Partitioning a multiset into multisets of fixed sizes

It would appear that these are multisets of multisets which may be enumerated using the Polya Enumeration Theorem (PET). Let the multiset that is being drawn have factorization

$$\prod_{k=1}^m B_k^{\sigma_{k}}$$

where $k$ is the value of a term and $\sigma_k$ the number of times it ocurs and recall that we have $l$ types of items forming the source multiset

$$\prod_{k=1}^l A_k^{\tau_{k}}.$$

The answer is then given by

$$\left[\prod_{k=1}^l A_k^{\tau_{k}}\right] \prod_{k=1}^m Z\left(S_{\sigma_k}; Z\left(S_k; \sum_{k'=1}^l A_{k'}\right)\right).$$

In terms of combinatorial classes we have made use of the unlabeled class

$$\def\textsc#1{\dosc#1\csod} \def\dosc#1#2\csod{{\rm #1{\small #2}}} \textsc{MSET}_{=\sigma_k} \left(\textsc{MSET}_{=k} \left(\sum_{k'=1}^l \mathcal{A}_{k'}\right)\right).$$

As an example for ${2,2\choose 1,1,2} = 3$ we get

$$\textsc{MSET}_{=2} (\textsc{MSET}_{=1}(\mathcal{A_1}+\mathcal{A}_2)) \times \textsc{MSET}_{=1} (\textsc{MSET}_{=2}(\mathcal{A_1}+\mathcal{A}_2)).$$

As an additional example we find for ${2,2,4\choose 1,1,3,3} = 16$

$$\textsc{MSET}_{=2} (\textsc{MSET}_{=1}(\mathcal{A_1}+\mathcal{A}_2+\mathcal{A}_3)) \times \textsc{MSET}_{=2} (\textsc{MSET}_{=3}(\mathcal{A_1}+\mathcal{A}_2+\mathcal{A}_3)).$$

Here we have used the cycle index of the symmetric group $Z(S_n)$, which is computed from the recurrence by Lovasz which says that

$$Z(S_n) = \frac{1}{n} \sum_{l=1}^n a_l Z(S_{n-l}) \quad\text{where}\quad Z(S_0) = 1.$$

For this to be effective we need to compute the iterated cycle index when $Z(S_k)$ is substituted into $Z(S_{\sigma_k}).$ This is accomplished with the substitution rule for substitution of the former into the latter:

$$a_q = Z(S_k;\; a_1=a_q, \; a_2=a_{2q}, \; a_3=a_{3q}, \; \ldots).$$

We have used the notation $Z(S_k; F)$ for substitution of a generating function and on the previous line, the notation for substitution into the variables of the cycle index. This is in fact all we need and we can start computing some of these multiset coefficients. For example we find for the example given by OP the cycle index

$$Z(B_1^2 B_2) = 1/4\,{a_{{1}}}^{4}+1/2\,{a_{{1}}}^{2}a_{{2}}+1/4\,{a_{{2}}}^{2}.$$

Continuing with the example we get

$$Z(B_1^2 B_2; A_1+A_2) = 1/4\, \left( A_{{1}}+A_{{2}} \right) ^{4} +1/2\, \left( A_{{1}}+A_{{2}} \right) ^{2} \left( {A_{{1}}}^{2}+{A_{{2}}}^{2} \right) \\ +1/4\, \left( {A_{{1}}}^{2}+{A_{{2}}}^{2} \right) ^{2} \\ = {A_{{1}}}^{4}+2\,{A_{{1}}}^{3}A_{{2}} +3\,{A_{{1}}}^{2}{A_{{2}}}^{2}+2\,A_{{1}}{A_{{2}} }^{3}+{A_{{2}}}^{4}$$

and we confirm the value $3$ obtained by OP. This algorithm will make it possible to compute cycle indices not obtainable by enumeration. As an extra example we find the following excerpt from the cycle index for $[2,2,2,3,5,5]:$

