`2n` boys are randomly divided into two subgroups containint `n` boys each. The probability that eh two tallest boys are in different groups is `n//(2n-1)` b. `(n-1)//(2n-1)` c. `(n-1)//4n^2` d. none of these

Options

A. `n//(2n - 1)`
B. `(n-1)(2n-1)`
C. `(n-1)//4n^(2)`
D. none of these

Answer

A

Explanation

The total number of ways in which 2n boys can be divided into two equal groups is
`=((2n)!)/((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
`=((2n - 2)!)/(((n - 1)!)^(2)2!)`
Two tallest boys can be put in different groups in `.^(2)C_(1)` ways. Hence, the required probability is
`= (2((2n-2)!)/(((n - 1)!)^(2)2!))/(((2n)!)/((n!)^(2)2!))`
`= (n)/(2n - 1)`