<p>In a certain algebraical exercise book there are 4 examples on arithmetical progressions, 5 examples on permutation and combination, and 6 examples on binomial theorem. Find the number of ways a teacher can select for his pupils at least one but not more than 2 examples from each of these sets.</p>
Step-by-Step Solution
Key Concept: For each topic, we must select EITHER exactly 1 example OR exactly 2 examples (not 0, not 3+). Use (nC₁ + nC₂) for each set independently, then multiply by the multiplication principle.
<p><strong>Step 1: Understand the constraint</strong></p><p>From each set, teacher must select: exactly 1 example OR exactly 2 examples (not 0, and not 3 or more).</p><p><strong>Step 2: Count choices from each set independently</strong></p><ul><li>From 4 AP examples: ⁴C₁ (select 1) + ⁴C₂ (select 2) = (4 + 6) ways</li><li>From 5 P&C examples: ⁵C₁ (select 1) + ⁵C₂ (select 2) = (5 + 10) ways</li><li>From 6 Binomial examples: ⁶C₁ (select 1) + ⁶C₂ (select 2) = (6 + 15) ways</li></ul><p><strong>Step 3: Apply multiplication principle</strong></p><p>Since selection from each set is independent, total ways = (number of ways from AP) × (number of ways from P&C) × (number of ways from Binomial)</p><p><strong>∴ Answer:</strong> (⁴C₁ + ⁴C₂)(⁵C₁ + ⁵C₂)(⁶C₁ + ⁶C₂) = (10)(15)(21) = <strong>3150</strong></p>
Correct Answer: (⁴C₁ + ⁴C₂)(⁵C₁ + ⁵C₂)(⁶C₁ + ⁶C₂)