<s>[INST] 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 $ 
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
 [/INST] This  also has  a  very simple  algebraic  proof. Suppose  we seek  to
evaluate
$$S_n = \sum_{k=0}^n {n\choose k} (-1)^k {3n-3k\choose n}.$$
Introduce
$${3n-3k\choose n} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{2n-3k+1}} \frac{1}{(1-z)^{n+1}} \; dz.$$
We get for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{2n+1}} \frac{1}{(1-z)^{n+1}} 
\sum_{k=0}^n {n\choose k} (-1)^k z^{3k}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{2n+1}} \frac{1}{(1-z)^{n+1}} 
(1-z^3)^n
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{2n+1}} \frac{1}{1-z} 
(1+z+z^2)^n
\; dz.$$
Now residues sum  to zero so we  have from the poles at  zero, one and
infinity
$$S_n - 3^n + 
\mathrm{Res}_{z=\infty} \frac{1}{z^{2n+1}} \frac{1}{1-z} 
(1+z+z^2)^n = 0.$$
Note that the residue at infinity is
$$-\mathrm{Res}_{z=0} \frac{1}{z^2} z^{2n+1}
\frac{1}{1-1/z} (1+1/z+1/z^2)^n
\\= -\mathrm{Res}_{z=0}  z^{2n}
\frac{1}{z-1} (1+1/z+1/z^2)^n
\\ = -\mathrm{Res}_{z=0}
\frac{1}{z-1} (z^2+z+1)^n = 0.$$
This leaves
$$\bbox[5px,border:2px solid #00A000]{S_n = 3^n.}$$
</s>