Surjections and splitting a set into three groups

1. Let the target be $\{1,2,3\}$. There are $3^n$ maps. Let $A_i$ be the set of maps never taking the value $i$; $|A_i|=2^n$, $|A_i\cap A_j|=1$ for $i\ne j$, and $|A_1\cap A_2\cap A_3|=0$ (as $n\ge1$). By inclusion-exclusion the number of maps missing no value is
$$3^n-3\cdot2^n+3.$$

2. A surjection $f$ gives the partition $\{f^{-1}(1),f^{-1}(2),f^{-1}(3)\}$, and each partition arises from exactly $3!=6$ surjections. Hence
$$S(n,3)=\frac{3^n-3\cdot2^n+3}6,$$
which is an integer, so $6\mid3^n-3\cdot2^n+3$.

3. A map onto $\{1,2\}$ is surjective unless it is constant: $2^n-2$ surjections, and each partition into $2$ groups comes from $2$ surjections. So $S(n,2)=\frac{2^n-2}2=2^{n-1}-1$.

Check. For $n=3$: $\frac{27-24+3}6=1$; $n=4$: $\frac{81-48+3}6=6$; $n=7$: $\frac{2187-384+3}6=301$.

Remark. In general, the number of surjections onto a $k$-set is $\sum_{j=0}^k(-1)^j\binom kj(k-j)^n=k!\,S(n,k)$.

Lesson. Count labelled objects (surjections) by inclusion-exclusion, then divide by the symmetry of the labels.