The sum of $$\binom{n}{0}\binom{3n}{2n}-\binom{n}{1}\binom{3n-3}{2n-3}+\binom{n}{2}\binom{3n-6}{2n-6}-\cdots \cdots \cdots $$
$\bf{My\; Try::}$ We can write above sum as $$\sum^{n}_{k=0}(-1)^k\binom{n}{k}\binom{3n-3k}{2n-3k} = \sum^{n}_{k=0}(-1)^k\binom{n}{k}\binom{3n-3k}{n}$$
So sum $$ = \sum^{n}_{k=0}(-1)^k\cdot \frac{n!}{k!\cdot (n-k)!}\times \frac{(3n-2k)!}{n!\cdot (2n-3k)!} = \sum^{n}_{k=0}(-1)^k \cdot \frac{(3n-2k)!}{k!\cdot (n-k)!\cdot (2n-3k)!}$$
Now i did not understand how can i solve it after that , help Required, Thanks