<p>A regular polygon of 10 sides is constructed. In how many ways can 3 vertices be selected so that no two vertices are consecutive?</p>
Step-by-Step Solution
Key Concept: Transform the non-consecutive selection problem into distributing identical objects by removing selected vertices first, then counting gaps. After choosing 3 vertices from 10, we have 7 remaining vertices that create 7 gaps (in a circle) where we must place our 3 selected vertices such that each gap contains at most 1 selected vertex.
<p><strong>Step 1:</strong> Fix the circular arrangement of 10 vertices. We need to select 3 vertices such that no two are adjacent (consecutive).</p><p><strong>Step 2:</strong> After selecting 3 vertices, 7 vertices remain. In a circular arrangement, these 7 remaining vertices create exactly 7 gaps (spaces between consecutive remaining vertices).</p><p><strong>Step 3:</strong> To ensure no two selected vertices are consecutive, we must place exactly one selected vertex in each of 3 different gaps. This ensures at least one unselected vertex separates any two selected vertices.</p><p><strong>Step 4:</strong> The number of ways to choose 3 gaps from 7 available gaps is C(7,3) = 7!/(3!·4!) = (7·6·5)/(3·2·1) = 210/3 = 70.</p><p><strong>Step 5:</strong> However, we must account for circular symmetry. In a regular polygon, rotations are considered identical. We divide by 10 to account for rotational equivalence when applicable, but since we're counting labeled selections (distinguishing which specific vertices are chosen), we use a different approach: employ the standard circular non-consecutive formula: C(n-k, k) for n vertices where n ≥ 2k. Here: C(10-3, 3) = C(7, 3) = 35... This needs correction.</p><p><strong>Step 5 (Corrected):</strong> For circular arrangements, the correct formula for selecting k non-consecutive items from n items is: [n/(n-k)]·C(n-k, k) for n ≥ 2k. With n=10, k=3: [10/7]·C(7,3) = [10/7]·35 = 50.</p><p>∴ Answer: <strong>50</strong></p>
Correct Answer: 50