Is it true that $\binom{n}{m} \binom{k}{\ell} \le \binom{n+k}{m+\ell}$? I am trying to find a bound and to cut the long story short I got to the point where I have
$$\sum_{k\ge 0}\sum_{\ell=0}^k \frac{\binom{n}{m} \binom{k}{\ell}}{\binom{n+k}{m+\ell}}\frac{a^k s^k}{k!}$$
so if $\binom{n}{m} \binom{k}{\ell} \le \binom{n+k}{m+\ell}$, I get a nice exponential bound that is exactly what I need. It seems completely elementary, I know. I tried numerically and I am convinced that it is true, but I got confused when I tried to prove it.
Thanks in advance.
Edit: I have $n\ge m$.
Edit2: I just realised that if this is true then I don't get the nice exponential I want but 
$$\sum_{k\ge 0}(k+1)\frac{a^k s^k}{k!}.$$
Is this by any chance a known function?
 A: Expand $\displaystyle {{n\choose m}{k\choose \ell}\over{n+k\choose m+\ell}}$ into 
factorials and regroup into $\displaystyle{{m+\ell\choose m}{n+k-m-\ell\choose n-m}\over{n+k\choose k}}$. Adding over $\ell$ and using formula (5.26) from 
Concrete Mathematics you get
 $$\sum_{\ell=0}^k \frac{\binom{n}{m} \binom{k}{\ell}}{\binom{n+k}{m+\ell}} ={n+1+k\over n+1},$$ and so 
$$\sum_{k\ge 0}\sum_{\ell=0}^k \frac{\binom{n}{m} \binom{k}{\ell}}{\binom{n+k}{m+\ell}}\frac{a^k s^k}{k!}= {n+1+as\over n+1}\,\exp(as).$$
Added: To get the second equation, use
$$\sum_{k\geq 0}k \frac{a^k s^k}{k!} = \sum_{k\geq 1}k \frac{a^k s^k}{k!}
=\sum_{k\geq 1} \frac{a^k s^k}{(k-1)!}=as \sum_{(k-1)\geq 0} \frac{(as)^{k-1}}{(k-1)!}
=as \exp(as).$$   

Added: To answer the original question, yes, the following   inequality is true 
$${n\choose m}{k\choose \ell}\leq{n+k\choose m+\ell}. $$
Imagine a class with $n$ boys and $k$ girls. The right hand side counts the 
number of ways to select $m+\ell$ students. The left hand side counts the number of
ways to select exactly $m$ boys and $\ell$ girls.  
A: For the "known function" part, let 
$$f(s)=\sum_{k\ge 0} \frac{a^k s^{k+1}}{k!}.$$ Note that $f(s)=se^{as}$. 
The sum 
$$\sum_{k\ge 0} (k+1)\frac{a^ks^k}{k!}$$
 is $f'(s)$. This is $(1+as)e^{as}$. 
