For each element \( x \in \{1, 2, ..., n\} \), we must assign it to the sets \( A, B, C \) such that the condition \( (A\cap B) \subseteq C \subseteq (A\cup B) \) is satisfied. Let's evaluate the choices for each element \( x \): Case 1: \( x \notin A \) and \( x \notin B \). Then \( x \notin A \cap B \) and \( x \notin A \cup B \). Thus, \( C \) cannot contain \( x \) (1 way). Case 2: \( x \in A \) and \( x \notin B \). Then \( x \notin A \cap B \) but \( x \in A \cup B \). Here, \( x \) can either be in \( C \) or not in \( C \) (2 ways). Case 3: \( x \notin A \) and \( x \in B \). Similar to Case 2, \( x \) can either be in \( C \) or not in \( C \) (2 ways). Case 4: \( x \in A \) and \( x \in B \). Then \( x \in A \cap B \). To satisfy the condition, \( C \) MUST contain \( x \) (1 way). Total ways to assign each element is \( 1 + 2 + 2 + 1 = 6 \) ways. Since there are \( n \) independent elements, the total number of triples is \( 6^n \).