TITLE: Finding the amount partitions with gives sizes of a multiset
QUESTION [5 upvotes]: A multiset $A$ contains $E$ positive integers. The multiplicity of each element is $r_i \; i=1,\ldots ,N$.
$A$ is partitioned in $M$ (we do not necessary have $M=N$) ordinary sets (where elements are therefore not repeated). The size of the $i$-th set $C_i$ is given and equal to $c_i$. 
Is there a way to count the number of possible partitions knowing $N, M, r_i, c_i$ with or without consider sets order?
I think the answer to this question may be found starting from here and maybe by employing the partial Bells polynomials .
Another way to interpret the problem can be the following: find the number of binary $N \times M$ matrices with fixed rows and columns sums with no column and row full of zeros (a similar problem has been answered here ) .

REPLY [2 votes]: We         cite         the        following         MSE         link
I as well  as this
MSE    link   II.
With  the present  question we  use  the interpretation  that we  have
multisets of  sets i.e. while the  source variables form sets  and not
multisets, these  sets can occur  multiple times.  Using  the notation
that was presented there we obtain the closed form for the case of the
sets being ordered
$$\left[\prod_{k=1}^l A_k^{\tau_{k}}\right]
\prod_{k=1}^m
Z\left(P_k; \sum_{k'=1}^l A_{k'}\right)^{\sigma_k}.$$
In terms  of combinatorial classes we  have made use of  the unlabeled
class
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}
\textsc{SEQ}_{=\sigma_k}
\left(\textsc{SET}_{=k}
\left(\sum_{k'=1}^l \mathcal{A}_{k'}\right)\right).$$
Note that  the cycle  index will create  the intermediate  sets during
evaluation. We have for the unordered case
$$\left[\prod_{k=1}^l A_k^{\tau_{k}}\right]
\prod_{k=1}^m
Z\left(S_{\sigma_k};
Z\left(P_k; \sum_{k'=1}^l A_{k'}\right)\right).$$
Again,  in terms  of combinatorial  classes we  have made  use of  the
unlabeled class
$$\textsc{MSET}_{=\sigma_k}
\left(\textsc{SET}_{=k}
\left(\sum_{k'=1}^l \mathcal{A}_{k'}\right)\right).$$
Here  we have  used  the  recurrence by  Lovasz  for  the cycle  index
$Z(P_n)$ of the  set operator $\textsc{SET}_{=n}$ on  $n$ slots, which
is
$$Z(P_n) = \frac{1}{n} \sum_{l=1}^n (-1)^{l-1} a_l Z(P_{n-l})
\quad\text{where}\quad
Z(P_0) = 1.$$
This  recurrence  lets us  calculate  the  cycle index  $Z(P_n)$  very
easily.
 Example  as per request. Using  the notation from the  link we
get  for the  pairing  $A_1 A_2  A_3^2 A_4^2$  and  $B_1^2 B_2^2$  the
combinatorial class
$$\textsc{MSET}_{=2}
(\textsc{SET}_{=1}(\mathcal{A_1}+\mathcal{A}_2
+\mathcal{A}_3+\mathcal{A}_4))
\times \textsc{MSET}_{=2}
(\textsc{SET}_{=2}(\mathcal{A_1}+\mathcal{A}_2
+\mathcal{A}_3+\mathcal{A}_4)).$$
This yields the two cycle indices
$$Y(B_1^2) = 1/2\,{a_{{1}}}^{2}+1/2\,a_{{2}}$$
and
$$Y(B_2^2) = 1/8\,{a_{{1}}}^{4}-1/4\,{a_{{1}}}^{2}a_{{2}}
+3/8\,{a_{{2}}}^{2}-1/4\,a_{{4}}.$$
Multiply to get the cycle index
$$Y(B_1^1 B_2^2) = 1/16\,{a_{{1}}}^{6}-1/16\,{a_{{1}}}^{4}a_{{2}}
+1/16\,{a_{{1}}}^{2}{a_{{2}}}^{2}
\\ -1/8\,{a_{{1}}}^{2}a_{{4}}+3/16\,{a_{{2}}}^{3}-1/8\,a_{{2}}a_{{4}}.$$
Substitute $A_1+A_2+A_3+A_4$ to get
$$Y(B_1^1 B_2^2; A_1+A_2+A_3+A_4)
= 1/16\,
\left( A_{{1}}+A_{{2}}+A_{{3}}+A_{{4}} \right) ^{6}
\\-1/16\,
\left( A_{{1}}+A_{{2}}+A_{{3}}+A_{{4}} \right) ^{4}
\left( {A_{{1}}}^{2}+{A_{{2}}}^{2}+{A_{{3}}}^{2}+{A_{{4}}}^{2} \right)
\\ +1/16\,
\left( A_{{1}}+A_{{2}}+A_{{3}}+A_{{4}} \right) ^{2}
\left( {A_{{1}}}^{2}+{A_{{2}}}^{2}+{A_{{3}}}^{2}+{A_{{4}}}^{2} \right) ^{2}
\\ -1/8\,
\left( A_{{1}}+A_{{2}}+A_{{3}}+A_{{4}} \right) ^{2}
\left( {A_{{1}}}^{4}+{A_{{2}}}^{4}+{A_{{3}}}^{4}+{A_{{4}}}^{4} \right)
\\+3/16\,
\left( {A_{{1}}}^{2}+{A_{{2}}}^{2}+{A_{{3}}}^{2}+{A_{{4}}}^{2} \right) ^{3}
\\ -1/8\,
\left( {A_{{1}}}^{2}+{A_{{2}}}^{2}+{A_{{3}}}^{2}+{A_{{4}}}^{2} \right)
\left( {A_{{1}}}^{4}+{A_{{2}}}^{4}+{A_{{3}}}^{4}+{A_{{4}}}^{4} \right).$$
Expand to obtain
$$\cdots+10\,A_{{1}}{A_{{2}}}^{2}A_{{3}}{A_{{4}}}^{2}
+3\,A_{{1}}{A_{{2}}}^{2}{A_{{4}}}^{3}
+A_{{1}}A_{{2}}{A_{{3}}}^{4}
+6\,A_{{1}}A_{{2}}{A_{{3}}}^{3}A_{{4}}
\\+10\,A_{{1}}A_{{2}}{A_{{3}}}^{2}{A_{{4}}}^{2}
+6\,A_{{1}}A_{{2}}A_{{3}}{A_{{4}}}^{3}
+A_{{1}}A_{{2}}{A_{{4}}}^{4}+\cdots$$
We see that for four  types of variables with multiplicities $1,1,2,2$
creating a multiset of sets  where the sets have cardinality $1,1,2,2$
the  total  number  of  configurations   is  ten.  These  are  clearly
determined by the pairs and we get:

 $\{\{A_1, A_3\},\{A_2, A_3\}\}$  
 $\{\{A_1, A_4\},\{A_2, A_4\}\}$  
 $\{\{A_1, A_3\},\{A_2, A_4\}\}$  
 $\{\{A_1, A_4\},\{A_2, A_3\}\}$  
 $\{\{A_1, A_2\},\{A_3, A_4\}\}$  
 $\{\{A_1, A_3\},\{A_3, A_4\}\}$  
 $\{\{A_1, A_4\},\{A_3, A_4\}\}$  
 $\{\{A_2, A_3\},\{A_3, A_4\}\}$  
 $\{\{A_2, A_4\},\{A_3, A_4\}\}$  
 $\{\{A_3, A_4\},\{A_3, A_4\}\}.$