Question

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

• A. $\frac{n}{2n-1}$
• B. $\frac{n-1}{2n-1}$
• C. $\frac{2n-1}{4n^2}$
• D. None of these

Answer 1

The correct option is A $\frac{n}{2n-1}$

Number of ways of forming two groups $=\frac{\left(2n\right)!}{n!n!}$
Leaving two tallest boys we can divide $2n-2$ boys into two groups in $\frac{(2n-2)!}{(n-1)!(n-1)!}$.
But the two tallest boys can be in any of the groups, each in different.
So favorable number of cases $\frac{2(2n-2)!}{(n-1)!(n-1)!}=\frac{n}{(2n-1)}$