<p>The number of ways to choose a subset of two elements <math>{a, b}</math> from the set <math>{1, 2, 3, 4, \ldots, 49, 50}</math> such that <math>|a - b| \leq 5</math> is</p>
Step-by-Step Solution
Key Concept: We need to count ordered pairs (a,b) where a < b from {1,2,...,50} with b - a ≤ 5. For each value of the difference d = b - a (from 1 to 5), count how many pairs exist and sum them.
<p><strong>Step 1:</strong> Set up the problem. We need unordered pairs {a, b} from {1, 2, ..., 50} where |a - b| ≤ 5. Without loss of generality, assume a < b, so we need b - a ≤ 5, meaning b - a ∈ {1, 2, 3, 4, 5}.</p><p><strong>Step 2:</strong> Count pairs for each difference d = b - a.</p><p><strong>For d = 1:</strong> Pairs (a, a+1) where a ∈ {1, 2, ..., 49}. Count = 49</p><p><strong>For d = 2:</strong> Pairs (a, a+2) where a ∈ {1, 2, ..., 48}. Count = 48</p><p><strong>For d = 3:</strong> Pairs (a, a+3) where a ∈ {1, 2, ..., 47}. Count = 47</p><p><strong>For d = 4:</strong> Pairs (a, a+4) where a ∈ {1, 2, ..., 46}. Count = 46</p><p><strong>For d = 5:</strong> Pairs (a, a+5) where a ∈ {1, 2, ..., 45}. Count = 45</p><p><strong>Step 3:</strong> Add all cases. Total = 49 + 48 + 47 + 46 + 45</p><p><strong>Step 4:</strong> Calculate the sum. This is the sum of integers from 45 to 49, which equals:</p><p>Sum = (49 + 48 + 47 + 46 + 45) = (45 + 46 + 47 + 48 + 49)</p><p>Using the formula for sum of arithmetic sequence: Sum = 5 × (45 + 49)/2 = 5 × 94/2 = 5 × 47 = 235 + 10 = 245</p><p>∴ Answer: A</p>
Correct Answer: A