<p>A man <i>X</i> has 7 friends, 4 of them are ladies and 3 are men. His wife <i>Y</i> also has 7 friends, 3 of them are ladies and 4 are men. Assume <i>X</i> and <i>Y</i> have no common friends. Then the total number of ways in which <i>X</i> and <i>Y</i> together can throw a party inviting 3 ladies and 3 men, so that 3 friends of each of <i>X</i> and <i>Y</i> are in this party, is</p>
Step-by-Step Solution
Key Concept: We need to partition 3 ladies and 3 men into contributions from X's and Y's friends such that exactly 3 friends from each person are invited. This requires considering all valid distributions where ladies and men contributions from X and Y sum to 3 each.
<p><strong>Step 1:</strong> Set up the problem. X has 4 ladies and 3 men. Y has 3 ladies and 4 men. We need 3 ladies, 3 men total, with exactly 3 friends from X and exactly 3 friends from Y.</p><p><strong>Step 2:</strong> Let x₁ = ladies from X, x₂ = men from X. Then y₁ = ladies from Y, y₂ = men from Y. Constraints: x₁ + x₂ = 3, y₁ + y₂ = 3, x₁ + y₁ = 3, x₂ + y₂ = 3.</p><p><strong>Step 3:</strong> From x₁ + x₂ = 3 and x₁ + y₁ = 3, we get x₂ = y₁. From x₁ + y₁ = 3 and y₁ + y₂ = 3, we get x₁ = y₂. Also x₂ + y₂ = 3 confirms consistency.</p><p><strong>Step 4:</strong> Since x₁ + y₁ = 3 and x₁ ≤ 4, y₁ ≤ 3, the possible cases are:<br/>• Case 1: x₁ = 0, y₁ = 3 (so x₂ = 3, y₂ = 0)<br/>• Case 2: x₁ = 1, y₁ = 2 (so x₂ = 2, y₂ = 1)<br/>• Case 3: x₁ = 2, y₁ = 1 (so x₂ = 1, y₂ = 2)<br/>• Case 4: x₁ = 3, y₁ = 0 (so x₂ = 0, y₂ = 3)</p><p><strong>Step 5:</strong> Calculate for each case:</p><p>Case 1: C(4,0)×C(3,3)×C(3,3)×C(4,0) = 1×1×1×1 = 1</p><p>Case 2: C(4,1)×C(3,2)×C(3,2)×C(4,1) = 4×3×3×4 = 144</p><p>Case 3: C(4,2)×C(3,1)×C(3,1)×C(4,2) = 6×3×3×6 = 324</p><p>Case 4: C(4,3)×C(3,0)×C(3,0)×C(4,3) = 4×1×1×4 = 16</p><p><strong>Step 6:</strong> Total = 1 + 144 + 324 + 16 = 485</p><p><strong>∴ Answer:</strong> B</p>
Correct Answer: B