Sets and Relations Part 1

Concept:The problem involves finding the number of onto functions from a finite set $A$ to a finite set $B$. The number of onto functions is given by total functions minus into (non-onto) functions. First, determine the elements of $A$ and $B$ by solving the given conditions.Formula:Number of onto functions from $A$ to $B$:$n(B)^{n(A)} - \binom{n(B)}{1} \cdot (n(B)-1)^{n(A)} + \binom{n(B)}{2} \cdot (n(B)-2)^{n(A)} - \cdots$For $n(B)=2$, onto functions = $2^{n(A)} - \binom{2}{1} \cdot 1^{n(A)}$.Solution:Set $A$:Condition: $\bigl| |x-3| - 3 \bigr| \le 1$, $x \in \mathbb{Z}$.This implies $|x-3| - 3 \in [-1,1]$, so $|x-3| \in [2,4]$.Hence $x-3 \in [-4,-2] \cup [2,4]$, giving $x \in [-1,1] \cup [5,7]$.For integers: $x = -1, 0, 1, 5, 6, 7$.Thus $n(A) = 6$.Set $B$:Condition: $\frac{(x-2)(x-4)}{x-1} \log_e |x-2| = 0$, $x \in \mathbb{R} \setminus \{1,2\}$.The product is zero when any factor is zero.1) $x-2=0 \Rightarrow x=2$, but $x \neq 2$ → discard.2) $x-4=0 \Rightarrow x=4$ → valid.3) $\log_e |x-2|=0 \Rightarrow |x-2|=1 \Rightarrow x=3$ or $x=1$. $x=1$ is excluded, so $x=3$ is valid.Thus $B = \{3,4\}$, so $n(B)=2$.Number of onto functions $f: A \to B$:Total functions = $2^6 = 64$.Into (non-onto) functions: only one element of $B$ is used. Choose that element in $\binom{2}{1}=2$ ways, and then map all 6 elements of $A$ to it: $1^6=1$ each. So into functions = $2$.Onto functions = $64 - 2 = 62$.Answer:62 (Option D)