$$\displaystyle\sum_{\substack{i,j=0 \\ i \neq j}}^{n}$$  $$^n C_{i}$$  $$^n C_{j}$$ is equal to 

We need to evaluate $$\displaystyle\sum_{\substack{i,j=0 \\ i \neq j}}^{n} \binom{n}{i}\binom{n}{j}$$.

$$\sum_{\substack{i,j=0 \\ i \neq j}}^{n} \binom{n}{i}\binom{n}{j} = \sum_{i=0}^{n}\sum_{j=0}^{n} \binom{n}{i}\binom{n}{j} - \sum_{i=0}^{n} \binom{n}{i}^2$$

$$\sum_{i=0}^{n}\sum_{j=0}^{n} \binom{n}{i}\binom{n}{j} = \left(\sum_{i=0}^{n}\binom{n}{i}\right)\left(\sum_{j=0}^{n}\binom{n}{j}\right) = 2^n \cdot 2^n = 2^{2n}$$

By Vandermonde's identity (or the Cauchy product):

$$\sum_{i=0}^{n} \binom{n}{i}^2 = \sum_{i=0}^{n} \binom{n}{i}\binom{n}{n-i} = \binom{2n}{n}$$

$$\sum_{\substack{i,j=0 \\ i \neq j}}^{n} \binom{n}{i}\binom{n}{j} = 2^{2n} - \binom{2n}{n}$$

Therefore, the correct answer is Option A: $$2^{2n} - \binom{2n}{n}$$.