<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>
<p>\(^{15}C_{10} \times 10!\)</p>
<p>\(^{15}C_{10} \times ^{10}P_{10}\)</p>
<p>\(^{15}C_{10}\)</p>
<p>\(^{24}C_{10} \times 10!\)</p>
Step-by-Step Solution
Key Concept: Arrange 10 persons first, then place them in 24 seats with non-consecutive constraint. This is equivalent to choosing 10 non-consecutive positions from 24, which requires placing 10 persons among 14 gaps created by leaving mandatory spaces between them.
<p><strong>Step 1:</strong> Understand the constraint. We need 10 persons in 24 seats such that no two sit consecutively (at least one empty seat between any two persons).</p><p><strong>Step 2:</strong> First, select 10 non-consecutive positions from 24. Imagine first placing 10 persons in a row: _ P _ P _ P _ ... _ P _ (this creates 11 gaps: before first, between each pair, and after last).</p><p><strong>Step 3:</strong> To ensure non-consecutiveness in our row of 24, we need at least 1 empty seat between consecutive persons. Think of it as: we have 10 persons and must place 9 mandatory spaces between them. This accounts for 10 + 9 = 19 seats.</p><p><strong>Step 4:</strong> We have 24 - 19 = 5 remaining empty seats to distribute freely among the 11 available gaps (before first person, between persons, after last person). This is equivalent to choosing 5 gaps from 11 gaps, which is C(11,5).</p><p><strong>Step 5:</strong> Alternatively, use the standard formula: number of ways to choose k non-consecutive items from n items = C(n-k+1, k) = C(24-10+1, 10) = C(15, 10).</p><p><strong>Step 6:</strong> Since the 10 persons are distinct, arrange them in the selected 10 seats in 10! ways.</p><p><strong>Step 7:</strong> Total arrangements = C(15,10) × 10! = C(15,5) × 10! = 3003 × 10!</p><p>∴ Answer: <strong>B (3003 × 10!)</strong></p>
Correct Answer: B