Closed form for $\sum_{k=0}^{l}\binom{k}{n}\binom{k}{m}$ Does there exist any closed form for the following sum?

$$\sum_{k=0}^{l}\binom{k}{n}\binom{k}{m}$$

Where 
$l \in \mathbb N$ and $m,n \in \mathbb Z$

My try:
$$ \sum_{k=\max\left(m,n\right)}^{l}\binom{k}{n}\binom{k}{m}=\sum_{k=0}^{l}\binom{k}{k-n}\binom{k}{k-m}$$$$=\left(-1\right)^{\left(-n-m\right)}\sum_{k=0}^{l}\binom{-n-1}{k-n}\binom{-m-1}{k-m}$$$$=\left(-1\right)^{\left(-n-m\right)}\sum_{k=0}^{l}\binom{-n-1}{-1-k}\binom{-m-1}{k-m}$$$$=\left(-1\right)^{\left(-n-m\right)}\binom{-n-m-2}{-m-1}$$
$$=\left(-1\right)^{\left(-n-m\right)}\binom{-n-m-2}{-n-1}=\left(-1\right)^{\left(-m-1\right)}\binom{m}{-n-1}$$$$=\left(-1\right)^{\left(-m-1\right)}\binom{m}{m+n+1}=\left(-1\right)^{n}\binom{n}{m+n+1}$$
I'm not sure whether it's right, so can someone verify the solution, and if it's not right then please provide a closed form (of course if that's exist).
 A: We present a proof of the identity by @Diger, which should be  considered
a starting point for additional simplification. We seek to show that
$$\sum_{k=0}^l {k\choose m} {k\choose n} =
\sum_{k=0}^n (-1)^k {l+1\choose m+k+1} {l-k\choose n-k}.$$
The RHS is
$$[z^n] \sum_{k=0}^n (-1)^k {l+1\choose m+k+1} 
z^k (1+z)^{l-k}.$$
The coefficient extractor enforces the range:
$$[z^n] \sum_{k\ge 0} (-1)^k {l+1\choose l-m-k} 
z^k (1+z)^{l-k}
\\ = [z^n] (1+z)^l [w^{l-m}] (1+w)^{l+1}
\sum_{k\ge 0} (-1)^k w^k
z^k (1+z)^{-k}
\\ = [z^n] (1+z)^l [w^{l-m}] (1+w)^{l+1}
\frac{1}{1+wz/(1+z)}
\\ = [z^n] (1+z)^{l+1} [w^{l-m}] (1+w)^{l+1}
\frac{1}{1+z+wz}
\\ = [z^n] (1+z)^{l+1} [w^{l-m}] (1+w)^{l+1}
\frac{1}{1+z(1+w)}
\\ = [z^n] (1+z)^{l+1} [w^{l-m}] 
\sum_{k\ge 0} (-1)^k z^k (1+w)^{k+l+1}
\\ = [z^n] (1+z)^{l+1}
\sum_{k\ge 0} (-1)^k z^k {k+l+1\choose l-m}.$$
This is
$$\bbox[5px,border:2px solid #00A000]{
\sum_{k=0}^n (-1)^k {l+1\choose n-k} {k+l+1\choose l-m}.}$$
The LHS is 
$$\sum_{k\ge 0} [[0\le k\le l]] [z^m] (1+z)^k [w^n] (1+w)^k
\\ = [z^m] [w^n]  \sum_{k\ge 0} (1+z)^k  (1+w)^k
[v^l] \frac{v^k}{1-v}
\\ = [z^m] [w^n] [v^l] \frac{1}{1-v} 
\sum_{k\ge 0} (1+z)^k  (1+w)^k v^k
\\ = [z^m] [w^n] [v^l] \frac{1}{1-v} 
\frac{1}{1-(1+z)(1+w)v}
\\ = [z^m] [w^n] 
[v^l] \frac{1}{v-1} 
\frac{1/(1+z)/(1+w)}{v-1/(1+z)/(1+w)}.$$
The inner term is
$$\mathrm{Res}_{v=0} \frac{1}{v^{l+1}}
\frac{1}{v-1} \frac{1/(1+z)/(1+w)}{v-1/(1+z)/(1+w)}.$$
