Help Show Binomial Identity: $\sum_{j=0}^{n} {n \choose j}{m+j \choose n} = \sum_{j=0}^{n} {n \choose j}{m \choose j}2^j$ I have been trying to solve this problem that I found in my old course notes for some time, but I have not been successful. Can anyone suggest a strategy or provide a hint? $\displaystyle\sum_{j=0}^{n} {n \choose j}{m+j \choose n} = \displaystyle\sum_{j=0}^{n} {n \choose j}{m \choose j}2^j$. 
 A: To sum up, the necessary hints are that:
$$ \binom{m+j}{n} = \sum\limits_{i=0}^{j}\binom{j}{i}\binom{m}{n-i},\tag{1}$$
$$ \binom{n}{i}\binom{i}{k}=\binom{n}{k}\binom{n-k}{i-k},  \tag{2}$$
$$ \sum\limits_{j=0}^{n}\sum\limits_{i=0}^{j} f(i,j) = \sum\limits_{i=0}^{n}\sum\limits_{j=i}^{n} f(i,j), \quad\text{and} \tag{3}$$
$$ \sum\limits_{y=0}^{x}\binom{x}{y} = 2^x. \tag{4}$$
Then, 
\begin{align}
\sum\limits_{j=0}^{n}\binom{n}{j}\binom{m+j}{n} &\stackrel{(1)}= \sum\limits_{j=0}^{n}\binom{n}{j}\sum\limits_{i=0}^{j}\binom{j}{i}\binom{m}{n-i},\\
&= \sum\limits_{j=0}^{n}\sum\limits_{i=0}^{j}\binom{n}{j}\binom{j}{i}\binom{m}{n-i},\\
&\stackrel{(2)}= \sum\limits_{j=0}^{n}\sum\limits_{i=0}^{j}\binom{n}{i}\binom{n-i}{j-i}\binom{m}{n-i},\\ 
&\stackrel{(3)}= \sum\limits_{i=0}^{n}\sum\limits_{j=i}^{n}\binom{n}{i}\binom{n-i}{j-i}\binom{m}{n-i},\\
&= \sum\limits_{i=0}^{n} \binom{n}{i}\binom{m}{n-i} \sum\limits_{j=i}^{n}\binom{n-i}{j-i}.\\ 
\end{align}
Define $x=n-i$ and $y=j-i$, we have 
\begin{align}
\sum\limits_{j=0}^{n}\binom{n}{j}\binom{m+j}{n} &= \sum\limits_{x=0}^{n}\binom{n}{x}\binom{m}{x} \sum\limits_{y=0}^{x}\binom{x}{y} \\
&\stackrel{(4)}= \sum\limits_{x=0}^{n}\binom{n}{x}\binom{m}{x} 2^x \\
&\stackrel{(j=x)}= \sum\limits_{j=0}^{n}\binom{n}{j}\binom{m}{j} 2^j. \tag*{$\blacksquare$}
\end{align}
