prove simple binomial sum, combinatorics I want to prove that:
$$\large\sum_{i = 1}^{n} \binom{n}{i}\binom{n}{i-1} = \binom{2n}{n-1}$$
On the right hand side we simply have the coefficient of $x^{n-1}$ of the term $(1+x)^{2n}$
But on the other hand,
$$(1+x)^{2n} = (1+x)^{n}(1+x)^{n} = \large\sum_{i = 0}^{n}\binom{n}{i}x^i\sum_{j = 0}^{n}\binom{n}{j}x^j$$
and the coefficient here of $x^{n-1}$ is when $j = n-1-i$ and obviously $i$ cant be higher than $n-1$ so overall:
$$\large\sum_{i=0}^{n-1}\dbinom{n}{i}\dbinom{n}{n-i-1}$$
How do we continue from here?

Let $k=i+1$. Then, $i=k-1$, and  
$$
\sum_{i=0}^{n-1}\binom{n}{i}\binom{n}{n-i-1}=\sum_{k=1}^{n}\binom{n}{k-1}\binom{n}{n-k}=\sum_{k=1}^{n}\binom{n}{k-1}\binom{n}{k},
$$
where the last equality follows since $\binom{n}{n-k}=\frac{n!}{k!(n-k)!}=\binom{n}{k}$.
