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

$\frac{n}{2n-1}$

The total number of ways of dividing 2n boys equally into a first and then a second subgroup is

$ {^{2n}C}_n × {^nC}_n = {^{2n}C}_n =\frac{2n!}{n!n!}$

There are 2 ways in which the two tallest boys lie in different groups and corresponding to each way the remaining (2n-2) boys can be divided into two groups is ${^{2n-2}C}_{n-1}.$

∴ Favourable number of ways $= 2 × {^{2n-2}C}_{n-1}$

Required probability $=\frac{2 × {^{2n-2}C}_{n-1}}{\frac{(2n)!}{n!n!}}=\frac{n}{2n-1}$