2n boys are divided into two equal subgroups, find the probability that the two tallest are in the same subgroup 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?
 A: It's not that hard.  Say the tallest boys are A and B, and the groups are picked by lining them all up and dividing the first $n$ from the second $n$.  Then WLOG A is in the first $n$.  Then it is equally likely for B to be in any of the remaining $2n-1$ spots, of which $n-1$ would place him in the same group as A.  So the probability is $\boxed{\frac{n-1}{2n-1}}$.
A: You should have written $(2n)!$ since your denominator should be 
$$\frac{1}{2}\binom{2n}{n} = \frac{1}{2} \frac{(2n)!}{n!n!}$$
while $2n! = 2 \cdot n!$.  Otherwise, your answer is correct.
\begin{align*}
\frac{\binom{2n - 2}{n - 2}}{\frac{(2n)!}{n!n!2!}} & = \frac{(2n - 2)!}{(n - 2)!n!} \cdot \frac{n!n!2!}{(2n)!}\\
& = \frac{\color{orange}{(2n - 2)!}}{\color{green}{(n - 2)!}\color{blue}{n!}} \cdot \frac{\color{blue}{n!}n(n - 1)\color{green}{(n - 2)!} \cdot 2}{(2n)(2n - 1)\color{orange}{(2n - 2)!}}\\
& = \frac{\color{red}{2n}(n - 1)}{\color{red}{2n}(2n - 1)}\\
& = \frac{n - 1}{2n - 1}
\end{align*}
Notice that the simplified form agrees with the answer provided by Cyclicduck.
