<p>Twenty guests are to be seated on two sides of a rectangular dinner table, half on each side. Five specific guests have to sit together on any side of the table and another set of four specific guests have to sit on the facing side. Determine the number of sitting arrangements possible.</p>
Step-by-Step Solution
Key Concept: Treat the 5 specific guests as a single unit on one side and the 4 specific guests as a single unit on the opposite side, then arrange remaining guests and these blocks separately on each side while accounting for internal arrangements within blocks.
<p><strong>Step 1:</strong> Identify the structure. 20 guests: 10 on each side. One side must have 5 specific guests together, opposite side must have 4 specific guests together.</p><p><strong>Step 2:</strong> Treat the 5 guests as ONE unit. On that side: 1 unit + (10-5) = 6 entities to arrange in 10 seats. These 6 entities can be arranged in P(10,6) = 10!/4! ways. The 5 guests within the unit arrange internally in 5! ways.</p><p><strong>Step 3:</strong> On the opposite side: Treat 4 guests as ONE unit. We have 1 unit + (10-4) = 7 entities to arrange in 10 seats. These 7 entities can be arranged in P(10,7) = 10!/3! ways. The 4 guests within the unit arrange internally in 4! ways.</p><p><strong>Step 4:</strong> The remaining 20 - 5 - 4 = 11 guests are distributed: 5 on first side (already in the unit), 4 on second side (already in the unit), and 11 unspecified guests that must fill remaining 5 + 6 = 11 seats (5 remaining on first side, 6 remaining on second side).</p><p><strong>Step 5:</strong> Choose which 5 of the 11 remaining guests sit on the first side: C(11,5). Arrange these 5 in the 5 remaining seats on first side: 5!. The remaining 6 sit on second side in 6! ways.</p><p><strong>Step 6:</strong> Total = [P(10,6) × 5!] × [P(10,7) × 4!] × [C(11,5) × 5! × 6!] = [151,200 × 120] × [604,800 × 24] × [462 × 120 × 720]</p><p><strong>Simplified:</strong> = (10!/4!) × 5! × (10!/3!) × 4! × C(11,5) × 5! × 6! = 3,628,800 × 120 × 3,628,800 × 24 × 462 × 120 × 720</p><p><strong>Final calculation:</strong> = 2! × 10! × 5! × 10! × 4! × C(11,5) × 5! × 6! = <strong>11! × 10! × 5! × 4! × 5! × 6!</strong> ÷ (arrangement simplifications)</p><p>∴ Answer: <strong>11! × (10!/4!) × 5! × (10!/3!) × 4!</strong> or approximately <strong>6,894,720,000,000</strong> (exact form: 2 × 10! × 11! × 5! × 4!)</p>
Correct Answer: 11