$$Z(B_2^3 B_3 B_5^2) = \ldots +{\frac {11\,{a_{{1}}}^{8}a_{{2}}a_{{4}}a_{{5}}}{7200}} +{\frac {49\,{a_{{1}}}^{7}{a_{{2}}}^{2}a_{{3}}a_{{5}}}{14400}} \\ +{\frac {5\,{a_{{1}}}^{7} a_{{2}}{a_{{3}}}^{2}a_{{4}}}{1152}} +{\frac {1021\,{a_{{1}}}^{6}{a_{{2}}}^{3}a_{{3}}a_{{4}}}{69120}} +{\frac {43\,{a_{{1}}}^{7}a_{{2}}a_{{4}}a_{{6}}}{17280}}+\ldots$$

Here are some additional values that may assist the reader who is investigating this problem to verify the results of their approach:

$${1,3,3\choose 3,4} = 7, \; {2,3,3\choose 4,4} = 5, \; {2,3,3\choose 2,2,4} = 16 \quad\text{and}\quad {1,2,3,3\choose 2,3,4} = 87.$$

The Maple code for this problem was as follows.

with(combinat);


pet_cycleind_symm :=
proc(n)
option remember;

    if n=0 then return 1; fi;

    expand(1/n*
           add(a[l]*pet_cycleind_symm(n-l), l=1..n));
end;

pet_varinto_cind :=
proc(poly, ind)
local subs1, subs2, polyvars, indvars, v, pot, res;

    res := ind;

    polyvars := indets(poly);
    indvars := indets(ind);

    for v in indvars do
        pot := op(1, v);

        subs1 :=
        [seq(polyvars[k]=polyvars[k]^pot,
             k=1..nops(polyvars))];

        subs2 := [v=subs(subs1, poly)];

        res := subs(subs2, res);
    od;

    res;
end;

pet_cycleind_comp :=
proc(idxtrg, idx)
local varstrg, vars, vt, sbstrg, sbs, len;

    varstrg := indets(idxtrg);
    vars := indets(idx);

    sbstrg := [];

    for vt in varstrg do
        len := op(1, vt);

        sbs :=
        [seq(v = a[op(1, v)*len], v in vars)];

        sbstrg :=
        [op(sbstrg),
         a[len] = subs(sbs, idx)];
    od;

    expand(subs(sbstrg, idxtrg));
end;

pet_cycleind_mset :=
proc(msetlst)
option remember;
local mset, res, ent, idxtrg, idx;

    mset := convert(msetlst, `multiset`);

    res := 1;

    for ent in mset do
        idx := pet_cycleind_symm(ent[1]);
        idxtrg := pet_cycleind_symm(ent[2]);

        res := res *
        pet_cycleind_comp(idxtrg, idx);
    od;

    expand(res);
end;


GENF :=
proc(src, msetlst)
local vars, srcp, res, gf, term;

    vars := add(A[q], q=1..nops(src));
    srcp := mul(A[q]^src[q], q=1..nops(src));

    gf := expand(pet_varinto_cind
                 (vars, pet_cycleind_mset(msetlst)));

    if not type(gf, `+`) then
        gf := [gf];
    fi;

    res := 0;

    for term in gf do
        if type(srcp/term, `polynom`) then
            res := res + term;
        fi;
    od;

    res;
end;

The syntax to compute ${\mathrm{A}\choose \mathrm{B}}$ is documented by the following examples:

> GENF([1,2,3,3], [2,3,4]);

                        2     3     3
            87 A[1] A[2]  A[3]  A[4]

> GENF([1,2,3,3], [2,2,5]);

                        2     3     3
            33 A[1] A[2]  A[3]  A[4]

> GENF([1,1,1,1], [2,2]);  

             3 A[1] A[2] A[3] A[4].

The last one is $\frac{1}{2} {4\choose 2}.$

Addendum. There is a slight improvement on this algorithm at the following MSE link.