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Find the number of surjective (onto) functions $f$ from $\{1,2,3,\ldots,20\}$ to $\{1,2,3,\ldots,20\}$ such that $f(k)$ is a multiple of 3 whenever $k$ is a multiple of 4.
Multiples of 4 in $\{1,\ldots,20\}$: $\{4,8,12,16,20\}$ — 5 elements. Their images must be multiples of 3 in $\{1,\ldots,20\}$: $\{3,6,9,12,15,18\}$ — 6 elements.
Assign the 5 multiples-of-4 to 5 of the 6 multiples-of-3 (injectively, since $f$ is onto): $6 \times 5 \times 4 \times 3 \times 2 = 6!/(6-5)! = 720$... but we need surjection on all 20 elements.
The remaining 15 domain elements map to the remaining 15 codomain elements (after fixing the 5 images above): $(15)!$ ways. But 1 multiple-of-3 is unused by the multiples-of-4, so it must be covered by remaining 15 elements.
Total $= 6! \times (15)! = (15)! \times 6!$
Using $l = L + \frac{TL}{AY} = L + kT$:
- $l_1 = L + kT_1$ and $l_2 = L + kT_2$ [cite: 506]
- Solving for $L$ by eliminating $k$: $L = \frac{T_2l_1 - T_1l_2}{T_2 - T_1}$[cite: 515].
The word Philosophy comes from the Greek words philos (loving) and sophia (wisdom), meaning love of wisdom. It originated with ancient Greek thinkers who sought rational explanations for existence, knowledge, and ethics.
Correct Answer: A — Love of Wisdom.
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