TITLE: Number of ways to partition $40$ balls with $4$ colors into $4$ baskets
QUESTION [11 upvotes]: Suppose there are $40$ balls with $10$ red, $10$ blue, $10$ green, and $10$ yellow. All balls with the same color are deemed identical. Now all balls are supposed to be put into $4$ identical baskets, such that each basket has $10$ balls. What is the number of ways to partition these balls?
I tried this problem, but it seems very complicated to correctly formulate, because the number of a particular color in a basket determines the partition of other baskets. I wonder someone can help figure out a quick and clean way to solve this problem?

REPLY [4 votes]: Consider the problem of $4n$ balls with  $n$ balls of each of the four
colors  being distributed  into four  indistinguishable baskets  where
each basket holds exactly $n$ balls.  The naive approach here would be
to use the Polya Enumeration Theorem (twice). Surprisingly enough this
is sufficient to compute the initial segment of the sequence using the
recurrence  by Lovasz  for the  cycle index  $Z(S_n)$ of  the multiset
operator      $\def\textsc#1{\dosc#1\csod}     \def\dosc#1#2\csod{{\rm
#1{\small #2}}} \textsc{MSET}_{=n}$ on $n$ slots, which is
$$Z(S_n) = \frac{1}{n} \sum_{l=1}^n  a_l Z(S_{n-l})
\quad\text{where}\quad
Z(S_0) = 1.$$
This  recurrence  lets us  calculate  the  cycle  index $Z(S_n)$  very
easily. The answer is then given by
$$[A^n B^n C^n D^n] Z(S_4)(Z(S_n)(A+B+C+D)).$$
Using Maple  and a reasonable  amount of computational  resources this
yields the sequence
$$1, 17, 93, 465, 1746, 5741, 16238, 41650, 97407, 212412, 434767, 
\\ 845366, 1569344, 2801696, 4828140, 8069053, 
\\ 13114785, 20796651, 32242621, 48986553, 73052382, 107114645, 
\\ 154621230, 220021932, 308940815,\ldots$$
In particular the value for $n=10$ is given by
$$212412.$$
This is  OEIS A253259 where  we discover a
variation  of the  problem definition  that confirms  the  validity of
these results. Oddly enough no recurrence relation or other indication
of how these numbers were computed is given in the OEIS entry. Perhaps
we will see a recurrence now that there are enough test data to verify
its correctness, if indeed it exists.
The Maple code for the above is quite straightforward.

with(combinat);

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_symm :=
proc(n)
local l;
option remember;

    if n=0 then return 1; fi;

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

V :=
proc(n)
    option remember;
    local comb, gf, var;

    comb := pet_varinto_cind(A+B+C+D, pet_cycleind_symm(n));
    gf := 
    expand(pet_varinto_cind(comb, pet_cycleind_symm(4)));

    for var in [A,B,C,D] do
        gf := coeff(gf, var, n);
    od;

    gf;
end;

Addendum. I realized I had  overlooked an additional OEIS link to
A257463  when  I  wrote the  above  several
hours ago. It  proposes a simple algorithm which  uses the isomorphism
between the  problem and factorizations  of $p_1^n p_2^n  p_3^n p_4^n$
into four factors $q$ all of which have $\Omega(q) = n.$ The algorithm
generates  all  of these  using  the  observation  that uniqueness  of
factorizations  can  be  guaranteed   by  generating  the  factors  in
non-increasing  order.    It  uses  memoization  to   speed  this  up.
Nonetheless  when I  pasted it  verbatim into  Maple and  tried  it on
several test  cases it performed very  poorly compared to  what I have
above. I will therefore keep the post for now. The quest continues.
Addendum II. As  per request by @WillOrrick I am  posting the code
for the general problem of $k$ colors.

V :=
proc(n, k)
    option remember;
    local base, comb, gf, var;

    base := add(Q[p], p=1..k);

    comb := pet_varinto_cind(base, pet_cycleind_symm(n));
    gf :=
    expand(pet_varinto_cind(comb, pet_cycleind_symm(k)));

    for var in [seq(Q[p], p=1..k)] do
        gf := coeff(gf, var, n);
    od;

    gf;
end;

We thus obtain for five colors the sequence
$$1, 73, 1417, 19834, 190131, 1398547, 8246011, 40837569, 174901563,
\\ 664006236, 2274999093, 7139338769, 20758868781, 56466073587,
\\ 144806582536, 352420554194, 818441723112, 1822255658908,\ldots$$
Addendum III. Granting Maple several hours of computation time and
5GB of memory we get for four colors:
$$1, 17, 93, 465, 1746, 5741, 16238, 41650, 97407, 212412, 434767, 
\\ 845366, 1569344, n2801696, 4828140, 8069053, 13114785, 20796651, 
\\ 32242621, 48986553, 73052382, 107114645, 154621230, 220021932, 
\\ 308940815, 428492880, 587520315, 797019526, 1070458096, 1424339518, 
\\ 1878618620, 2457435561, 3189651885, 4109787687, 5258703597,\ldots$$
This confirms the generating function by @WillOrrick.