With this problem the salient feature is that the first guess which is
$$\sum_{q=m}^n {n\choose q} (-1)^{m-q} (n-q)^k$$
does not produce the correct answer. The underlying poset has nodes
for each subset $P$ of the set $Q$ of $n$ cells where the node
represents cells from $P$ plus possibly additional cells being empty,
ordered by the superset relation with the node of all cells $Q$ being
empty (which does not contain any configurations) being at the bottom.
(With $P_1$ a superset of $P_2$ the configurations of the former
constitute a subset of the latter.) Now a configuration that has
exactly $p$ empty cells where $m\le p\le n$ receives total weight (sum
of the weights of all nodes where it is included)
$$\sum_{q=m}^p {p\choose q} (-1)^{m-q}.$$
While this yields a weight of one for $p=m$ it is not equal to zero
for $p\gt m$ and hence cannot be used to count configurations with
exactly $m$ empty cells. A better approach is to choose the $m$ empty
cells first and use a poset where the nodes $P$ represent the extra
empty cells in addition to the $m$ already selected, which no longer
participate in the inclusion-exclusion. This yields
$${n\choose m} \sum_{q=0}^{n-m} {n-m\choose q} (-1)^q
(n-m-q)^k.$$
Here we are interested in the count of zero empty extra cells. The
single node with $q=n-m$ is at the bottom. With tbis approach the
weight of a configuration on the remaining $n-m$ cells with exactly
$p$ extra empty cells where $0\le p\le n-m$ is given by
$$\sum_{q=0}^{p} {p\choose q} (-1)^q.$$
This evaluates to one when $p=0$ and is zero otherwise, which is
precisely the weights that we require for this problem. Observe that
we have a Stirling number here which can be seen by writing
$$\frac{n!}{m!} \frac{1}{(n-m)!}
\sum_{q=0}^{n-m} {n-m\choose q} (-1)^{n-m-q} q^k$$
which yields
$$\bbox[5px,border:2px solid #00A000]{
\frac{n!}{m!} {k\brace n-m}.}$$
If we had not been asked to use inclusion-exclusion the answer would
have been ${n\choose m} \times {k\brace n-m} (n-m)!$ by inspection.
