Is this algorithm for Exact Three Cover sub-exponential, because I find $length(s)/3$ combinations for $C$?
No. Exponential time is sometimes used to mean running times in $2^{\Theta(n)}$, and other times it is used to mean running times in $2^{\Theta(n^k)}$ for some constant $k&gt;0$.

Your algorithm is exponential w.r.t. the second definition and superexponential w.r.t. the first.
In fact your algorithm takes time $2^{\Theta(n^3)}$, where $n = |S|$ (notice that $\frac{n!}{3! \cdot (n-3)!} = \frac{n(n-1)(n-2)}{6} =\Theta(n^3)$).