Let $$S = \{x : x \in \mathbb{R} \text{ and } (\sqrt{3} + \sqrt{2})^{x^2-4} + (\sqrt{3} - \sqrt{2})^{x^2-4} = 10\}$$. Then $$n(S)$$ is equal to

We need to find $$n(S)$$ where $$S = \{x : x \in \mathbb{R} \text{ and } (\sqrt{3}+\sqrt{2})^{x^2-4} + (\sqrt{3}-\sqrt{2})^{x^2-4} = 10\}$$.

Note that $$(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2}) = 3 - 2 = 1$$, so $$(\sqrt{3}-\sqrt{2}) = \frac{1}{\sqrt{3}+\sqrt{2}}$$.

Let $$t = (\sqrt{3}+\sqrt{2})^{x^2-4}$$. Then $$(\sqrt{3}-\sqrt{2})^{x^2-4} = \frac{1}{t}$$, and the equation becomes:

$$t + \frac{1}{t} = 10$$

$$t^2 - 10t + 1 = 0$$

$$t = \frac{10 \pm \sqrt{96}}{2} = 5 \pm 2\sqrt{6}$$

Since $$(\sqrt{3}+\sqrt{2})^2 = 3 + 2\sqrt{6} + 2 = 5 + 2\sqrt{6}$$, one solution is $$t = 5 + 2\sqrt{6} = (\sqrt{3}+\sqrt{2})^2$$, which gives $$x^2 - 4 = 2$$ and hence $$x^2 = 6$$, i.e., $$x = \pm\sqrt{6}$$.

Similarly, $$t = 5 - 2\sqrt{6} = (\sqrt{3}+\sqrt{2})^{-2}$$ leads to $$x^2 - 4 = -2$$, so $$x^2 = 2$$ and $$x = \pm\sqrt{2}$$.

Thus, $$S = \{-\sqrt{6}, -\sqrt{2}, \sqrt{2}, \sqrt{6}\}$$, so $$n(S) = 4$$.

$$n(S) = 4$$, which matches Option B. Therefore, the answer is Option B.