Consider two sets A = { x in mathbb{Z} : |(| | vedclass

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(A) First, we find the elements of set $A$: $|(|x - 3| - 3)| \leq 1 \implies -1 \leq |x - 3| - 3 \leq 1$ $2 \leq |x - 3| \leq 4$ This implies $2 \leq

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$62$

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(A) First, we find the elements of set $A$: $|(|x - 3| - 3)| \leq 1 \implies -1 \leq |x - 3| - 3 \leq 1$ $2 \leq |x - 3| \leq 4$ This implies $2 \leq x - 3 \leq 4$ or $-4 \leq x - 3 \leq -2$ $5 \leq x \leq 7$ or $-1 \leq x \leq 1$ Since $x \in \mathbb{Z}$, $A = \{-1, 0, 1, 5, 6, 7\}$. The number of elements in $A$ is $n(A) = 6$. Next, we find the elements of set $B$: $\frac{(x - 2)(x - 4)}{x - 1} \log_{e}(|x - 2|) = 0$ This implies $(x - 2)(x - 4) = 0$ or $\log_{e}(|x - 2|) = 0$. If $(x - 2)(x - 4) = 0$, then $x = 2$ or $x = 4$. Since $x \in \mathbb{R} - \{1, 2\}$, we accept $x = 4$. If $\log_{e}(|x - 2|) = 0$, then $|x - 2| = 1$, so $x - 2 = 1$ or $x - 2 = -1$. This gives $x = 3$ or $x = 1$. Since $x \in \mathbb{R} - \{1, 2\}$, we accept $x = 3$. Thus, $B = \{3, 4\}$, and the number of elements in $B$ is $n(B) = 2$. The number of onto functions from a set with $n$ elements to a set with $m$ elements is given by $m^n - \binom{m}{1}(m-1)^n + \binom{m}{2}(m-2)^n - \dots$ For $n = 6$ and $m = 2$, the number of onto functions is $2^6 - 2 = 64 - 2 = 62$.

Consider two sets $A = \{ x \in \mathbb{Z} : |(| x - 3| - 3)| \leq 1 \}$ and $B = \{ x \in \mathbb{R} - \{1, 2\} : \frac{(x - 2)(x - 4)}{x - 1} \log_{e}(|x - 2|) = 0 \}$. Then the number of onto functions $f: A \rightarrow B$ is equal to:

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