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