Question

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

• A. $\frac{1}{2}$
• B. $\frac{n}{2n-1}$
• C. $\frac{(n-1)}{(2n-1)}$
• D. $\frac{2n}{2n-1}$

Answer 1

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

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