Let us have a set $A$ of $A=n$ objects. I would like to know the number of ways to partition the set into $k$ nonempty subsets. This, I believe, is given by the Stirling numbers of the second kind, $S(n,k)$. However, I have an additional constraint roughly saying that each of the subsets must have certain minimum number of elements. Formally: there are given natural numbers $l_1, ..., l_k$ and it must be possible to label the $k$ subsets of $A$ by symbols $A_i, i = 1,...,k$ such that $A_i \ge l_i$. Naturally, it holds that $l_i \ge 1 \forall i$ and $\sum\limits_i l_i \le n$. Any ideas how I could compute that?

1$\begingroup$ Do you count labelled (according to these constraints) partitions or those which may be labelled, regardless by how many ways? $\endgroup$ – Fedor Petrov Mar 14 '18 at 13:11

1$\begingroup$ The later. But seeing how to compute labeled would probably be also insightful. $\endgroup$ – user1747134 Mar 14 '18 at 13:40
For fixed $l_1,\dots,l_k$ and counting the labelled partitions onto $A_1,\dots,A_k$, $A_i\geqslant l_i$, the exponential generating function is $\prod_k \sum_{m\geqslant l_i} \frac{x^m}{m!}=\prod_k (e^x1x^2/2!\dotsx^{l_i1}/(l_i1)!)$. Unlikely this has closed form.

2$\begingroup$ Not sure if this is counted as a closedform but still: $$\prod_{i=1}^k \sum_{m\geq l_i}\frac{x^m}{m!}=\prod_{i=1}^k \frac{x^{l_i}}{(l_i1)!} \int_0^1 e^{x(1y)} y^{l_i1} dy.$$ $\endgroup$ – Max Alekseyev Mar 15 '18 at 16:09
Stirling numbers of second kind can be expressed as the sum: $$S(n,k) = \sum_{{j_1 + j_2 + \dots + j_n = k\atop j_1 + 2 j_2 + \dots + nj_n = n}} \frac{n!}{j_1!1!^{j_1}j_2!2!^{j_2}\cdots j_n!n!^{j_n}},\qquad(\star)$$ which essentially sums over the partitions of $n$ into $k$ parts representing the subset sizes (with $j_i$ being the multiplicity of size $i$).
Let's assume that $l_1\geq l_2\geq \dots\geq l_k$. It is possible to label the subsets as required if and only if for each $i=1,2,\dots,k$, the following inequality holds: $$\sum_{t=l_i}^n j_t\geq i.$$ (Equivalently, this can be viewed as the Young diagram of the partition encoded by $j$'s containing the Young diagram of the partition $(l_1,l_2,\dots,l_k)$). Restricting the sum $(\star)$ to such partitions will yield the number of required set partitions.

$\begingroup$ This does not seem to be tractable to compute, or is it? $\endgroup$ – user1747134 Mar 16 '18 at 10:32

$\begingroup$ @user1747134: The number of terms is at most $p_k(n)$, i.e., the number of partitions of $n$ into $k$ parts. $\endgroup$ – Max Alekseyev Mar 16 '18 at 14:09