how to calculate $\sum^{n}_{k=m}\binom{k}{m}\binom{n}{k}$? 
For natural numbers $m \leq n$ calculate (i.e. express by a simple formula
  not containing a sum)
  $$\sum^{n}_{k=m}\binom{k}{m}\binom{n}{k}.$$

I searched and answer is probably
$=\binom{n}{m}\sum^{n}_{k=m}\binom{n-m}{n-k}$
$=\binom{n}{m}\sum^{n-m}_{k=0}\binom{n-m}{k}$
$=\binom{n}{m}2^{n-m}$
Q
1: Is the answer correct?
2: Why take $\binom{n}{m}$ in front of sum?
3: How to transform the first formula to become second formula?
 A: It is sufficient to answer (3).

We obtain for integral $0\leq m\leq n$:
  \begin{align*}
\color{blue}{\sum_{k=m}^n\binom{k}{m}\binom{n}{k}}&=\sum_{k=m}^n\binom{n}{m}\binom{n-m}{n-k}\tag{1}\\
&=\binom{n}{m}\sum_{k=m}^n\binom{n-m}{n-k}\tag{2}\\
&=\binom{n}{m}\sum_{k=0}^{n-m}\binom{n-m}{n-(k+m)}\tag{3}\\
&=\binom{n}{m}\sum_{k=0}^{n-m}\binom{n-m}{k}\tag{4}\\
&=\binom{n}{m}\sum_{k=0}^{n-m}\binom{n-m}{k}1^k1^{n-m-k}\\
&=\binom{n}{m}(1+1)^{n-m}\tag{5}\\
&\,\,\color{blue}{=\binom{n}{m}2^{n-m}}
\end{align*}
  and the claim follows.

Comment:


*

*In (1) we use the binomial identity $\binom{k}{m}\binom{n}{k}=\binom{n}{m}\binom{n-m}{n-k}$.

*In (2) we factor out $\binom{n}{m}$ which does not depend on the index $k$.

*In (3) we shift the index to start with $k=0$. To compensate the index-shift $k\to k-m$ we replace in the summand $k$ with $k+m$.

*In (4) we use the binomial identity $\binom{p}{q}=\binom{p}{p-q}$.

*In (5) we apply the binomial theorem.
A: $$\binom{k}m\binom{n}k=\frac{k!}{m!(k-m)!}\frac{n!}{k!(n-k)!}=\frac{n!}{m!(k-m)!(n-k)!}=$$$$\frac{n!(n-m)!}{m!(n-m)!(k-m)!(n-k)!}=\binom{n}{m}\binom{n-m}{n-k}$$
$\binom{n}{m}$ can be taken in front because it does not depend on index $k$.
