Seeking a combinatorial proof $\sum _{k=0}^{2n} (-1)^k \binom{4n-2k}{2n-k}\binom{2k}{k}=\binom{2n}{n}\times 2^{2n}$ I would appreciate if somebody could help me with the following problem
Q: Seeking a combinatorial proof $(\binom{n}{k}=\frac{n!}{k! (n-k)!}
)$
$$\sum _{k=0}^{2n} (-1)^k \binom{4n-2k}{2n-k}\binom{2k}{k}=\binom{2n}{n}\times
 2^{2n}$$

Since
$$ \sum_{k\geq 0}\binom{2k}{k}x^k = \frac{1}{\sqrt{1-4x}}, \tag{1}$$
your sum can be read as a convolution, i.e. as the coefficient of $x^{2n}$ in the product between $\frac{1}{\sqrt{1-4x}}$ and $\frac{1}{\sqrt{1+4x}}$, but, due to $(1)$:
$$ [x^{2n}]\frac{1}{\sqrt{1-16 x^2}} = [x^n]\frac{1}{\sqrt{1-16 x}} = \binom{2n}{n}\cdot 4^n,\tag{2}$$
hence:
$$ \sum_{k=0}^{2n}\binom{4n-2k}{2n-k}\binom{2k}{k}(-1)^k = \binom{2n}{n}\cdot 2^{2n}. \tag{3}$$
