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

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

To determine the count of many-one functions \( f: A \to B \) where \( 1 \in f(A) \), with \( A = \{1, 2, 3, 4\} \) and \( B = \{1, 4, 9, 16\} \), the subsequent steps are taken:

Each of the 4 elements in set \( A \) can be assigned to any of the 4 elements in set \( B \). Consequently, the total number of possible functions \( f: A \to B \) is \( 4^4 = 256 \).

Next, we subtract the functions where \( 1 otin f(A) \), meaning no element from \( A \) maps to 1. For these functions, the elements of \( A \) can only map to 4, 9, or 16. This gives each of the 4 elements in \( A \) 3 possible choices, resulting in \( 3^4 = 81 \) such functions.

Therefore, there are 81 functions where \( 1 otin f(A) \). The count of functions where \( 1 \in f(A) \) is calculated as \( 256 - 81 = 175 \).

However, the requirement is for many-one functions, which implies that at least two elements in \( A \) map to the same element in \( B \). Functions that are one-to-one do not satisfy this condition. The number of one-to-one functions, where each of the 4 elements in \( A \) maps uniquely to one of the 4 elements in \( B \), is equivalent to the number of permutations of 4 elements taken 4 at a time, which is \( P(4,4) = 4! = 24 \).

Thus, the number of many-one functions where \( 1 \in f(A) \) is found by subtracting the one-to-one functions from the total functions where \( 1 \in f(A) \): \( 175 - 24 = 151 \).

Hence, the number of many-one functions \( f: A \to B \) where \( 1 \in f(A) \) is 151.