TITLE: How many ways to get an odd number of each color in each bin?
QUESTION [6 upvotes]: Suppose you have an odd number of white balls and the same number of blacks balls. 

How many different ways are there of putting the balls into bins so
  that you have an odd number of each color in each bin?

For example, if you have $3$ white and $3$ black there are $2$ different ways. You either put them all in one bin or one white and one black in each of $3$ bins.  For $5$ white and $5$ white balls there are $4$ different ways.  These are:
(wwwwwbbbbb)
(wwwbbb)(wb)(wb)
(wwwb)(wbbb)(wb)
(wb)(wb)(wb)(wb)(wb)

REPLY [5 votes]: The cycle  index $Z(S_n)$  of the  symmetric group  (multiset operator
$\def\textsc#1{\dosc#1\csod}     \def\dosc#1#2\csod{{\rm     #1{\small
#2}}}\textsc{MSET}$) has $Z(S_0)=1$ and the recurrence
$$Z(S_n) = \frac{1}{n}\sum_{l=1}^n a_l Z(S_{n-l}).$$ 
Extracting coefficients from this Maple will produce 
$$1, 2, 4, 12, 32, 85, 217, 539, 1316, 3146, 7374, 16969, 38387, 85452, 
\\ 187456, 405659, 866759, 1830086, 3821072, 7894447, 16148593, 
\\ 32723147, 65719405, 130871128, 258513076, 506724988,
\ldots$$ 
where we have used memoization. 
The repertoire here was 
$$f(W, B) = \sum_{p_1=0}^q \sum_{p_2=0}^q W^{2p_1+1} B^{2p_2+1},$$ 
the substitution $a_l = f(W^l, B^l)$  and the coefficient being extracted 
$$\sum_{k=1}^{2q+1} [W^{2q+1}] [B^{2q+1}] Z(S_k)(f(W,B)).$$
The Maple code runs as follows.

X :=
proc(n, q, q1, q2)
option remember;

    if n = 0 then
        if q1 = 0 and q2 = 0 then
            return 1;
        else
            return 0;
        fi;
    fi;

    add(add(add(X(n-l, q, q1-(2*p1+1)*l, q2-(2*p2+1)*l),
                p2=0..floor((q2/l-1)/2)),
            p1=0..floor((q1/l-1)/2)),
        l=1..n)/n;
end;

R := q -> add(X(k, q, 2*q+1, 2*q+1), k=1..2*q+1);