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Question

Let A=xW,thesetofwholenumbersandx<3,B=xN,thesetofnaturalnumbersand2x<4 and C=3,4, then how many elements will (AB)×C contain?


A

6

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B

8

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C

10

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D

12

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Solution

The correct option is B

8


Finding the number of elements in (AB)×C:

The given set is, A=xW,thesetofwholenumbersandx<3

A=0,1,2

Also,B=xN,thesetofnaturalnumbersand2x<4

B=2,3

Given, C=3,4

Therefore, (AB)=0,1,2,3

The number of elements in AB:

n(AB)=4

And, n(C)=2

So, the number of elements in (AB)×C:

n[(AB)×C]=4×2=8

Hence, the correct answer is option (B).


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