Permutations & Combinations
Functions and Counting
Grade 11
Question:
<p>Let \(A = \{x_1, x_2, \ldots, x_7\}\) and \(B = \{y_1, y_2, y_3\}\) be two sets containing seven and three distinct elements respectively. Then the total number of functions \(f: A \to B\) that are onto, if there exists exactly three elements \(x\) in \(A\) such that \(f(x) = y_2\), is equal to</p>
<p>\(14 \cdot {}^7C_2\)</p>
<p>\(16 \cdot {}^7C_3\)</p>
<p>\(12 \cdot {}^7C_3\)</p>
<p>\(14 \cdot {}^7C_3\)</p>
Step-by-Step Solution
Key Concept: Use inclusion-exclusion with a fixed constraint: first choose which 3 elements from A map to y₂, then distribute remaining 4 elements to y₁ and y₃ such that both are non-empty (onto condition).
<p><strong>Step 1:</strong> Since exactly 3 elements of A map to y₂, choose these 3 elements: C(7,3) = 35 ways.</p><p><strong>Step 2:</strong> The remaining 4 elements of A must map to B such that f is onto. Since y₂ is already covered, we need both y₁ and y₃ to appear in the images of these 4 remaining elements.</p><p><strong>Step 3:</strong> Each of the 4 remaining elements can map to either y₁ or y₃: 2⁴ = 16 total ways.</p><p><strong>Step 4:</strong> From these 16 ways, subtract cases where all 4 map to y₁ only (1 way) or all 4 map to y₃ only (1 way). Valid distributions: 16 - 2 = 14 ways.</p><p><strong>Step 5:</strong> Total number of onto functions = 35 × 14 = <strong>490</strong></p><p>∴ Answer: D</p>
Correct Answer: D