<p>\(2n\) boys are randomly divided into two subgroups containing \(n\) boys each. The probability that the two tallest boys are in different groups is</p>
Step-by-Step Solution
Key Concept: Once the first tallest boy is placed in a group (say Group A with n boys), there are n-1 remaining spots in Group A and n spots in Group B out of 2n-1 total remaining positions. The probability the second tallest boy goes to Group B is n/(2n-1).
<p><strong>Step 1:</strong> Fix the tallest boy in one group (say Group A). This is without loss of generality since we only care about whether the two tallest are separated.</p><p><strong>Step 2:</strong> Group A now has 1 boy and needs n-1 more boys. Group B needs n boys. There are 2n-1 remaining boys total to distribute.</p><p><strong>Step 3:</strong> For the second tallest boy to be in a different group (Group B), he must be selected for Group B. The probability is the number of spots available in Group B divided by total remaining positions: n/(2n-1)</p><p><strong>Step 4:</strong> Alternatively, the number of ways to put the 2 tallest boys in different groups divided by total ways to divide 2n boys into two groups of n each gives:</p><p>P = (n × n × C(2n-2, n-1)) / C(2n, n) = n/(2n-1)</p><p>∴ Answer: A</p>
Correct Answer: A