smc

We can start off by finding patterns in $S(n)$. When we calculate a few values we realize either from performing the calculation or because the calculation was performed in the exact same way that $S(n)=2^n-2((n_4)-(n_5)\dots (n_n))$. Rearranging the expression we realize that the terms aside from $2^{2015}$ are congruent to $0$ mod $12$(Just put the equation in terms of $2^{2015}$ and the four combinations excluded and calculate the combinations mod $12$). Using patterns we can see that $2^{2015}$ is congruent to $8$ mod $12$. Therefore $\boxed{8}$ is our answer.