2n boys are randomly divided into two subgroups containing n boys each. The probability that the two tallest boys are in different groups is

The total number of ways in which2nboys can be divided into two equal groups is

$=\frac{\left(2\mathrm{n}\right)!}{(\mathrm{n}!{)}^{2}2!}$

Now, the number of ways in which 2n - 2 boys other than the two tallest boys can be divided into two equal groups is 

$=\frac{(2\mathrm{n}-2)!}{\left(\right(\mathrm{n}-1)!{)}^{2}2!}$

Two tallest boys can be put in different groups in${ }^2{\mathrm{C}}_1$ ways.

Hence, the required probability is

$\begin{array}{c}=\frac{2\frac{(2\mathrm{n}-2)!}{\left(\right(\mathrm{n}-1)!{)}^{2}2!}}{\frac{\left(2\mathrm{n}\right)!}{(\mathrm{n}!{)}^{2}2!}}\\ =\frac{\mathrm{n}}{2\mathrm{n}-1}\end{array}$