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Let $A=\{x \in W$, the set of whole numbers and $x < 3\}$, $\mathrm{B}=\{\mathrm{x} \in \mathrm{N}$, the set of natural numbers and $2 \leq \mathrm{x} < 4\}$ and
$\mathrm{C}=\{3,4\}$, then how many elements will $(\mathrm{A} \cup \mathrm{B}) \times \mathrm{C}$
contain?
Options:
$\mathrm{C}=\{3,4\}$, then how many elements will $(\mathrm{A} \cup \mathrm{B}) \times \mathrm{C}$
contain?
Solution:
2533 Upvotes
Verified Answer
The correct answer is:
8
We have $\mathrm{A}=\{0,1,2\}$
$\mathrm{B}=\{2,3\}$
$\mathrm{C}=\{3,4\}$
$\mathrm{A} \cup \mathrm{B}=\{0,1,2,3\}$
$(\mathrm{A} \cup \mathrm{B}) \times \mathrm{C}=\{(0,3),(0,4),(1,3),(1,4) ;(2,3)$
$(2,4),(3,3) ;(3,4)\}$
$\therefore \mathrm{n}[(\mathrm{A} \cup \mathrm{B}) \times \mathrm{C}]=8$
$\mathrm{B}=\{2,3\}$
$\mathrm{C}=\{3,4\}$
$\mathrm{A} \cup \mathrm{B}=\{0,1,2,3\}$
$(\mathrm{A} \cup \mathrm{B}) \times \mathrm{C}=\{(0,3),(0,4),(1,3),(1,4) ;(2,3)$
$(2,4),(3,3) ;(3,4)\}$
$\therefore \mathrm{n}[(\mathrm{A} \cup \mathrm{B}) \times \mathrm{C}]=8$
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