The identity $\binom{n+k}{k} = \sum_i \binom{n}{i}\binom{k}{i}$

Question

I've been working out some representation theory problems and I've stumbled upon the following identity: $$ \binom{n+k}{k} = \sum_i \binom{n}{i}\binom{k}{i} $$

Proof: consider the subgroup $S_n\times S_k\leq S_{n+k}$. The index is equal to both sides: for the left hand side by simply considering the orders of the respective groups, for the right hand side by noting that a left coset is determined by choosing how many elements switch sides and where they land. (I've also checked this identity experimentally for $n,k$ up to 20.)

I was wondering if there is an easy, elementary combinatorial proof.

Answer 1

Choosing $k$ people from $n$ boys and $k$ girls. The group consist of $i$ boys and $k-i$ girls. Remember that $\binom{k}{i}=\binom{k}{k-i}$

Answer 2

$\binom{n+k}{k}$ is the number of group of $k$ people among a group of $n$ men and $k$ women. To perform that, you can either take $\binom{k}{k}$ women and $\binom{n}{0}$ men, or $\binom{k}{k-1}$ women and $\binom{n}{1}$ men or $\binom{k}{k-2}$ women and $\binom{n}{2}$ men... Therefore, you get $$\binom{n+k}{k}=\sum_{i=0}^k\binom{k}{k-i}\binom{n}{i}=\sum_{i=0}^k\binom{k}{i}\binom{n}{i}.$$