<p>Four numbers are chosen from \(\{1, 2, \ldots, 20\}\). What is the probability that the chosen numbers are in arithmetic progression (AP)?</p>
Step-by-Step Solution
Key Concept: Count APs with 4 terms from {1,2,...,20} by fixing first term and common difference, then divide by total ways to choose 4 numbers. An AP is determined by first term a and common difference d where a, a+d, a+2d, a+3d ≤ 20.
<p><strong>Step 1:</strong> Count total ways to choose 4 numbers from 20: C(20,4) = 20!/(4!·16!) = 4845</p><p><strong>Step 2:</strong> Count 4-term APs with first term a and common difference d. We need a, a+d, a+2d, a+3d all in {1,2,...,20}, so a ≥ 1 and a+3d ≤ 20, giving a ≤ 20-3d.</p><p><strong>Step 3:</strong> For each d, count valid values of a:</p><ul><li>d=1: a ∈ {1,2,...,17}, giving 17 APs</li><li>d=2: a ∈ {1,2,...,14}, giving 14 APs</li><li>d=3: a ∈ {1,2,...,11}, giving 11 APs</li><li>d=4: a ∈ {1,2,...,8}, giving 8 APs</li><li>d=5: a ∈ {1,2,...,5}, giving 5 APs</li><li>d=6: a ∈ {1,2}, giving 2 APs</li></ul><p><strong>Step 4:</strong> Total APs = 17+14+11+8+5+2 = 57</p><p><strong>Step 5:</strong> Probability = 57/4845 = 19/1615</p><p>∴ Answer: A</p>
Correct Answer: A