Having trouble understanding summation identity I'm having trouble understanding how the right side of this summation is equal to the left side. Any information would be helpful!
$$\sum_{k=0}^{n} {r \choose k}{s \choose n-k}={r+s \choose n}$$

This is known as Vandermonde's Identity. It says that for each way to choose $n$ items from a set of $r+s$ items, for some $k$ we must choose $k$ from the set of $r$ and $n-k$ from the set of $s$. This translates to
$$
\binom{r+s}{n}=\sum_{k=0}^n\binom{r}{k}\binom{s}{n-k}\tag{1}
$$

Another way to look at this is to look at
$$
(1+x)^{r+s}=(1+x)^r(1+x)^s\tag{2}
$$
Then, using the Binomial Theorem, we get
$$
\begin{align}
\sum_{n=0}^{r+s}\binom{r+s}{n}x^n
&=\sum_{i=0}^r\binom{r}{i}x^i\sum_{j=0}^s\binom{s}{j}x^j\\
&=\sum_{n=0}^{r+s}\sum_{k=0}^n\binom{r}{k}\binom{s}{n-k}x^kx^{n-k}\\
&=\sum_{n=0}^{r+s}\sum_{k=0}^n\binom{r}{k}\binom{s}{n-k}x^n\tag{3}
\end{align}
$$
Comparing coefficients of $x^n$ in $(3)$, we get $(1)$.