Residues sum to zero and the residue at infinity in $v$ is zero.
The contribution from minus the residue at $v=1/(1+z)/(1+w)$ is
$$- [z^m] (1+z)^{l+1} [w^n] (1+w)^{l+1} 
\frac{1/(1+z)/(1+w)}{1/(1+z)/(1+w)-1}
\\ = - [z^m] (1+z)^{l+1} [w^n] (1+w)^{l+1} 
\frac{1/(1+z)}{1/(1+z)-(1+w)}
\\ = [z^m] (1+z)^{l+1} [w^n] (1+w)^{l+1} 
\frac{1/(1+z)}{w+z/(1+z)}
\\ = [z^m] (1+z)^{l+1} [w^n] (1+w)^{l+1} 
\frac{1/z}{w(1+z)/z+1}.$$
Now with $l,m,n$ positive integers we must have $l\ge n,m$ or else 
there is no contribution to $k^\underline{m} k^\underline{n}.$
This means we continue with
$$[z^m] (1+z)^{l+1} 
\sum_{k=0}^n {l+1\choose k} 
\frac{1}{z} (-1)^{n-k} \frac{(1+z)^{n-k}}{z^{n-k}}
\\ = \sum_{k=0}^n (-1)^{n-k} {l+1\choose k} 
{l+1+n-k\choose m+1+n-k}.$$
This is
$$\bbox[5px,border:2px solid #00A000]{
\sum_{k=0}^n (-1)^{n-k} {l+1\choose k} 
{l+1+n-k\choose l-m}.}$$
We have the same closed form for LHS and RHS, thus proving the claim.
 For a full proof we also need to show that the contribution from 
$v=1$ is zero. We get
$$[z^m] [w^n] \frac{1/(1+z)/(1+w)}{1-1/(1+z)/(1+w)}
= [z^m] [w^n]  \frac{1}{(1+z)(1+w)-1}
\\ = [z^m] [w^n]  \frac{1}{z+w+zw}
= [z^{m+1}] [w^n] \frac{1}{1+w(1+z)/z}
\\ = [z^{m+1}] (-1)^n \frac{(1+z)^n}{z^n}
= (-1)^n {n\choose n+m+1} = 0.$$
A: I doubt there is closed form, but this is another identity which can be derived by contour integration
$$\sum_{k=0}^l {k \choose m} {k \choose n} = \sum_{k=0}^n (-1)^k {l+1 \choose m+k+1}{l-k \choose n-k} \, .$$
If you are interested I can write it down. It is useful when $l$ is large and either $m$ or $n$ is small.
edit: On part of your try the third row is still correct, while the fourth equality (first time no sum) is wrong.
A: I guess the best that you can get when $0 \le n \le m$ is in table III, page 15, eq. (4.9) of Gould's combinatorial identities:
$$\sum_{k=0}^{l}{k \choose n}{k \choose m} = \sum_{k=0}^{n}{n \choose k}{m \choose k}{l+k+1 \choose n+m+1}$$
I don't know whether that can be extended to $m,n \in \mathbb Z$.
As remarked there, the original source is “The class of the free metabelian group with exponent $p^2$”, by S. Bachmuth and H. Y. Mochizuki, Communications on Pure and Applied Math., Vol.
21 (1968), pp. 385-399.
A: Maxima says there is no closed form.
load(zeilberger);
GosperSum(binomial(k, n) * binomial(k, m), k, 0, l);

