<p>X has 7 friends, 3 of them are males (M) and 4 are females (L). Y has 7 friends, 4 of them are males (M) and 3 are females (L). A party of 3 from X's friends and 3 from Y's friends is to be arranged such that there are 3 males and 3 females in the party. The total number of ways of arranging the party is:</p>
Step-by-Step Solution
Key Concept: The constraint of 3 males and 3 females total must be satisfied by selecting from both X's and Y's friends independently, then combining cases where male counts from X and Y sum to 3. You must enumerate all valid distributions: (males from X, males from Y) can be (0,3), (1,2), (2,1), or (3,0).
<p><strong>Step 1:</strong> Identify the constraint. We need 3 males and 3 females total from 6 people (3 from X, 3 from Y).</p><p><strong>Step 2:</strong> Enumerate all valid distributions of males and females across X and Y:</p><p>• <strong>Case 1:</strong> 0 males from X, 3 females from X; 3 males from Y, 0 females from Y<br/>Ways = C(3,0)×C(4,3) × C(4,3)×C(3,0) = 1×4 × 4×1 = 16</p><p>• <strong>Case 2:</strong> 1 male from X, 2 females from X; 2 males from Y, 1 female from Y<br/>Ways = C(3,1)×C(4,2) × C(4,2)×C(3,1) = 3×6 × 6×3 = 324</p><p>• <strong>Case 3:</strong> 2 males from X, 1 female from X; 1 male from Y, 2 females from Y<br/>Ways = C(3,2)×C(4,1) × C(4,1)×C(3,2) = 3×4 × 4×3 = 144</p><p>• <strong>Case 4:</strong> 3 males from X, 0 females from X; 0 males from Y, 3 females from Y<br/>Ways = C(3,3)×C(4,0) × C(4,0)×C(3,3) = 1×1 × 1×1 = 1</p><p><strong>Step 3:</strong> Add all cases: 16 + 324 + 144 + 1 = <strong>485</strong></p><p>∴ Answer: D</p>
Correct Answer: D