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Question

If $n open parentheses A intersection B close parentheses equals 5 comma blank n open parentheses A intersection C close parentheses equals 7$ and $n open parentheses A intersection B intersection C close parentheses equals 3 comma$ then the minimum possible value of $n left parenthesis B intersection C right parenthesis$ is

0
1
3
2

The correct answer is: 3

Minimum possible value of $n left parenthesis B intersection C right parenthesis$ is $n open parentheses A intersection B intersection C close parentheses equals 3$

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