<p>The total number of ways of selecting two numbers from the set \(\{1, 2, 3, 4, \ldots, 3n\}\) so that their sum is divisible by 3 is equal to</p>
Step-by-Step Solution
Key Concept: Classify numbers by their remainder mod 3, then count pairs within and across classes whose sum ≡ 0 (mod 3). Only pairs from the same class or complementary classes (remainder 1 and 2) work.
<p><strong>Step 1: Classify numbers by remainder mod 3</strong></p><p>In {1, 2, 3, ..., 3n}:</p><ul><li>Remainder 0: {3, 6, 9, ..., 3n} → n numbers</li><li>Remainder 1: {1, 4, 7, ..., 3n-2} → n numbers</li><li>Remainder 2: {2, 5, 8, ..., 3n-1} → n numbers</li></ul><p><strong>Step 2: Find pairs with sum ≡ 0 (mod 3)</strong></p><p>For two numbers to have sum divisible by 3:</p><ul><li><strong>Both from remainder 0 class:</strong> (0 + 0 ≡ 0) → C(n,2) = n(n-1)/2</li><li><strong>Both from remainder 1 class:</strong> (1 + 1 ≡ 2, ✗) → 0 pairs</li><li><strong>Both from remainder 2 class:</strong> (2 + 2 ≡ 1, ✗) → 0 pairs</li><li><strong>One from remainder 1, one from remainder 2:</strong> (1 + 2 ≡ 0) → n × n = n²</li></ul><p><strong>Step 3: Calculate total</strong></p><p>Total = n(n-1)/2 + n² = (n² - n)/2 + n² = (n² - n + 2n²)/2 = (3n² - n)/2 = <strong>n(3n-1)/2</strong></p><p>∴ Answer: B</p>
Correct Answer: B