TITLE: Can this be proved using Combinatorics or generating functions?
QUESTION [10 upvotes]: What is the proof for this identity?
$\displaystyle\sum_{r=0}^{n}\binom{n}{r}\binom{m+r}{n}\ = \displaystyle\sum_{r=0}^{n}\binom{n}{r}\binom{m}{r}2^{r}$
I thought of changing $r$ to $n-r$ on left and right both. Then, I thought of building a set with atleast $n$ elements and atmost $2n$ elements with atmost $n$ elements from $m$ element set, and we can do this on LHS by choosing $n-r$ elements from n and out of remaining $r+m$ elements we can choose n more elements. On RHS, we can again choose $n-r$ elements from $n$ and $r$ elements from $m$, then remaining $r$ elements from n element set have 2 choices each. But I think it has a flaw. Can someone please help? If my proof is not correct, please suggest a combinatorial or a generating function proof.

REPLY [3 votes]: This is another version of Phicar’s solution.
There are $n$ students and $m$ professionals. All professionals are member of chess club while students can be in the club or not. Suppose that we want to choose $n$ people from the club’s members to join a competition.
In the LHS, $r$ students are in chess club, $\binom{n}{r}$. Therefore, there are $m+r$ chess club members to choose from, $\binom{m+r}{n}$. Hence $\sum_{r=0}^{n}{\binom{n}{r}\binom{m+r}{n}}$
In the RHS, $r$ professionals join the competition, $\binom{m}{r}$. Therefore, $n-r$ students must be members of the club and join the competition, $\binom{n}{n-r}=\binom{n}{r}$. The remaining $r$ students can be in the club or not, $2^{r}$. Hence $\sum_{r=0}^{n}{\binom{n}{r}\binom{m}{r}2^{r}}$