<p>The number of functions <i>f</i> from {1, 2, 3, …, 20} onto {1, 2, 3, …, 20} such that <i>f</i>(<i>k</i>) is a multiple of 3 whenever <i>k</i> is a multiple of 4, is</p>
Step-by-Step Solution
Key Concept: Identify which domain elements are restricted, which codomain elements they map to, and count valid onto functions by considering combinatorial arrangements of restricted and unrestricted mappings.
<p><strong>Step 1:</strong> Identify the constraint: <i>f</i>(<i>k</i>) must be a multiple of 3 whenever <i>k</i> is a multiple of 4.</p><p><strong>Step 2:</strong> The multiples of 4 in {1, 2, 3, …, 20} are: {4, 8, 12, 16, 20}. That's 5 elements.</p><p><strong>Step 3:</strong> The multiples of 3 in {1, 2, 3, …, 20} are: {3, 6, 9, 12, 15, 18}. That's 6 elements.</p><p><strong>Step 4:</strong> Since <i>f</i> is onto, we need <i>f</i>({4, 8, 12, 16, 20}) ⊆ {3, 6, 9, 12, 15, 18}.</p><p><strong>Step 5:</strong> For the function to be onto the entire codomain {1, 2, 3, …, 20}, all 6 multiples of 3 must be in the range. We can assign the 5 elements {4, 8, 12, 16, 20} to 6 possible values (multiples of 3) such that all 6 are covered. This can be done in <sup>6</sup>C₅ × 5! = 6 × 5! = 6! ways.</p><p><strong>Step 6:</strong> The remaining 15 elements (non-multiples of 4) can map to any element in the codomain in 15! ways to satisfy the onto condition.</p><p><strong>Total number of onto functions = 15! × 6!</strong></p><p>∴ Answer is (a).</p>
Correct Answer: A