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:

• A. \( 127 \)
• B. \( 139 \)
• C. \( 163 \)
• D. \( 151 \)

Hint:

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.

Answer 1

The Correct Option is A

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