Number of 2n-1 equal size partitions up to symmetry Consider the $K_{2n}$ (or just the set $\{1,\dots,2n\}$) with $S_{2n}$ acting on the vertices. Moreover consider a collection of 2n-1 partitions of the vertices into two equal sized sets (repeated partitions are allowed). 
How many different kinds of such collections can there be up to symmetry and how can you construct them? 
For example, in the $K_{6}$ the collections (writing just the first set of the partition):
$$\mathcal{S}=\{\{1,3,6\}, \{1,4,5\},\{1,4,6\},\{1,5,6\},\{1,5,6\}\}$$
and 
$$\mathcal{S'}=\{\{1,3,6\},\{1,4,5\},\{1,4,5\},\{1,4,6\},\{1,5,6\}\}$$
are different up to symmetry.
Are there papers about this topics or similar results?
Thanks in advance!
Richard
 A: NOTE to the reader of this post : this answers an interesting question but not the OP's question.  This question is still open.
There are $p = \frac{1}{2}{{2n} \choose n}$ possible distinct partitions - half of all the ${{2n} \choose n}$ partitions since each partition has a symmetric counterpart.
And each collection has $k = 2n - 1$ of them.  
Sets (or collections) that allow repetition are called multisets.  So really, this becomes the problem of finding out how many unique multisets of size $k$ you can make out of $p$ elements.
http://en.wikipedia.org/wiki/Multiset_coefficient#Counting_multisets
The answer is ${p + k - 1 \choose k}$.  Or to be specific, ${\frac{1}{2}{2n \choose n} + 2n - 2 \choose 2n - 1}$.
If you want to construct them all, to simplify things, you can give a number to each partition and represent them as the integer set $P = \{1, 2, \ldots p\}$.
Then you should be able to generate all multisets of size $k$ out of $p$ elements (though there are a whole lot of them).
