I am looking for a simple combinatorial proof of the binomial identity: $$\sum_{j=0}^n \binom{2j}{j}\binom{2n-2j}{n-j} = 4^n.\tag{1}$$
The standard way I know is to exploit the generating function: $$\sum_{j=0}^{+\infty}\binom{2j}{j}x^j = \frac{1}{\sqrt{1-4x}},$$ but I wonder if a simple combinatorial explanation of $(1)$ is known.
