<p>A badminton club has 10 couples as members. They meet to organise a mixed double match. If each wife refuse to partner as well as oppose her husband in the match then in how many different ways can the match be arranged?</p>
Step-by-Step Solution
Key Concept: This is a derangement problem combined with pairing: we need to find derangements D(10) where no wife is with or against her husband, then account for team formation and match arrangement.
<p><strong>Step 1: Understand the constraint</strong> Each of 10 wives must be paired with a husband other than her own AND must oppose (play against) a pair that doesn't include her own husband.</p><p><strong>Step 2: Form the pairs</strong> We need to partition 20 people into 4 pairs for a mixed doubles match (2 pairs per team). Each wife must partner with a different husband.</p><p><strong>Step 3: Key insight - Derangement approach</strong> First, arrange the 10 husbands in a sequence (10! ways). For each arrangement, assign the 10 wives such that wife i is NOT in position i (this is a derangement D₁₀). However, we need wives paired with different husbands AND opposing different pairs.</p><p><strong>Step 4: Calculate using inclusion-exclusion</strong> The number of ways to arrange 10 couples such that no wife partners with her husband is: <br>D(10) × (ways to partition into opposing teams)</p><p>Since we need 4 specific pairs with 2 vs 2 configuration: The derangement D(10) = 10! × (1 - 1/1! + 1/2! - 1/3! + ... + 1/10!)</p><p><strong>Step 5: Match arrangement</strong> Once valid pairs are formed, we need to select which 2 pairs form one team. This involves C(4,2)/2 = 3 ways (dividing by 2 since teams are interchangeable).</p><p><strong>Calculation:</strong> D(10) ≈ 1,334,961 (using derangement formula)</p><p>Total arrangements = D(10) × 2! × 2! × 3 = 1,334,961 × 4 × 3 ÷ 2</p><p>∴ <strong>Answer: 7,209,600 or equivalent derangement-based result depending on exact problem specification</strong></p>
Correct Answer: 7