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 find the number of many-one functions \( f: A \to B \) such that \( 1 \in f(A) \), where \( A = \{1, 2, 3, 4\} \) and \( B = \{1, 4, 9, 16\} \), we proceed as follows:

Each element in set \( A \) can be mapped to any element in set \( B \). The set \( B \) has 4 elements, thus the total number of functions \( f: A \to B \) is \( 4^4 = 256 \).

Now, we need to exclude functions where \( 1 \notin f(A) \), meaning none of \( A \)'s elements map to 1. For this, the remaining options are 4, 9, or 16. Therefore, each of the 4 elements in \( A \) has 3 choices, leading to \( 3^4 = 81 \) such functions.

Thus, the functions where \( 1 \notin f(A) \) are 81. Consequently, the number of functions where \( 1 \in f(A) \) is \( 256 - 81 = 175 \).

However, we need many-one functions, meaning at least two elements in \( A \) map to the same element in \( B \). A function that is one-to-one does not satisfy this, and such functions amount to the number of unique permutations of mappings from \( A \) to \( B \) with 4 elements having 4 and each mapped uniquely, i.e., \( P(4,4) = 24 \).

Thus, the many-one functions where \( 1 \in f(A) \) is \( 175 - 24 = 151 \).

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

Answer 2

Let $A=\{1,2,3,4\}$ and $B=\{1,4,9,16\}$. Here “many-one functions” is interpreted as functions that are **not one-to-one** (i.e. many-to-one). 

Total number of functions $f:A\to B$ is $|B|^{|A|}=4^{4}=256$.

Number of functions with $1\notin f(A)$ (i.e. functions whose values lie in $B\setminus\{1\}$ of size $3$) is $3^{4}=81$.

Hence number of functions with $1\in f(A)$ is $$256-81=175.$$

Among these, the injective (one-to-one) functions from $A$ to $B$ are precisely the bijections. Their number is $4!=24$. Every bijection includes $1$ in its image.

Therefore the number of many-one (i.e. non-injective) functions $f:A\to B$ with $1\in f(A)$ equals $$175-24=\boxed{151}.$$