prove
$\sum_{r=1}^{n}\binom{n}{r}\binom{n-1}{r-1}=\binom{2n-1}{n-1}$
what I have so far:
applying $\binom{n}{r}=\binom{n}{n-r}$ $$\sum_{r=1}^{n}\binom{n}{r}\binom{n-1}{r-1}=\sum_{r=1}^{n}\binom{n}{r}\binom{n-1}{n-r}$$ applying Vandermonde's Identity and subtracting off the difference $$\sum_{r=1}^{n}\binom{n}{r}\binom{n-1}{n-r}=\sum_{r=0}^{n}\binom{n}{r}\binom{n-1}{n-r}-\binom{n}{0}\cdot\binom{n-1}{n}=\binom{2n-1}{n}-\binom{n}{0}\cdot\binom{n-1}{n}=\binom{2n-1}{n-1}$$
Is this a valid proof? what happens when I try to evaluate $\binom{n}{0}\cdot\binom{n-1}{n}$? does this break the proof? any alternate proofs or guidance is appreciated.
No, it is fine. As Lord Shark the Unknown said, $\binom{n}{0}\cdot\binom{n-1}{n}=1\cdot 0$ since, by convention $\binom n m=0$ iff $m>n$. (For instance, there are zero ways to select $n$ items from $n-1$, after all.)
$\binom {2n-1}{n-1}$ counts the ways we may select $n-1$ items from a heap of $2n-1$. Now suppose this heap was actually subdivided into two, one of size $n$ and the other of size $n-1$, so when selecting the $n-1$ items from these heaps, we do so by taking some of them (say, $r$) from the first and the rest from the other. Count the distinct ways we can do this, thereby showing that ... $\therefore \quad \dbinom{2n-1}{n-1}=\mathop{\huge \sum}\limits_{r=1}^n\dbinom n r\dbinom{n-1}{r-1}$