<s>[INST] Prove that $\sum\limits_{k=0}^{m}\binom{m}{k}\binom{n+k}{m}=\sum\limits_{k=0}^{m}\binom{n}{k}\binom{m}{k}2^k$ Prove that 
$$\sum_{k=0}^{m}\binom{m}{k}\binom{n+k}{m}=\sum_{k=0}^{m}\binom{n}{k}\binom{m}{k}2^k$$
What should I do for this equation? Should I focus on proving $\binom{m}{k}\binom{n+k}{m}=\binom{n}{k}\binom{m}{k}2^k$?
 [/INST] I view that the purpose on the lefthand side of the eqn as: 
Let $A = {x_{1},x_{2},...,x_{n}}$ and $B={y_{1},y_{2},...,y_{m}}$.  where $n \ge m$
$ \left( \begin{array}{cc}
m  \\
0 \\
 \end{array} \right) $ 
$ \left( \begin{array}{cc}
n  \\
m \\
 \end{array} \right) $ accounts the number of ways to choose $m$ elements from $A$.
$ \left( \begin{array}{cc}
m  \\
1 \\
 \end{array} \right) $ 
$ \left( \begin{array}{cc}
n + 1  \\
m \\
 \end{array} \right) $ accounts the number of ways to choose $m$ elements from $A$ plus one element from $B$. ( from the set {${x_{1},x_{2},...,x_{n},y_{i}}$} )
Keep on going until the last term,
$ \left( \begin{array}{cc}
m  \\
m \\
 \end{array} \right) $ 
$ \left( \begin{array}{cc}
n + m  \\
m \\
 \end{array} \right) $ accounts the number of ways to choose $m$ elements from $A$ and $B$. ( from the set {${x_{1},x_{2},...,x_{n},y_{1},y_{2},...,y_{m}}$} ) 
$$ $$
Now for the right hand side of the eqn, count the same thing as before as :
Set the number of $x_{i}$'s that has been chosen is $j$, then the remaining $m-j$ are elements from $B$. That is
$ \left( \begin{array}{cc}
n  \\
j \\
 \end{array} \right) $ 
$ \left( \begin{array}{cc}
m  \\
m-j \\
 \end{array} \right) $ =  $ \left( \begin{array}{cc}
n  \\
j \\
 \end{array} \right) $ 
$ \left( \begin{array}{cc}
m  \\
j \\
 \end{array} \right) $ 
About the $2^k$ i am still confused, but i hope this will help a little..
</s>