<p>In how many ways can 3 boys and 15 girls sit together in a row such that between any 2 boys at least 2 girls sit?</p>
Step-by-Step Solution
Key Concept: Arrange the 15 girls first in a row (creating 16 gaps: before first girl, between consecutive girls, and after last girl), then choose 3 gaps with at least 2 girls between any two selected gaps. This ensures the constraint is automatically satisfied.
<p><strong>Step 1:</strong> Arrange 15 girls in a row: <strong>15!</strong> ways</p><p><strong>Step 2:</strong> When 15 girls sit in a row, they create 16 possible gaps (including ends): _G_G_G_..._G_</p><p><strong>Step 3:</strong> To ensure at least 2 girls sit between any two boys, we need to select 3 gaps from these 16 gaps such that no two selected gaps are adjacent or have only one gap between them. This is equivalent to choosing 3 gaps from the reduced set of 14 valid gap positions (gaps 1 through 14, skipping the constraint violations).</p><p><strong>Step 4:</strong> The valid number of ways to choose 3 gaps with the required spacing is <strong>{}^{14}C_3</strong> (choosing 3 positions from 14 where gaps are sufficiently spaced).</p><p><strong>Step 5:</strong> Arrange the 3 boys in the selected gaps: <strong>3!</strong> ways</p><p><strong>∴ Answer: \(3! \times 15! \times {}^{14}C_3\)</strong></p>
Correct Answer: \(3! \times 15! \times {}^{14}C_3\)