Question

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

Answer 1

Correct Answer - A
The total number of ways of dividing 2n boys equally into a first and then a second subgroup is
`.^(2n)C_(n)xx .^(n)C_(n)=.^(2n)C_(n)=(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)`.
`therefore` Favourable number of ways `=2xx .^(2n-2)C_(n-1)`
Required probability `=((2xx .^(2n-2)C_(n-1))/((2n)!))/(n!n!)=(n)/(2n-1)`