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Question

In the above question, the number of families which buy none of $A comma B$ and $C$ is

(a) 4000
3300
4200
5000

The correct answer is: (a) 4000

We have,
Required number of families
$equals n left parenthesis A ´ intersection B ´ intersection C ´ right parenthesis$
$equals n left parenthesis A union B union C right parenthesis ´$
$equals N minus n left parenthesis A union B union C right parenthesis$
$equals 10000 minus left curly bracket n open parentheses A close parentheses plus n open parentheses B close parentheses plus n open parentheses C close parentheses minus n left parenthesis A intersection B right parenthesis right curly bracket$
$negative n open parentheses B intersection C close parentheses minus n open parentheses A intersection C close parentheses plus n left parenthesis A intersection B intersection C right parenthesis right curly bracket$
$equals 10000 minus 4000 minus 2000 minus 1000 plus 500 plus 300 plus 400 minus 200$
$equals 4000$

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