Permutations & Combinations
Onto functions
Grade 11
Question:
<p>Let \(A = \{x_1, x_2, x_3, \ldots, x_7\}\), \(B = \{y_1, y_2, y_3\}\). The total number of functions \(f: A \to B\) that are on to and there are exactly three elements \(x\) in \(A\) such that \(f(x) = y_2\) is equal to</p>
<p>490</p>
<p>510</p>
<p>630</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: Use Surjection condition: First choose which 3 elements of A map to y₂, then distribute remaining 4 elements between y₁ and y₃ such that both are covered (inclusion-exclusion on remaining distribution).
<p><strong>Step 1:</strong> Choose 3 elements from A that map to y₂: C(7,3) = 35 ways</p><p><strong>Step 2:</strong> Remaining 4 elements must map to {y₁, y₃} such that BOTH y₁ and y₃ are covered (onto condition for these 4 elements).</p><p><strong>Step 3:</strong> Total ways to distribute 4 elements to 2 elements: 2⁴ = 16</p><p><strong>Step 4:</strong> Subtract cases where all 4 go to y₁ only: 1 way. Subtract cases where all 4 go to y₃ only: 1 way.</p><p><strong>Step 5:</strong> Valid distributions = 2⁴ - 2 = 16 - 2 = 14</p><p><strong>Step 6:</strong> Total functions = C(7,3) × 14 = 35 × 14 = 490</p><p>∴ Answer: C</p>
Correct Answer: C