Is there a formula for the number of k-partitions of a set? In how many ways can a set be partitioned into k non-empty subsets?
 A: If the set is non empty, it can be calculated with a recursive formula.
Let $S_{n,k}$ be the number of $k$-partitions of a set $X$ ($n\ge k \ge 1$).
Simple observations: $S_{n,1}=1,\enspace S_{n,n}=1$. 
Let us now suppose $n>k>1$. Let $x_0\in X$, and consider a $k$-partition $\mathcal P$ of $X$.


*

*Either $\{x_0\}$ is a member of $\mathcal P$. Then $\mathcal P'=\mathcal P\setminus\bigl\{\{x_0\}\bigr\} $ is a $k-1$-partition of $X'=X\setminus \{x_0\}$. These are $S_{n-1,k-1}$ in number.

*Or $\{x_0\}$ is a not a member of $\mathcal P$. Then $\mathcal P$ induces a $k$-partition of $X'$, by taking the intersections of its members with $X'$. Conversely, from a $k$-partition of $X'$, you get a  $k$-partition of $X$ by  adjoining $x_0$ to one of its members. As there are $p$ choices for this adjunction, these partitions are $kS_{n-1,k}$ in number.


Conclusion: we obtain the following recurrence relation:
$$S_{n,k}=S_{n-1,k-1}+kS_{n-1,k}.$$
We can compute the values of the $S_{n,k}$s in a way very similar to Pascal's triangle. Here are the first values:
$$ \begin{array}{r|*{8}{c}}
k = &1 & 2 & 3 & 4 & 5 & 6 & 7 \\
\hline
n = 1 & 1 \\
2 & 1 & 1 \\
3 & 1 & 3 & 1 \\
4 & 1 & 7 & 6 & 1\\
5 & 1 & 15 & 25 & 10 & 1 \\
6 & 1 & 31 & 90 & 65 & 15 & 1\\
7 & 1 & 63 & 301 & 350 & 140 & 21 & 1\\
8 & 1 & 127 & 969 & 1701 & 1050 & 266 & 28 & 1
  \end{array}$$
A: Following the above notation, the number of non-empty $k$-partitions in a set of $n$ elements is:
$$S_{n,k} = \frac{1}{(k-1)!}\sum_{j=0}^{k-1}{(-1)^{j}\binom{k-1}{j}}(k-j)^{n-1}$$
For example:
$$S_{n,2} = \sum_{j=0}^{1}{(-1)^{j}\binom{1}{j}(2-j)^{n-1}} = 2^{n-1} - 1$$
$$S_{n,3} = \frac{1}{2}\sum_{j=0}^{2}{(-1)^{j}\binom{2}{j}}(3-j)^{n-1} = \frac{3^{n-1}-2^{n}+1}{2}$$
and so on.
The following is just an anecdotal extra, but I remember writing this formula when I played Final Fantasy VIII Remastered a while ago to calculate the number of ways to assign GF to the three active party members, which turned out to be a problem equivalent to finding out the number of distinct 3-partitions of the set of acquired GFs.
