Key Concepts and Formulas

• Total Number of Functions: For sets $A$ and $B$ with $|A|=m$ and $|B|=n$, the total number of functions $f: A \rightarrow B$ is $n^m$.
• One-to-One (Injective) Function: A function $f: A \rightarrow B$ is injective if distinct elements in $A$ map to distinct elements in $B$. The number of injective functions from a set of size $m$ to a set of size $n$ ($m \le n$) is $P(n, m) = \frac{n!}{(n-m)!}$.
• Many-One Function: A function that is not one-to-one.
• Range of a Function ($f(A)$): The set of all output values of the function. The condition $1 \in f(A)$ means that the element $1$ from the codomain $B$ must be an image of at least one element from the domain $A$.
• Complementary Counting: The number of elements satisfying a property is the total number of elements minus the number of elements not satisfying the property.

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Step-by-Step Solution

We are given sets $A = \{1, 2, 3, 4\}$ and $B = \{1, 4, 9, 16\}$. We need to find the number of many-one functions $f: A \rightarrow B$ such that $1 \in f(A)$. We have $|A| = 4$ and $|B| = 4$.

Step 1: Calculate the Total Number of Functions from A to B For each of the 4 elements in set $A$, there are 4 possible choices in set $B$ to which it can be mapped. Since these choices are independent for each element in $A$, the total number of functions is $|B|^{|A|}$. $\text{Total functions} = |B|^{|A|} = 4^4 = 256$

Step 2: Calculate the Number of Functions where $1 \in f(A)$ The condition $1 \in f(A)$ means that the element $1$ from the codomain $B$ must be the image of at least one element from the domain $A$. It is easier to calculate the complement: the number of functions where $1 \notin f(A)$, and then subtract this from the total number of functions. If $1 \notin f(A)$, it means that none of the elements in $A$ can be mapped to $1 \in B$. Therefore, the image of each element in $A$ must be chosen from the set $B \setminus \{1\} = \{4, 9, 16\}$. This reduced codomain has $4-1=3$ elements. The number of functions where $1 \notin f(A)$ is $3^{|A|}$. $\text{Functions where } 1 \notin f(A) = 3^4 = 81$ Now, we can find the number of functions where $1 \in f(A)$ using complementary counting: $\text{Functions where } 1 \in f(A) = (\text{Total functions}) - (\text{Functions where } 1 \notin f(A))$ $= 256 - 81 = 175$ Let $N_{\text{range } 1 \in B} = 175$.

Step 3: Calculate the Number of One-to-One (Injective) Functions from A to B A one-to-one function maps distinct elements of $A$ to distinct elements of $B$. The number of one-to-one functions from a set of size $m$ to a set of size $n$ is given by the permutation formula $P(n, m) = \frac{n!}{(n-m)!}$. Here, $m=|A|=4$ and $n=|B|=4$. $\text{Number of one-to-one functions} = P(4, 4) = \frac{4!}{(4-4)!} = \frac{4!}{0!} = 4! = 24$ Let $N_{\text{one-to-one}} = 24$.

Step 4: Determine if One-to-One Functions Satisfy $1 \in f(A)$ Since $|A|=|B|=4$, any one-to-one function from $A$ to $B$ must also be surjective (onto). This means that the range of every one-to-one function is the entire codomain, $f(A) = B = \{1, 4, 9, 16\}$. Therefore, all 24 one-to-one functions automatically satisfy the condition $1 \in f(A)$.

Step 5: Calculate the Number of Many-One Functions where $1 \in f(A)$ We are looking for functions that are many-one AND satisfy $1 \in f(A)$. The set of all functions satisfying $1 \in f(A)$ can be divided into two disjoint subsets:

1. One-to-one functions that satisfy $1 \in f(A)$.
2. Many-one functions that satisfy $1 \in f(A)$.

We want to find the size of the second subset. $\text{Many-one functions with } 1 \in f(A) = (\text{Functions with } 1 \in f(A)) - (\text{One-to-one functions with } 1 \in f(A))$ From Step 2, we know there are 175 functions where $1 \in f(A)$. From Step 4, we know all 24 one-to-one functions satisfy $1 \in f(A)$. $\text{Number of many-one functions with } 1 \in f(A) = 175 - 24 = 151$

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Common Mistakes & Tips

• Confusing Many-One and One-to-One: Ensure a clear understanding of the definitions. Many-one functions have at least two domain elements mapping to the same codomain element.
• Overlooking the $1 \in f(A)$ Condition: This condition must be applied to the final set of many-one functions. Complementary counting is crucial for the "at least one" aspect of this condition.
• Special Case $|A|=|B|$: For equal-sized domain and codomain, injective functions are also surjective (bijective). This means all injective functions will cover the entire codomain, thus satisfying any condition about elements being present in the range.

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Summary

The problem requires us to count many-one functions from $A$ to $B$ with the specific condition that the element $1$ must be in the range of the function. We first calculated the total number of functions and then used complementary counting to find the number of functions where $1$ is indeed in the range ($175$). Next, we determined the number of one-to-one functions ($24$). Since $|A|=|B|$, all one-to-one functions are bijections and thus their range is the entire codomain, meaning they all satisfy the $1 \in f(A)$ condition. Finally, to find the many-one functions satisfying the condition, we subtracted the one-to-one functions (which are not many-one) from the total functions satisfying the condition ($175 - 24 = 151$).

The final answer is $\boxed{151}$.