This gives for $p=2:$
$${2\brace 1} n {m-1\choose l-1}
+ {2\brace 2} n (n-1) {m-2\choose l-2}
\\ = n {m-1\choose l-1}
+ n (n-1) {m-2\choose l-2}.$$
confirming the result from the first answer. We get for $p=3$
$${3\brace 1} n {m-1\choose l-1}
+ {3\brace 2} n (n-1) {m-2\choose l-2}
+ {3\brace 3} n (n-1) (n-2) {m-3\choose l-3}
\\ = n {m-1\choose l-1}
+ 3 n (n-1) {m-2\choose l-2}
+ n (n-1) (n-2) {m-3\choose l-3}.$$
