<p>A man X has 7 friends, 4 of them are ladies and 3 are men. His wife Y also has 7 friends, 3 of them are ladies and 4 are men. Assume X and Y have no common friends. Then, the total number of ways in which X and Y together can throw a party inviting 3 ladies and 3 men, so that 3 friends of each of X and Y are in this party, is</p>
Step-by-Step Solution
Key Concept: The constraint '3 friends of each of X and Y' means we must partition the 3 ladies and 3 men such that exactly 3 come from X's friends and exactly 3 come from Y's friends. This requires careful casework based on how ladies and men are distributed between X's and Y's friend groups.
<p><strong>Step 1: Identify the constraint structure</strong></p><p>We need 3 ladies and 3 men total, with exactly 3 friends from X and exactly 3 friends from Y (since no common friends exist).</p><p><strong>Step 2: Determine possible distributions</strong></p><p>If we take 'a' ladies and 'b' men from X's friends, then: a + b = 3 and (3-a) + (3-b) = 3. This gives a + b = 3 and 6 - a - b = 3, confirming our constraint.</p><p>Since X has 4 ladies and 3 men, and Y has 3 ladies and 4 men, we have two cases:</p><p><strong>Case 1: 2 ladies and 1 man from X; 1 lady and 2 men from Y</strong></p><p>Number of ways = C(4,2) × C(3,1) × C(3,1) × C(4,2)</p><p>= 6 × 3 × 3 × 6 = 324</p><p><strong>Case 2: 1 lady and 2 men from X; 2 ladies and 1 man from Y</strong></p><p>Number of ways = C(4,1) × C(3,2) × C(3,2) × C(4,1)</p><p>= 4 × 3 × 3 × 4 = 144</p><p><strong>Step 3: Add both cases</strong></p><p>Total = 324 + 144 = 468</p><p>∴ Answer: D</p>
Correct Answer: D