<p>In how many ways can 10 people take seats in 24 fixed seats so that out of every pair of seats equidistant from the beginning and end at least one seat is empty?</p>
Step-by-Step Solution
Key Concept: Recognize that seats equidistant from ends form complementary pairs (1st-24th, 2nd-23rd, etc.), creating 12 pairs. The constraint means from each pair, at most one can be occupied, requiring us to select which seats to use before arranging people.
<p><strong>Step 1:</strong> Identify the structure. The 24 seats form 12 complementary pairs: (1,24), (2,23), (3,22), ..., (12,13). The constraint states that from each pair, at most one seat can be occupied.</p><p><strong>Step 2:</strong> Since we need to seat 10 people and each pair contributes at most 1 occupied seat, we must select exactly 10 pairs from the 12 available pairs, and from each selected pair, choose which of its 2 seats to occupy.</p><p><strong>Step 3:</strong> Number of ways to choose 10 pairs from 12: C(12,10) = C(12,2) = 66</p><p><strong>Step 4:</strong> For each chosen pair, decide which seat (left or right) is occupied: 2^10 ways</p><p><strong>Step 5:</strong> Arrange 10 people in the 10 selected seats: 10! ways</p><p><strong>Step 6:</strong> Total = C(12,10) × 2^10 × 10! = 66 × 1024 × 3,628,800</p><p><strong>Calculation:</strong> 66 × 1024 × 10! = 67,584 × 10!</p><p>∴ <strong>Answer: 66 × 2^10 × 10! or 67,584 × 10!</strong></p>
Correct Answer: 66