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

The total number of ways in which 2n boys can be divided into two qual groups is

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

Now, the number of ways in which 2n boys can be divided into two equal group is

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

Two tallest boys can be put in different groups in ${}^2{C}_1$ ways,Hence,the required probability is

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