Question

Let $A,B,C$ be finite sets. Suppose that $n\left(A\right)=10,n\left(B\right)=15,n\left(C\right)=20,n(A\cap B)=8$ and $n(B\cap C)=9$. Then the maximum possible value of $n(A\cup B\cup C)$ is

Answer 1

$n(A\cup B\cup C)=n\left(A\right)+n\left(B\right)+n\left(C\right)-n(A\cap B)-n(B\cap C)-n(C\cap A)+n(A\cap B\cap C)=28-\left[n\right(C\cap A)-n(A\cap B\cap C\left)\right]$
We know that $n(C\cap A)\ge n(A\cap B\cap C)\Rightarrow n(C\cap A)-n(A\cap B\cap C)\ge 0$
$\therefore$ Maximum possible value of $n(A\cup B\cup C)$ is $28$