<p>Number of ways in which 200 people can be divided in 100 couples is</p>
<p>\(\dfrac{(200)!}{2^{100}(100)!}\)</p>
<p>\(1 \times 3 \times 5 \times \cdots \times 199\)</p>
<p>\(\left(\dfrac{101}{2}\right)\left(\dfrac{102}{2}\right)\cdots\left(\dfrac{200}{2}\right)\)</p>
<p>\(\dfrac{(200)!}{(100)!}\)</p>
Step-by-Step Solution
Key Concept: When dividing 2n people into n indistinguishable couples, we must account for overcounting by dividing by n! (since the order of couples doesn't matter) and by 2^n (since each couple's internal order doesn't matter).
<p><strong>Step 1:</strong> Select 2 people from 200 for first couple: C(200,2) ways</p><p><strong>Step 2:</strong> Select 2 people from remaining 198 for second couple: C(198,2) ways</p><p><strong>Step 3:</strong> Continue this process for all 100 couples: C(200,2) × C(198,2) × C(196,2) × ... × C(2,2)</p><p><strong>Step 4:</strong> Since the 100 couples are indistinguishable (unordered), divide by 100!</p><p><strong>Step 5:</strong> Total ways = [C(200,2) × C(198,2) × ... × C(2,2)] / 100! = 200!/(2^100 × 100!)</p><p><strong>Derivation:</strong> This equals (200!)/(2^100 × 100!) because when we multiply all binomial coefficients, we get 200! in numerator and 2!^100 × 100! in denominator.</p><p>∴ Answer: 200!/(2^100 × 100!)</p>
Correct Answer: AB