<p>In how many ways can 5 boys and 3 girls sit in a row so that no two girls are together?</p>
Step-by-Step Solution
Key Concept: First arrange 5 boys in a row creating 6 gaps (before first boy, between consecutive boys, after last boy), then select and arrange 3 girls in these 6 gaps to ensure no two girls sit together.
<p><strong>Step 1:</strong> Arrange 5 boys in a row. This can be done in <strong>5! ways</strong>.</p><p><strong>Step 2:</strong> When 5 boys are arranged, they create <strong>6 gaps</strong>: one before the first boy, one after each boy, and one after the last boy. These gaps are: _B_B_B_B_B_</p><p><strong>Step 3:</strong> To ensure no two girls sit together, we must place each girl in a different gap. We need to choose 3 gaps out of 6 available gaps and arrange 3 girls in them.</p><p><strong>Step 4:</strong> This can be done in <strong>⁶P₃ ways</strong> (permutation since order matters and we select 3 from 6 gaps).</p><p><strong>Step 5:</strong> By multiplication principle, total number of ways = <strong>5! × ⁶P₃</strong></p><p><strong>Calculation:</strong> 5! × ⁶P₃ = 120 × (6 × 5 × 4) = 120 × 120 = 14,400</p><p>∴ Answer: <strong>⁶P₃ × 5!</strong> or <strong>14,400</strong></p>
Correct Answer: \({}^6P_3 \times 5!\)