How many ways can you partition a set with $2n$ elements into two sets? Suppose $M=\{x_1,\ldots,x_{2n}\}$. How many ways can we partition this into two sets s.t. their cardinality equals $n$? Recall how to make a partition of a set.
My work so far:
This is a combinatorial problem, and I think this cooks down to how many ways you can choose $n$ elements out of $2n$ without overcounting and then split it up into two sets. With this reasoning I've checked that it works for $|M|\leq 6$, but I don't know if this is right in general. Expressing it mathematically we get: $$\dfrac{^{2n}\mathrm{C}_n}{2}=\binom{2n}{n}:2=\dfrac{2n!}{(2n-n)!n!2}=\dfrac{2n!}{(n!)^22}$$
Can someone explain why this is correct or incorrect?
 A: First, choose $k$ items for first set. The remaining items will be in second set. Number of options of choosing $k$ items from $2n$ items is $\displaystyle {2n\choose k}$. Now we need to sum all the options, i.e $k$ between $1$ and $2n-1$, thus the number of ways is $$\sum_{k=1}^{2n-1}{2n\choose k}=\sum_{k=0}^{2n}{2n\choose k}-{2n\choose 0}-{2n\choose 2n}=2^{2n}-2=4^n-2$$
This number should be divided by $2$ as we are actually double counting the same partition, hence number of ways is $2^{2n-1}-1$.
A: Each element goes in set $1$ or set $2$. We have two choices for each element, leading to $2^{2n}$ choices in total. We need to remove a factor of $2$ because we don't care about which partition is which (presumably), so $2^{2n-1}$. Then we need to remove the possibility of one of the partitions being empty. There's only one way to choose that, so the answer is $2^{2n-1}-1$.
To compare to Galc127's answer, this is:
$$2^{2n-1} - 1 = (4^n - 2)/2$$
The factor of $2$ has been taken out because the two partitions are indistinguishable. With $m$ partitions, we would divide by $m!$ instead.
A: $M$ has $2^{2n}$ subsets and each subset $A\subseteq M$ induces a pair $\{A,A^c\}$. 
This pair is a partition of $M$ if $A\notin\{\varnothing,M\}$, so at first sight there are $2^{2n}-2$ partitions of $M$ that have $2$ elements.
Now observe that the sets $A$ and $A^c$ induce the same pair. 
Repairing this double counting we eventually find: $$\frac12(2^{2n}-2)=2^{2n-1}-1$$ partitions of $M$ that have two elements.

If you would be working under the (not mentioned) extra condition that both elements of the partition must have equal cardinality then we must look at pairs $\{A,A^c\}$ where $|A|=n$. 
There are $\binom{2n}{n}$ to choose a subset $A\subset M$ with $|A|=n$ so at first sight there are $\binom{2n}{n}$ partitions of $M$ that have two elements with equal cardinality.
Again there is double counting so eventually we find: $$\frac12\binom{2n}{n}$$ partitions of $M$ that have two elements with equal probability.
A: Partition of Set of $2^{2n } $ elements in to two sets. It depends on the way you can write $ 2n $ as sum of two numbers. I don't see why one partition cant be empty. It is about making a subset of cardinality $k$, where $k$ can be $0 \, to (2n/2).$. $ \, \sum_{k=0}^n {2n \choose k}$
A: Call the sets $A,\,A^c:=M\backslash A$. Choose whether $x_1\in A$ etc. in $2^{2n}$ ways. The symmetry $A\to A^c$ halves this to $2^{2n-1}$. If you insist neither set be empty, the answer is $2^{2n-1}-1$.
