<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: Break the problem into cases based on how many ladies and men are selected from each person's friend group, ensuring the constraint of selecting exactly 3 friends from each.
<p><strong>Step 1:</strong> X has 4 ladies and 3 men. Y has 3 ladies and 4 men.</p><p><strong>Step 2:</strong> We need 3 ladies and 3 men total, with 3 friends from X and 3 friends from Y.</p><p><strong>Step 3:</strong> Possible distributions:</p><p>Case 1: 2 ladies and 1 man from X; 1 lady and 2 men from Y</p><p>Ways = C(4,2) × C(3,1) × C(3,1) × C(4,2) = 6 × 3 × 3 × 6 = 324</p><p>Case 2: 1 lady and 2 men from X; 2 ladies and 1 man from Y</p><p>Ways = C(4,1) × C(3,2) × C(3,2) × C(4,1) = 4 × 3 × 3 × 4 = 144</p><p>Case 3: 3 ladies and 0 men from X; 0 ladies and 3 men from Y</p><p>Ways = C(4,3) × C(3,0) × C(3,0) × C(4,3) = 4 × 1 × 1 × 4 = 16</p><p>Case 4: 0 ladies and 3 men from X; 3 ladies and 0 men from Y</p><p>Ways = C(4,0) × C(3,3) × C(3,3) × C(4,0) = 1 × 1 × 1 × 1 = 1</p><p><strong>Total = 324 + 144 + 16 + 1 = 485</strong></p><p>∴ Answer is C.</p>
Correct Answer: C