binomials product alternating sum calculation I need to somehow prove that $\sum\limits_{k = 0}^{n - 1} {n \choose k} {3 n - k - 1 \choose 2 n - k}(-1)^k = (-1)^{n + 1} {2 n - 1 \choose n}$.
I didn't manage to do it using induction or any combinatorial ideas. Could someone help me?

Suppose we seek to verify that
$$\sum_{k=0}^{n-1} 
{n\choose k} {3n-k-1\choose 2n-k} (-1)^k
= (-1)^{n+1} {2n-1\choose n}$$
which is the same as
$$\sum_{k=0}^{n} 
{n\choose k} {3n-k-1\choose 2n-k} (-1)^k
= \left((-1)^n+(-1)^{n+1}\right) {2n-1\choose n} = 0.$$
Introduce
$${3n-k-1\choose 2n-k} = {3n-k-1\choose n-1} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n}} 
(1+z)^{3n-k-1} \; dz.$$

This yields for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n}} 
(1+z)^{3n-1} 
\sum_{k=0}^n {n\choose k} (-1)^k \frac{1}{(1+z)^k} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n}} 
(1+z)^{3n-1} \left(1-\frac{1}{1+z}\right)^n \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n}} 
(1+z)^{3n-1} \frac{z^n}{(1+z)^{n}} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} 
(1+z)^{2n-1} \; dz = 0.$$
