Prove that $\binom{n}{0}\cdot \binom{2n}{n}-\binom{n}{1}\cdot \binom{2n-2}{n}+\binom{n}{2}\cdot \binom{2n-4}{n}+........... = 2^n$ 
Prove that $$\binom{n}{0}\cdot \binom{2n}{n}-\binom{n}{1}\cdot \binom{2n-2}{n}+\binom{n}{2}\cdot \binom{2n-4}{n}+........... = 2^n$$

$\bf{My\; Try::}$ Coefficient of $x^n$ in $$\left[\binom{n}{0}(1+x)^{2n}-\binom{n}{1}(1+x)^{2n-2}+\binom{n}{2}(1+x)^{2n-4}-........\right]$$
So Coefficient of $x^n$ in $$\left[(1+x)^2-1\right] = x^n\left(2+x\right)^n$$
So Coefficient of $x^n$ in $$x^n\left[\binom{n}{0}2^n+\binom{n}{1}2^{n-1}x+.......+\binom{n}{n}x^n\right] = 2^n$$
My Question is can we solve it Using Combinotarial way.
If yes then plz explain here.
Thanks
 A: You can indeed prove it combinatorially. Suppose that you $n$ pairs of shoes, and you want to know how many ways there are to pick one shoe from each pair. Clearly the answer is $2^n$, but we can also use an inclusion-exclusion argument to calculate it. Number the pairs of shoes from $1$ through $n$, and let $\mathscr{A}_k$ be the collection of sets of $n$ shoes that do not include either shoe from pair $k$. The number in which we’re interested is then also
$$\binom{2n}n-\left|\bigcup_{k=1}^n\mathscr{A}_k\right|\;,$$
since there are altogether $\binom{2n}n$ sets of $n$ shoes that can be drawn from the collection of $n$ pairs.
If $\varnothing\ne J\subseteq[n]$, the sets of $n$ shoes in $\left|\bigcap_{k\in J}\mathscr{A}_k\right|$ are the ones drawn from the $n-|J|$ pairs whose numbers are not in $J$, so by the inclusion-exclusion principle we have
$$\begin{align*}
\left|\bigcup_{k=1}^n\mathscr{A}_k\right|&=\sum_{\varnothing\ne J\subseteq[n]}(-1)^{|J|-1}\left|\bigcap_{k\in J}\mathscr{A}_k\right|\\
&=\sum_{\varnothing\ne J\subseteq[n]}(-1)^{|J|-1}\binom{2(n-|J|)}n\\
&=\sum_{k=1}^n(-1)^{k-1}\binom{n}k\binom{2n-2k}n\;,
\end{align*}$$
since there are $\binom{n}k$ subsets of $[n]$ of cardinality $k$, and hence
$$\begin{align*}
2^n&=\binom{2n}n-\sum_{k=1}^n(-1)^{k-1}\binom{n}k\binom{2n-2k}n\\
&=\binom{2n}n+\sum_{k=1}^n(-1)^k\binom{n}k\binom{2n-2k}n\\
&=\sum_{k=0}^n(-1)^k\binom{n}k\binom{2n-2k}n\;.
\end{align*}$$
