2n boys are randomly divided into two sub-groups containing n boys each. Find the probability that two TOP RANK-HOLDER are in differed groups.	Select:
A) (n-1) / 2n
B) (n-1) / (2n-1)
C) n / (2n-1)
D) None of these
E) Cannot be determined
Explanation :   Without restriction   No. of ways to divide 2n boys into 2 equal groups having n boys each = 2nCn x nCn.   With restriction,   No. of ways to have two top ranks-holder in different group = 2n-2Cn-1 x n-1Cn-1 x 2!.   ( i.e separate two top rankers and then make two groups out of (2n - 2) boys.   The required probability   = Selections with restriction / Selections without restriction .   = (2n-2Cn-1 x n-1Cn-1 x 2! ) / (2nCn x nCn ).   = { (2n-2)! / [ (n-1)! x (n-1)! ] } x 2! / (2n! / ( n! x n! ) ) .   = (n-1) / 2n - 1.   Hence, the probability is (n-1) / (2n-1).   Answer : (n-1) / (2n-1)
<result>B</result>
2n boys are randomly divided into two sub-groups containing n boys each. Find the probability that two TOP RANK-HOLDER are in differed groups.