It's time to solve combinatorics 2

Note that $\omega+1=e^{\pm i\pi/3}$ , a primitive sixth root of unity. Thus $S_n=(\omega+1)^n$ can attain $\boxed{6}$ values, namely, all sixth roots of unity.

$S_n=\displaystyle \sum_{ k=0}^{n}\dbinom{n}{k}{\omega}^k = (1 + \omega)^n = (- \omega^2)^n\ (\because 1 + \omega + \omega^2 = 0)$

check the cases ans will be $\boxed{6}$ .