Question

Consider that $n\left(S\right)$ represented the number of elements in set S. If $n(A\cup B\cup C)=40,n(A\cap B^{\prime }\cap C^{\prime })=5,n(B\cap A^{\prime }\cap C^{\prime })=10,n(C\cap B^{\prime }\cap A^{\prime })=6$ then number of element which belongs to at least two of the set is

• A. Less than $19$
• B. More than $21$ but less than $40$
• C. $19$
• D. $20$

Answer 1

The correct option is A Less than $19$

Given,

$n(A\cup B\cup C)=40,n(A^{\prime }\cap B^{\prime }\cap C^{\prime })=5$

$n(B\cap A^{\prime }\cap C^{\prime })=10,n(C\cap B^{\prime }\cap A^{\prime })=6$

Then number of element Which belongs to at least two of the sets is-

$\le n(A\cup B\cup C)-n(A^{\prime }\cap B^{\prime }\cap C^{\prime })-n(B\cap A^{\prime }\cap C^{\prime })-n(C\cap B^{\prime }\cap A^{\prime })$

$\le 40-5-10-6$

$\le 19$