Stirling numbers of the second kind with max cardinality We know that Stirling numbers of the second kind is the number of ways to partition a set of $n$ elements into $m$ nonempty sets.
My question is what's the number ways if the max cardinality of all the partitioned sets is $k$.
 A: I Googled restricted partitions stirling and the third hit was http://dlmf.nist.gov/26.9
Looks pretty thorough.
A: I you want for an exact closed formula, I suspect that it would be fairly complex, if it can be found - perhaps by generating functions.
If you are interested in approximate/asymptotic formulas for large values of $n, m$, (and $n/m \gtrsim 3$) you can try the following trick. 
There are in total $m^n$ ways of placing distinct objects in $m$ distinct cells  But we want the subset that consist of those configurations that have from 1 to $k$ objects in each cell. 
To estimate the relative size of this subset, we notice that the experiment of placing randomly $n$ distinct objects (equiprobably) in $m$ distinct cells is asymptotically equivalent to throwing $m$ independent Poisson variables with mean $\lambda = n/m$. 
The probability that this configuration fits our restricted subset is
$ \displaystyle \left( \sum_{j=1}^k e^{-\lambda} \frac{\lambda^j}{j!} \right)^m $
From this we can estimate our desider number, multiplying this expression by $m^n$ ,and dividing by $m!$ if we want to consider the cells as undistinguishable, as Stirling numbers do. So
$\displaystyle  S_k(n,m) \approx \frac{m^n}{m!} \left( e^{-\lambda} \sum_{j=1}^k  \frac{\lambda^j}{j!} \right)^m 
 $ 
with $\lambda = n/m$. 
BTW setting $k = \infty$ we have an approximation of the standard Stirling numbers of the second kind (further simplified-approximated by the Stirling approximation of factorials).
$\displaystyle  S(n,m) \approx \left( \frac{m}{e} \right)^{n-m} \; \frac{ \left( e^{\lambda} -1 \right)^m} {\sqrt{2 \pi \; m}}
 $ 
A: 
Let $a_m(n,k)$ be the number of set partitions of $n$ elements into $m$ non-empty partitions with maximum size $k$ each. Then the following recurrence relation is valid for $k\geq 1$:
  \begin{align*}
a_k(n+1,m)&=ma_k(n,m)+a(n,m-1)-\binom{n}{k}a(n-k,m-1)\qquad n,m\geq 1
\end{align*}
  with boundary conditions
  \begin{align*}
a_k(n,m)&=a_k(n,m)=0\, \qquad\qquad n<m,n>km\\
a_k(n,n)&=1\, \qquad\qquad\qquad\qquad\quad  n\geq 0\\
a_k(n,1)&=
\begin{cases}
1&\qquad\qquad\qquad\quad 1\leq n\leq k\\
0&\qquad\qquad\qquad\quad n>k
\end{cases}\\
a_k(0,m)&=a_k(n,0)=0\qquad\qquad\ \,   n,m>0\\
\end{align*}

A proof can be found in this answer.
