Probability of dividing boys in 2 groups A group of $2n$ boys is to be divided into two groups of $n$ boys . What is the probability that the two tallest boys are in different groups ? 
This is how I attempted it: 
The probability that the two boys are in same group can be obtained as follows:
First we separate those two particular boys leaving us with $2n-2$ boys. We then form a group of $n$ boys not containing the two particular boys giving us a group $n-2$ boys in which the two boys can be accommodated. The probability of forming such groups is $\frac{\binom{2n-2}{n}}{\binom{2n}{n}}$. Thus the actual probability of forming the groups with the two boys in different groups is $1- \frac{\binom{2n-2}{n}}{\binom{2n}{n}}$ . However there seems a problem with this. Could you please point out where I was wrong ? 
Also , please note that I already know the correct solution to this problem. I just wanted to correct my mistake. 
Thanks for your help !
 A: Given that $2n$ boys are randomly divided into two subgroups containing $n$ boys each.
So, we can choose $n$ boys from $2n$ boys in $\dfrac{\dbinom{2n}{n}}{2!}=\dfrac{(2n)!}{2!\times(n!)^2}$ ways.
We need to find the probability of two tallest boys in different groups.
So, if we seperate $2$ tallest boys from $2n$ boys we have $2n-2$ boys.
Now we can arrange the group of boys in $\dfrac{\dbinom{2n-2}{n-1}}{2!}=\dfrac{(2n-2)!}{((n-1)!)^2\times 2!}$ ways
Now we can separate these $2$ tallest boys into $2$ different groups in $\dbinom{2}{1}$ ways.
So, in total we have $\dfrac{(2n-2)!}{((n-1)!)^2\times 2!}\times \dbinom{2}{1}$ ways
The required probability is $\dfrac{\dfrac{(2(n-1))!\times2}{((n-1))^2\times2!}}{\dfrac{(2n)!}{(n!)^2\times2!}}=\dfrac{n}{2n-1}$
A: Let us assume that the tallest boys are Andrew and Bruce. The configurations of this kind
$$ (A\text{ together with }n-1\text{ other people })\quad (B\text{ together with }n-1\text{ other people }) $$
are $\binom{2n-2}{n-1}$ (it is enough to select Andrew's mates), while the configurations of this kind
$$ (A,B\text{ together with }n-2\text{ other people })\quad (n\text{ other people }) $$
are $\binom{2n-2}{n-2}=\binom{2n-2}{n}$ (it is enough to select Andrew and Bruce's mates). The wanted probability is so
$$ \frac{\binom{2n-2}{n-1}}{\binom{2n-2}{n-1}+\binom{2n-2}{n}}=\frac{\binom{2n-2}{n-1}}{\binom{2n-1}{n-1}}=\color{red}{\frac{n}{2n-1}}. $$
