How would you use Stirling numbers to partition a n unique objects, into m unique subsets, with minimum cardinality of 2 for each subset? For example, how would one go about partitioning 10 numbered balls into 4 numbered boxes, such that each box contains at least 2 balls?
 A: We have using combinatorial classes as in Analytic Combinatorics by
Flajolet and Sedgewick the following class
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}
\textsc{SEQ}_{=m}(\textsc{SET}_{\ge 2}(\mathcal{Z})).$$
This gives the exponential generating function
$$F(z) = (\exp(z)-z-1)^m.$$
To relate this to Stirling numbers we write
$$n! [z^n] (\exp(z)-z-1)^m
= n! [z^n] \sum_{r=0}^m {m\choose r} (\exp(z)-1)^{m-r}
(-1)^r z^r
\\ = n! \sum_{r=0}^{\min(n,m)} {m\choose r} 
(-1)^r [z^{n-r}] (m-r)! \frac{(\exp(z)-1)^{m-r}}{(m-r)!}
\\ = n! \sum_{r=0}^{\min(n,m)} {m\choose r} 
(-1)^r \frac{(m-r)!}{(n-r)!}  {n-r\brace m-r}
\\ = m! \sum_{r=0}^{\min(n,m)} {n\choose r} (-1)^r
{n-r\brace m-r}.$$
If we look this up in the OEIS we find OEIS
A200091 which says "The number of ways of
putting $n$ labeled items into $k$  labeled boxes so that each box
receives at least 2 objects" so we have  the right answer.

Note that the closely related class (boxes are not labeled i.e. 
indistinguishable)
$$\textsc{SET}_{=m}(\textsc{SET}_{\ge 2}(\mathcal{Z}))$$
has EGF by the same computation
$$G(z) = \frac{1}{m!} (\exp(z)-z-1)^m$$
which yields for the coefficients
$$\sum_{r=0}^{\min(n,m)} {n\choose r} (-1)^r
{n-r\brace m-r}$$
which is OEIS A009299 "associated Stirling
numbers of the second kind."

What we have here is an  inclusion-exclusion (PIE) argument where $r$
gives the number of  singletons (single element in a box). The nodes $Q$
of the underlying poset represent  distributions that have the elements
of $Q$ as singletons plus  possibly more. So the Hasse diagram of the
poset has as nodes the subsets of  $[n]$ of cardinality at most $m$ and
is ordered by set inclusion. The weight on a node $Q$ is  $(-1)^{|Q|}.$ A
distribution that has set of singletons $P$ exactly is  thus included
in all nodes $Q\subseteq P$  for a total weight of
$$\sum_{Q\subseteq P} (-1)^{|Q|} 
= \sum_{r=0}^{|P|} {|P|\choose r} (-1)^r = 0.$$
This is zero because $|P|\ge 1.$ On the other hand a distribution that
has no singletons which is what we seek is only included in
$Q=\emptyset$ for a weight of $(-1)^{|\emptyset|} = 1$, precisely as
required. The total count of all weights is
$$\sum_{Q\subseteq [n], |Q|\le m} (-1)^{|Q|} {n-|Q|\brace m-|Q|}
= \sum_{r=0}^{\min(n,m)} {n\choose r} (-1)^r {n-r\brace m-r}$$
which  concludes  the argument.  Here  we  have counted  any  possible
distribution of the remaining $n-|Q|$ balls into the remaining $m-|Q|$
boxes at  a node, with  no box empty  (Stirling numbers of  the second
kind). We can get the count for the first class from the count for the
second by multiplying by $m!.$
