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

$\begin{array}{l} n(A \cup B \cup C)=n(A)+n(B)+n(C)-n(A \cap B)-n(B \cap C)-n(C \cap A) +n(A \cap B \cap C) \\ = 10+15+20-8-9-n(C \cap A)+n(A \cap B \cap C) \\ \because n(C \cap A) \geqslant n(A \cap B \cap C)\end{array}$

$\begin{array}{c}\Rightarrow \quad n(C \cap A)-n(A \cap B \cap C) \geqslant 0 \\ \therefore n(A \cap B \cap C) \leq 28 \\ \begin{aligned} Now, n(A \cup B) & =n(A)+n(B)-n(A \cap C) \\ & =10+15-8 \\ & =17\end{aligned}\end{array}$

$\begin{array}{l} \text { Similarly, } n(B \cup C)=26 \\ \because n(A \cup B \cup C) \geqslant n(A \cup C) \\ n(A \cup B \cup C) \geqslant 17 \& n(A \cup B \cup C) \geqslant 26 \\ \therefore 26 \leqslant n(A \cup B \cup C) \leqslant 28 \end{array}$

$\text { Possible values } 26 \text { or } 27 \text { on } 28 .$