<p>A man has 7 relatives, 4 of them are ladies and 3 gentlemen; his wife has 7 relatives, 3 of them are ladies and 4 gentlemen. In how many different ways can they invite a dinner party of 3 ladies and 3 gentlemen so that there are 3 of the man's relatives and 3 of the wife's relatives?</p>
Step-by-Step Solution
Key Concept: The constraint requires exactly 3 from man's relatives AND 3 from wife's relatives. You must distribute 3 ladies and 3 gentlemen across two groups simultaneously, ensuring the total from each side equals 3.
<p><strong>Step 1: Identify the constraint structure</strong></p><p>We need 3 ladies and 3 gentlemen total, with exactly 3 from man's relatives and 3 from wife's relatives.</p><p><strong>Step 2: Set up cases based on gender distribution from man's side</strong></p><p>If man contributes <em>x</em> ladies and (3-<em>x</em>) gentlemen, then wife must contribute (3-<em>x</em>) ladies and <em>x</em> gentlemen.</p><p><strong>Step 3: List all valid cases</strong></p><ul><li><strong>Case 1:</strong> Man gives 0 ladies, 3 gentlemen → Wife gives 3 ladies, 0 gentlemen<br>C(4,0) × C(3,3) × C(3,3) × C(4,0) = 1 × 1 × 1 × 1 = 1</li><li><strong>Case 2:</strong> Man gives 1 lady, 2 gentlemen → Wife gives 2 ladies, 1 gentleman<br>C(4,1) × C(3,2) × C(3,2) × C(4,1) = 4 × 3 × 3 × 4 = 144</li><li><strong>Case 3:</strong> Man gives 2 ladies, 1 gentleman → Wife gives 1 lady, 2 gentlemen<br>C(4,2) × C(3,1) × C(3,1) × C(4,2) = 6 × 3 × 3 × 6 = 324</li><li><strong>Case 4:</strong> Man gives 3 ladies, 0 gentlemen → Wife gives 0 ladies, 3 gentlemen<br>C(4,3) × C(3,0) × C(3,0) × C(4,3) = 4 × 1 × 1 × 4 = 16</li></ul><p><strong>Step 4: Add all cases</strong></p><p>Total = 1 + 144 + 324 + 16 = <strong>485</strong></p>
Correct Answer: 485