If 2n boys are divided into two equal subgroups randomly, find the probability that the two tallest boys are in the same subgroup.

Here is what I though:  

There are $$\frac{2n!}{n!\cdot n!\cdot2!}$$ ways to divide 2n boys into two subgroups.  

The tallest two will be in one group, thus the rest of boys have $$\begin{pmatrix}2n-2 \\n-2\end{pmatrix}$$ possibilities.  

So my answer is $$\frac{\begin{pmatrix}2n-2 \\n-2\end{pmatrix}}{\frac{2n!}{n!\cdot n!\cdot2!}}$$  

I'm not really sure how to simplify this, and it seems too complicated to be the right answer...  

Is there anywhere I made a mistake or any other easier way to do this problem?