Given: ${\left(\sqrt{3}+\sqrt{2}\right)}^x+{\left(\sqrt{3}-\sqrt{2}\right)}^x=10$

Now, let ${\left(\sqrt{3}+\sqrt{2}\right)}^x=t\Rightarrow {\left(\sqrt{3}-\sqrt{2}\right)}^x=\frac{1}{t}$

So, ${\left(\sqrt{3}+\sqrt{2}\right)}^x+{\left(\sqrt{3}-\sqrt{2}\right)}^x=10$

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

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

$\Rightarrow t=5\pm 2\sqrt{6}$

$\Rightarrow t={\left(\sqrt{3}-\sqrt{2}\right)}^2, {\left(\sqrt{3}+\sqrt{2}\right)}^2$

Now, taking ${\left(\sqrt{3}+\sqrt{2}\right)}^x={\left(\sqrt{3}-\sqrt{2}\right)}^2={\left(\sqrt{3}+\sqrt{2}\right)}^{-2}$ we get, $x=-2$

And equating ${\left(\sqrt{3}+\sqrt{2}\right)}^x={\left(\sqrt{3}+\sqrt{2}\right)}^2$ we get, $x=2$

Hence, the number of elements in $S$ is $2$