<p>Eighteen guests have to be seated, half on each side of a long table. Four particular guests desire to sit on one particular side and three others on the other side. Determine the number of ways in which the sitting arrangements can be made.</p>
Step-by-Step Solution
Key Concept: First fix the constrained guests (4 on side A, 3 on side B), then distribute remaining 11 guests by choosing 5 more for side A. Finally, arrange all 9 guests on each side independently.
<p><strong>Step 1:</strong> Total guests = 18, seated 9 on each side of table.</p><p><strong>Step 2:</strong> Constraint: 4 particular guests want side A, 3 particular guests want side B. This leaves 18 - 4 - 3 = 11 guests to be distributed.</p><p><strong>Step 3:</strong> Side A will have 4 (fixed) + 5 (to choose) = 9 guests. Side B will have 3 (fixed) + 6 (remaining) = 9 guests. Choose which 5 of the 11 remaining guests go to side A: <strong>11C5</strong> ways.</p><p><strong>Step 4:</strong> Arrange the 9 guests on side A in their 9 seats: <strong>9!</strong> ways.</p><p><strong>Step 5:</strong> Arrange the 9 guests on side B in their 9 seats: <strong>9!</strong> ways.</p><p><strong>Step 6:</strong> Total arrangements = 11C5 × 9! × 9!</p><p>∴ Answer: <strong>11C5 × 9! × 9!</strong></p>
Correct Answer: 11C5 × 9! × 9!