<p>20 persons are sitting in a particular arrangement around a circular table. Three persons are to be selected for leaders. The number of ways of selection of three persons such that no two were sitting adjacent to each other is</p>
Step-by-Step Solution
Key Concept: Use the complementary counting approach: Total ways minus adjacent cases, OR directly count non-adjacent selections by first removing selected persons and arranging gaps. For circular arrangements, fix one person to break rotational symmetry, then count non-adjacent pairs from remaining persons.
<p><strong>Step 1:</strong> For selecting k non-adjacent persons from n persons arranged in a circle, the formula is: <strong>n/(n-k) × C(n-k, k)</strong> when n ≥ 2k.</p><p><strong>Step 2:</strong> Here n = 20, k = 3. Check: 20 ≥ 2(3) = 6 ✓</p><p><strong>Step 3:</strong> Apply formula: (20/(20-3)) × C(20-3, 3) = (20/17) × C(17, 3)</p><p><strong>Step 4:</strong> Calculate C(17, 3) = (17 × 16 × 15)/(3 × 2 × 1) = 4080/6 = 680</p><p><strong>Step 5:</strong> Final answer = (20/17) × 680 = (20 × 680)/17 = 13600/17 = <strong>800</strong></p><p><strong>Alternative approach:</strong> If we think of it as selecting 3 persons such that at least one person sits between any two selected persons: After selecting 3, we have 17 remaining persons and 3 selected persons. We need to arrange 3 selected persons among 17 gaps, which gives C(17, 3) × (3!/3) for circular = 680 × 2 = 1360... (This requires careful verification of the exact formula used)</p><p>∴ Answer: <strong>D (800)</strong></p>
Correct Answer: D