Question:
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:
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.
Updated On: Feb 5, 2025
- \( 127 \)
- \( 139 \)
- \( 163 \)
- \( 151 \)
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The Correct Option is A
Solution and Explanation
Since \( 1 \in f(A) \), we need to assign the element 1 of \( B \) to one of the elements of \( A \). This can be done in \( 4 \) ways.
After assigning 1 to one of the elements of \( A \), the remaining elements of \( B \) (i.e., \( 4, 9, 16 \)) can be assigned to the other three elements of \( A \). Each of the three remaining elements of \( A \) can be assigned to one of the three remaining elements of \( B \), and there are no restrictions on this assignment.
Thus, the total number of many-one functions is:
\[
4 \times 3^3 = 127.
\]
% Topic - Counting functions
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