Let $A = \{1, 6, 11, 16, \dots\}$ and $B = \{9, 16, 23, 30, \dots\}$ be the sets consisting of the first $2025$ terms of two arithmetic progressions. Then $n(A \cup B)$ is

$A$ and $B$ are two sets having $3$ and $6$ elements respectively. Consider the following statements. Statement $(I)$: Minimum number of elements in $A \cup B$ is $6$. Statement $(II)$: Maximum number of elements in $A \cap B$ is $3$. Which of the following is correct?

If $A = \{x \in \mathbb{R} : |x - 2| > 1\}$,$B = \{x \in \mathbb{R} : \sqrt{x^2 - 3} > 1\}$,$C = \{x \in \mathbb{R} : |x - 4| \geq 2\}$,and $\mathbb{Z}$ is the set of all integers,then the number of subsets of the set $(A \cap B \cap C)^c \cap \mathbb{Z}$ is .... .

For three non-impossible events $A$,$B$,and $C$,$P(A \cap B \cap C) = 0$,$P(A \cup B \cup C) = \frac{3}{4}$,$P(A \cap B) = \frac{1}{3}$,and $P(C) = \frac{1}{6}$. The probability that exactly one of $A$ or $B$ occurs but $C$ does not occur is:

Let $\bigcup_{i=1}^{50} X_{i} = \bigcup_{i=1}^{n} Y_{i} = T$,where each $X_{i}$ contains $10$ elements and each $Y_{i}$ contains $5$ elements. If each element of the set $T$ is an element of exactly $20$ of the sets $X_{i}$ and exactly $6$ of the sets $Y_{i}$,then $n$ is equal to:

In a city,$25\%$ of families have a telephone and $15\%$ have a car,while $65\%$ of families have neither a telephone nor a car. If $2000$ families have both a car and a telephone,then consider the following statements:
$1.$ $10\%$ of families have both a car and a telephone.
$2.$ $35\%$ of families have either a car or a telephone.
$3.$ $40,000$ families live in the city.
Which of these statements are true?