<p>In how many ways can 10 persons take seats in a row of 24 fixed seats so that no two persons take consecutive seats?</p>
Step-by-Step Solution
Key Concept: First select 10 non-consecutive seats from 24 using the gap method: arrange 14 empty seats creating 15 gaps, then choose 10 gaps for persons. Then arrange the 10 persons in the selected seats.
<p><strong>Step 1:</strong> Use the gap method to select 10 non-consecutive seats from 24.</p><p>When we place 10 persons with no two consecutive, we first imagine 10 persons separated by gaps. We need 14 empty seats (24 - 10 = 14) to create spacing.</p><p><strong>Step 2:</strong> Arrange 14 empty seats in a row: _ E _ E _ E _ ... _ E _</p><p>This creates 15 possible gaps (before first empty, between empties, after last empty) where we can place our 10 persons.</p><p><strong>Step 3:</strong> Choose 10 gaps from these 15 gaps for the 10 persons: \(\binom{15}{10}\) ways.</p><p><strong>Step 4:</strong> Arrange the 10 distinct persons in the 10 selected seats: \(10!\) ways.</p><p><strong>Step 5:</strong> Total ways = \(\binom{15}{10} \times 10!\)</p><p>Note: \(\binom{15}{10} = \binom{15}{5} = 3003\), so answer = \(3003 \times 10! = 3003 \times 3628800\)</p>
Correct Answer: \(\binom{15}{10} \times 10!\)