Procreation with several genders Most mathematicians are aware that our species consists of two genders, 
denoted for simplicity by the multisets $\lbrace X,X\rbrace$ and $\lbrace X,Y\rbrace$, with 
offspring given by $\lbrace A,B\rbrace$ for $A\in\lbrace X,X\rbrace$ 
and $B\in \lbrace X,Y\rbrace$.
I am asking for the existence of combinatorial structures generalizing this construction. 
More precisely, given a finite set $\mathcal C$ of $n$ distinct elements
($n=2$ and $\mathcal C=\lbrace X,Y\rbrace$ in the above example) and an integer $k$, we denote by $\mathcal M_k$ the set of all multisets containing exactly 
$k$ not necessarily distinct elements of $\mathcal C$. 
A gender partition of a subset $\mathcal S\subset \mathcal M_k$ is a partition of $\mathcal S$ into $k$ non-empty parts $\mathcal G_1,\dots,\mathcal G_k$ called genders such that the following two conditions hold: 
(i) Given $(g_1,\dots,g_k)\in\mathcal G_1\times \dots
\times \mathcal G_k$, every element $(x_1,\dots,x_k)\in g_1\times \dots\times g_k$ gives rise to a multiset $\lbrace x_1,\dots,x_k\rbrace$ which is in $\mathcal S$.
(ii) Every multiset $g\in\mathcal S$ is of the form $\lbrace x_1,\dots,x_k\rbrace$
for $(x_1,\dots,x_k)\in g_1\times \dots\times g_k$ where $g_i\in
\mathcal G_i$ are suitable elements.
Examples with $\mathcal C=\lbrace X,Y\rbrace$ are:
(a) $\mathcal S= \lbrace X,X\rbrace\cup \lbrace X,Y\rbrace$ with $\mathcal G_i$ 
given by singletons.
(b) $\mathcal S= \lbrace \lbrace X,X\rbrace,\lbrace Y,Y\rbrace\rbrace
\cup \lbrace X,Y\rbrace$ with $\mathcal G_1=\lbrace \lbrace X,X\rbrace,\lbrace Y,Y\rbrace\rbrace$ and $\mathcal G_2$ consisting of $\lbrace X,Y\rbrace$.
(c) An example with $k=3$ (easily generalizable to arbitrary values of $k$)
is given by $\mathcal S=\lbrace \lbrace X,X,X\rbrace,\lbrace X,X,Y\rbrace,
\lbrace X,Y,Y\rbrace\rbrace$ with $\mathcal G_i$ given by singletons.
More examples of gender partitions $\mathcal S=\mathcal G_1\cup
\dots\cup \mathcal G_k$ are fairly easy to construct. (And there are fairly easy
notions for "products", "quotients", one can split an element of $\mathcal C$ 
into several new elements, etc.)
The following additional condition is more difficult to satisfy:
Call a gender partition $\mathcal S=\mathcal G_1\cup \dots\cup\mathcal G_k$
balanced if $\mathcal S$ admits a stationary probability measure 
$\mu$ giving equal weight $\frac{1}{k}$ to all genders $\mathcal G_i$.
A probability measure $\mu$ on $\mathcal S$ is stationary if 
the probabiliy $\mu(\lbrace x_i,\dots,x_k\rbrace)$ of every offspring 
of $(g_1,\dots,g_k)$ (with respect to uniform choices for $x_i\in g_i$)
is proportional to $\prod_{i=1}^k\mu(g_i)$. (Stationary probability 
measures exist always and are unique if $\mathcal S$ is
minimal in some sense.)
Example: The examples (a) and (b) above are balanced, (c) is not balanced.
Question: Produce other examples of balanced gender partitions.
Is there for example a balanced gender partition for $k=3$?
Remark: One can also consider probabilities on offsprings 
which depend on the choice of $x_i\in g_i$. Example (c) is not balanced
even in this more general framework.
Variation: Instead of working with multisets, one can also work with 
sequences of length $k$. An offspring of $k$ sequences
$g_1=(g_1(1),\dots,g_1(k)),\dots,g_k=(g_k(1),\dots,g_k(k))$ with 
$g_i\in\mathcal G_i$ is then given by 
$g_1(\sigma(1)),\dots,g_k(\sigma(k))$ where $\sigma$ is a (not necessarily arbitrary) permutation of 
$\lbrace 1,\dots,k\rbrace$.
 A: So, I hope I understand the definitions correctly.  Here's a way to construct an example with $k = 3$ genders (say A, B and C) using $n = 9$ sex chromosomes, which I will take to be the elements of $\mathbb{Z}/9\mathbb{Z}$: start with all 165 multisets of chromosomes unused.  Choose any unused multiset $\{x, y, z\}$, and add it, together with the multisets $\{x + 3, y + 3, z + 3\}$ and $\{x + 6, y + 6, z + 6\}$, to gender A, add the multisets $\{x + 1, y + 1, z + 1\}, \{x + 4, y + 4, z + 4\}, \{x + 7, y + 7, z + 7\}$ to gender B and add the multisets $\{x + 2, y + 2, z + 2\}, \{x + 5, y + 5, z + 5\}, \{x + 8, y + 8, z + 8\}$ to gender C.  Continue until every multiset of chromosomes is used up.  The resulting genders (each consisting of 55 multisets of chromosomes) are symmetric and so (if I'm not mistaken) are balanced.
It's easy to see how to construct a wide variety of similar gender partitions for given $k$ if we may choose $n$ appropriately.  These partitions have nice symmetry and use all possible multisets of chromosomes.  This says nothing at all about, say, constructing gender partitions for $n = 2$ (though I believe that I've confirmed by case analysis that there are no balanced gender partitions for $k = 3$ and $n = 2$).
