Let \( A = \{1, 2, 3, 4\} \) and \( B = \{1, 4, 9, 16\} \). Then the number of many-one functions \( f: A \to B \) such that \( 1 \in f(A) \) is equal to:
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When counting many-one functions, remember:
- A many-one function can map multiple elements of the domain to a single element of the codomain.
- Consider the restrictions (e.g., \(1 \in f(A)\)) and calculate accordingly, using the basic counting principle and permutations.
Given that \( 1 \in f(A) \), the element 1 from set \( B \) must be mapped to an element in set \( A \). There are \( 4 \) possible assignments for this element.
Following the assignment of 1, the remaining elements of \( B \), namely \( 4, 9, 16 \), must be mapped to the remaining three elements of \( A \). For each of these three elements in \( B \), any of the three available elements in \( A \) can be selected as its image, with no further constraints.
Consequently, the total count of many-one functions is calculated as:\[4 \times 3^3 = 127.\]% Topic - Counting functions
