Let $$S = x \in R : (\sqrt{3} + \sqrt{2})^x + (\sqrt{3} - \sqrt{2})^x = 10$$. Then the number of elements in $$S$$ is:

We need to find the set $$S = \{x \in \mathbb{R} : (\sqrt{3} + \sqrt{2})^x + (\sqrt{3} - \sqrt{2})^x = 10\}$$.

First, note that $$(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2}) = 3 - 2 = 1$$, so $$(\sqrt{3} - \sqrt{2}) = \frac{1}{\sqrt{3} + \sqrt{2}}$$ and hence $$(\sqrt{3} - \sqrt{2})^x = (\sqrt{3} + \sqrt{2})^{-x}$$.

Let $$t = (\sqrt{3} + \sqrt{2})^x$$. Since $$\sqrt{3} + \sqrt{2} > 1$$, it follows that $$t > 0$$. Substituting into the given equation gives $$t + \frac{1}{t} = 10$$.

Multiplying both sides by $$t$$ yields $$t^2 + 1 = 10t$$, or $$t^2 - 10t + 1 = 0$$.

Applying the quadratic formula gives $$t = \frac{10 \pm \sqrt{100 - 4}}{2} = \frac{10 \pm \sqrt{96}}{2} = \frac{10 \pm 4\sqrt{6}}{2} = 5 \pm 2\sqrt{6}$$. Both roots are positive since $$2\sqrt{6} \approx 4.899$$, so $$t_1 = 5 + 2\sqrt{6} > 0$$ and $$t_2 = 5 - 2\sqrt{6} \approx 0.101 > 0$$.

Observe that $$5 + 2\sqrt{6} = (\sqrt{3} + \sqrt{2})^2$$, because $$(\sqrt{3} + \sqrt{2})^2 = 3 + 2\sqrt{6} + 2 = 5 + 2\sqrt{6}$$; similarly, $$5 - 2\sqrt{6} = (\sqrt{3} - \sqrt{2})^2 = (\sqrt{3} + \sqrt{2})^{-2}$$. Hence, if $$(\sqrt{3} + \sqrt{2})^x = 5 + 2\sqrt{6}$$ then $$x = 2$$, and if $$(\sqrt{3} + \sqrt{2})^x = 5 - 2\sqrt{6}$$ then $$x = -2$$.

Therefore, $$S = \{-2, 2\}$$, which contains 2 elements. The correct answer is 2 (Option C).