TITLE: Counting set partitions of $\{1,2,...,n\}$ into exactly $k$ non-empty subsets with max size $m$
QUESTION [6 upvotes]: Let $a(n,k,m)$ denote the number of set partitions of $\{1,2,...,n\}$ into exactly $k$ non-empty subsets with max size $m$. Thus all subsets have number of elements $\leq m$. (Here size $m$ doesn't have to be attained by a subset, but if it makes the question easier, then that's fine).
The number with no restriction on part sizes is given by the Stirling number of the second kind $S(n,k)$.
I tried to construct a recurrence to the Stirling numbers but couldn't work in the constraint of the maximal size. 
$$a(n+1,k,m) = a(n,k-1,m) + k. a(n,k,m-1) + ?? $$
or
$$a(n+1,k,m) = a(n,k-1,m) + k. a(n,k,m) - ?? $$
with initial conditions $a(n,n,m) = 1, a(1,1,m)=1$ for $m\geq 1$ and $a(n,k,m) = 0$ for $m<n/k$.

REPLY [2 votes]: Let $a_m(n,k)$ be the number of set partitions of $n$ elements into $k$ non-empty partitions with maximum size $m$ each.
We can obtain the recurrence in Markus Scheuer's answer using elementary counting methods. We use a similar method described here used to obtain a recurrence for the Stirling numbers of the second kind. https://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind#Recurrence_relation)
We want to show that 
$$a_m(n+1,k)=ka_m(n,k)+a_m(n,k−1)−{n\choose m}a_m(n−m,k−1) \qquad n,k≥1$$
Proof:
A partition of $n+1$ objects into $k$ non-empty sets either has the $n+1$th object as a singleton or not. The number of ways that it is a singleton is given by $a_m(n,k-1)$ since this is the number of ways to partition the remaining $n$ objects into $k-1$ sets.
Next we need to count the other case where the $n+1$th object is in one of the $k$ subsets containing the other $n$ objects. This can be done in $ka_m(n,k)$ ways since $a_m(n,k)$ is the number of ways of partitioning $n$ objects into $k$ subjects and there are $k$ choices. However, this also includes cases where we have added the $n+1$th object to a subset with $m$ objects which we don't want to include. So we need to subtract the number of times this happens. This is the number of subsets with exactly $m$ objects in all of the $a_m(n,k)$ partitions of $n$ objects into $k$ subsets with max size $m$. To make a subset of size $m$, we choose $m$ objects from $n$ in ${n\choose m}$ ways. This occurs with multiplicity of the number of ways of putting the remaining $n-m$ objects into $k-1$ subsets (with max size $m$). This is $a_m(n-m,k-1)$ ways.
Putting this together we get the desired result.