If n(A∩B) = 10, n(B∩C) = 20 and n(A∩C) = 30, then the greater possible value of n(A ∩ B ∩ C) is…..
Solution
\[ n(A \cap B)=10 \]
\[ n(B \cap C)=20 \]
\[ n(A \cap C)=30 \]
Since
\[ A \cap B \cap C \subseteq A \cap B \]
therefore,
\[ n(A \cap B \cap C) \le 10 \]
Also,
\[ n(A \cap B \cap C) \le 20 \]
and
\[ n(A \cap B \cap C) \le 30 \]
Hence, the greatest possible value is the minimum of
\[ 10,\ 20,\ 30 \]
\[ =10 \]
Answer
\[ \boxed{10} \]