<p>If <i>n</i> is even, the number of ways of arranging <i>n</i> persons if each person has exactly one neighbor is</p>
<p>\({}^{n-1}P_{n/2} \cdot {}^{n+1}P_{n/2}\)</p>
<p>\({}^{n}P_{n/2} \cdot {}^{n+1}P_{n/2}\)</p>
<p>\({}^{n-1}P_{n/2} \cdot {}^{n}P_{n}\)</p>
<p>None of these</p>
Step-by-Step Solution
Key Concept: When each person has exactly one neighbor, we must form n/2 pairs that are isolated from each other. The key is counting ordered pairs of adjacent people and then arranging these independent pairs.
<p><strong>Step 1:</strong> Understand the constraint: each person has exactly one neighbor means we partition n people into n/2 indivisible pairs.</p><p><strong>Step 2:</strong> Arrange n persons into n/2 ordered pairs. First, partition n people into n/2 unordered pairs: this is n!/(2^(n/2) × (n/2)!). However, since pairs themselves are distinguishable in a line, we organize by treating each pair as a unit.</p><p><strong>Step 3:</strong> Alternatively, think of it directly: we form n/2 pairs from n people, then arrange these pairs. The number of ways to partition n people into n/2 distinguishable pairs is n!/2^(n/2), but we must account for pair positioning.</p><p><strong>Step 4:</strong> The correct formula is: arrange n/2 pairs in a line (which is (n/2)! ways) and within each pair, 2 people can be ordered (2^(n/2) ways). But more directly: form pairs, then arrange them linearly and internally = (n/2)! × 2^(n/2).</p><p>∴ Answer: <strong>D</strong> = <strong>(n/2)! × 2^(n/2)</strong> or equivalently <strong>n!/(2^(n/2))</strong> depending on how pairs are formed and ordered.</p>
Correct Answer: D