If n(A∩B) = 5, n(A∩C) = 7 and n(A∩B∩C) = 3, then the minimum possible value of n(B∩C) is….
Solution
\[ n(A \cap B)=5 \]
\[ n(A \cap C)=7 \]
\[ n(A \cap B \cap C)=3 \]
Since
\[ A \cap B \cap C \subseteq B \cap C \]
therefore,
\[ n(B \cap C) \ge 3 \]
The minimum possible value occurs when there are no extra elements in \(B \cap C\) other than the common elements of all three sets.
Hence,
\[ n(B \cap C)=3 \]
Answer
\[ \boxed{3} \]