Binomial coefficients / sums I need help solving this task so if anyone had a similar problem it would help me.
$\sum\limits_{k=m}^{n}{n \choose k}{k \choose m}={n \choose m}2^{n-m}$
I tried this:
$\sum\limits_{k=m}^{n}\frac{n!k!}{k!(n-k)!m!(k-m)!}={n \choose m}2^{n-m}\\\sum\limits_{k=m}^{n}\frac{n!}{(n-k)!m!(k-m)!}={n \choose m}2^{n-m}\\\frac{n!}{m!}\sum\limits_{k=m}^{n}\frac{1}{(n-k)!(k-m)!}={n \choose m}2^{n-m}$
Shift $z=k-m\rightarrow k=z+m$
$\frac{n!}{m!}\sum\limits_{z=0}^{n-m}\frac{1}{(n-(z+m))!(z+m-m)!}={n \choose m}2^{n-m}\\\frac{n!}{m!}\sum\limits_{z=0}^{n-m}\frac{1}{(n-z-m)!z!}={n \choose m}2^{n-m}$
And now with this part I don’t know what to do next.
I tried something this way.
$\frac{1}{(n-m)!}+\frac{1}{(n-1-m)!}+...+\frac{1}{n-m}!={n \choose m}2^{n-m}$
But now I don’t know how to calculate the expression on the left side.
Thanks in advance !

Almost there! $$\frac {n!}{m!} \sum_{z=0}^{n-m} \frac 1{(n-z-m)!z!}=\frac {n!} {(n-m)!m!} \sum_{z=0}^{n-m} \frac {(n-m)!}{(n-m-z)!z!} = \binom n m \sum_{z=0}^{n-m}\binom {n-m}z = \binom n m 2^{n-m}$$
