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
 A: Let's try to see this combinatorially in the form of picking marbles.
Suppose you have a $3\times n$ grid of marbles, and you want to choose $2n$
marbles out of them. The number of ways to do so is
$$\binom{3n}{2n}=\binom{n}{0}\binom{3n}{2n}.$$
The number of ways to do so with the additional requirement that we choose at least $k$ full rows can be computed as first choosing $k$ rows out of $n$, and then choosing $2n-3k$ marbles out of the remaining $3n-3k$ freely, so this number equals
$$\binom{n}{k}\binom{3n-3k}{2n-3k}.$$
We're taking an alternating sum over these terms, which suggests the inclusion-exclusion principle; the alternating sum counts the number of ways to pick $2n$ marbles without picking any full row. The only way to do so is by picking exactly $2$ marbles from each row, and there are $3$ ways to do so for each row. This shows that the alternating sum equals $3^n$.
A: 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
$$\bbox[5px,border:2px solid #00A000]{S_n = 3^n.}$$
A: \begin{align*}
&\binom{n}{0}\binom{3n}{2n}-\binom{n}{1}\binom{3n-3}{2n-3}+\binom{n}{2}\binom{3n-6}{2n-6}-\cdots \\
&=\binom{n}{0}\binom{3n}{n}-\binom{n}{1}\binom{3n-3}{n}+\binom{n}{2}\binom{3n-6}{n}-\cdots
\end{align*}
The right hand side is the coefficient of $x^n$ in 
\begin{align*}
&\binom{n}{0}(1+x)^{3n} - \binom{n}{1}(1+x)^{3n-3} + \cdots \\
&= (1+x)^{3n}\left(1- \frac{1}{(1+x)^3}\right)^n \\
&= (1+x)^{3n-3}(3x+3x^2+x^3)^n \\
&= x^n(1+x)^{3n-3}(3+3x+x^2)^n
\end{align*}
and hence equals $3^n$.
