Counting number of balanced two-way partition of the set

Question

Given a set with $2n$ elements, show that the number of balanced two-way partition of the set $$P(2n)=\frac{2n!}{2\times n!\times n!}$$

I'm getting is as P(2n)=${2n}\choose{n}$. But I'm getting different ans, Am I assuming something wrong.

Answer 1

As already noted in the comments, one of the mistakes in your answer is that you considered all partitions into two subsets, not just balanced partitions, i.e. partitions into subsets of equal size, and taking this into account improves your answer from $2^{2n}$ to $\binom{2n}n$. The missing factor of $2$ arises because conventionally, in partitioning sets into subsets, while one treats the elements of the set as distinguishable (in contrast to partitioning numbers, where only the size of the parts matters), one treats the subsets as indistinguishable, e.g. the partitions of $[1,2n]$ into $[1,n]$ and $[n+1,2n]$ and into $[n+1,2n]$ and $[1,n]$ are the same partition.