gives NON GOSPER SUMMABLE
A: This can be proved using simple terms, with telescoping series and reccurence.
Let $$U_{l,n,m}=\binom{l}{n}-\binom{m}{n}=\sum_{k=l}^{m+1} \binom{k}{n}-\binom{k-1}{n}=\sum_{k=l-1}^{m} \binom{k}{n-1}$$ with $n<=m$ and $l>=m$ and $\sum$ is decreasing.
Let $$V_{l,n}=\sum_{k=n}^{0} (-1)^{k}\binom{l}{k}=\sum_{k=n}^{0} (-1)^{k}\binom{l-1}{k}+\binom{l-1}{k-1}$$
This is a telescoping sum which equals $$V_{l,n}=\binom{l-1}{n}$$ with $k$ positive and $l>=n$ and $\sum$ is decreasing.
Now let $T_{l,n,m}=\sum_{k=l}^{m} \binom{k}{m} \binom{k}{n}$ which is our sum in question.
$$\begin{eqnarray*}
T_{l,n,m} & = & \sum_{k=l}^{m} \binom{k}{m} \binom{k}{n} \\
& = & \sum_{k=l}^{m} \binom{k}{m} \binom{l}{n} - \sum_{k=l-1}^{m} \binom{k}{m} \left(\binom{l}{n}-\binom{k}{n}\right) \\ 
& = & \sum_{k=l}^{m} \binom{k}{m} \binom{l}{n} - \sum_{k=l-1}^{m} \binom{k}{m} U_{l,n,k} \\
& = & \left( \sum_{k=l}^{m} \binom{k}{m} \right) \binom{l}{n} - \sum_{k=l-1}^{m} \binom{k}{m} \left(\sum_{i=l-1}^{k} \binom{i}{n-1} \right)\\
\end{eqnarray*}$$
Switching terms of this sum by writing $k$ in respect of $i$.
$$\begin{eqnarray*}
T_{l,n,m} & = & \binom{l+1}{m+1} \binom{l}{n} - \sum_{i=l-1}^{m} \binom{i}{n-1} \left(\sum_{k=i}^{m} \binom{k}{m} \right)\\
& = & \binom{l+1}{m+1} \binom{l}{n} - \sum_{i=l-1}^{m} \binom{i}{n-1}  \binom{i+1}{m+1} \\
\end{eqnarray*}$$
Until now there is no appearence of reccurence hence we try to write the sum in convenient form with $T$, so let:
$$S_{l,n,m}=\sum_{i=l}^{m-1} \binom{i}{n}  \binom{i+1}{m}$$ which means $$ T_{l,n,m} = \binom{l+1}{m+1} \binom{l}{n} - S_{l-1,n-1,m+1}$$
We return back to $S$ now:
$$\begin{eqnarray*}S_{l,n,m} & = & \sum_{i=l}^{m-1} \binom{i}{n}  \binom{i+1}{m}\\
& = & \sum_{i=l}^{m-1} \left(\binom{i+1}{n}-\binom{i}{n-1}\right)  \binom{i+1}{m} \\
& = & \sum_{i=l}^{m-1} \left(\binom{i+1}{n}-\binom{i+1}{n-1}+\binom{i}{n-2}\right)  \binom{i+1}{m} \\
...& = & \sum_{i=l}^{m-1} \left(\binom{i+1}{n}-\binom{i+1}{n-1}+...\binom{i+1}{0}\right)  \binom{i+1}{m} \\
& = & \sum_{i=l}^{m-1} \left(\binom{i+1}{n}\binom{i+1}{m}\right)-\sum_{i=l}^{m-1}\left(\binom{i+1}{n-1}\binom{i+1}{m}\right)+...\sum_{i=l}^{m-1}\left(\binom{i+1}{0}\binom{i+1}{m}\right)   \\
& = & \sum_{i=l+1}^{m} \left(\binom{i}{n}\binom{i}{m}\right)-\sum_{i=l+1}^{m}\left(\binom{i}{n-1}\binom{i}{m}\right)+...\sum_{i=l+1}^{m}\left(\binom{i}{0}\binom{i}{m}\right)   \\
& = & T_{l+1,n,m}-T_{l+1,n-1,m}+...T_{l+1,0,m}   \\
\end{eqnarray*}$$
So then, we plug directly in $T$ we get:
$$\begin{eqnarray*}
T_{l,n,m} & = & \binom{l+1}{m+1} \binom{l}{n} - S_{l-1,n-1,m+1} \\
& = & \binom{l+1}{m+1} \binom{l}{n} - T_{l,n-1,m+1}+T_{l,n-2,m+1}-...T_{l,0,m} \\
\end{eqnarray*}$$
The linear deployment of this formula leads to troubled understanding so we just suppose it's right for $n-1$ to $0$ then we prove for $n$.
$T_{l,0,m}$ is blatantly true for n=0.
$$\begin{eqnarray*}
T_{l,n,m} & = & \binom{l+1}{m+1} \binom{l}{n} - T_{l,n-1,m+1}+T_{l,n-2,m+1}-...T_{l,0,m} \\
& = & \binom{l+1}{m+1} \binom{l}{n} - \sum_{k=0} \binom{l-k}{n-1-k} \binom{l+1}{m+2+k} + \sum_{k=0} \binom{l-k}{n-2-k} \binom{l+1}{m+3+k} -... \end{eqnarray*}$$
We note that these sums are lined diagonally therefor we collect them horizontally.
$$\begin{eqnarray*} & = & \binom{l+1}{m+1} \binom{l}{n} - \sum_{k=0}^{n-1} (-1)^{k}\binom{l}{n-1-k} \binom{l+1}{m+2} + \sum_{k=0}^{n-2} (-1)^{k}\binom{l-1}{n-2-k} \binom{l+1}{m+3}-... \\
& = & \binom{l+1}{m+1} \binom{l}{n} - \binom{l+1}{m+2} V_{l,n-1} + \binom{l+1}{m+2} V_{l-1,n-2} - ... \\
T_{l,n,m} & = & \binom{l+1}{m+1} \binom{l}{n} - \binom{l+1}{m+2} \binom{l-1}{n-1} + \binom{l+1}{m+3} \binom{l-2}{n-2} - ...
\end{eqnarray*}$$

The proof by linear expansion takes a lot of space and deep simplifications by Writing $T_{n-1}$ in terms of $T_{n-2},T_{n-3},...T_0$, so the same for $T_{n-2}$ yet for $T_{n-3}$ to finally land in multiple vandermonde correlative forms summed up togather, it's more complicated to be written in latex, I found this transformation about three and half years ago when trying to reduce a PE problem.
