Let \( n(A)=4 \) and \( n(A \cup B)=10 \). (a) What are the possible values of \( n(B) \) ? (b) If \( A \cap B=\varnothing \), what is the only possible value of \( n(B) \) ? (a) The possible value(s) for \( n(B) \) is/are \( \square \). (Use a comma to separate answers as needed. Use ascending order.

To solve the problem, we can use the principle of set theory. (a) We know that \( n(A \cup B) = n(A) + n(B) - n(A \cap B) \). Given that \( n(A) = 4 \) and \( n(A \cup B) = 10 \), we can rearrange the equation: \[ 10 = 4 + n(B) - n(A \cap B) \] This simplifies to: \[ n(B) = 6 + n(A \cap B) \] Since \( n(A \cap B) \) can take values from \( 0 \) (if sets A and B are disjoint) to \( 4 \) (if all elements of A are also in B), the possible values of \( n(A \cap B) \) can range from \( 0 \) to \( 4 \). This means: - If \( n(A \cap B) = 0 \), then \( n(B) = 6 \). - If \( n(A \cap B) = 1 \), then \( n(B) = 7 \). - If \( n(A \cap B) = 2 \), then \( n(B) = 8 \). - If \( n(A \cap B) = 3 \), then \( n(B) = 9 \). - If \( n(A \cap B) = 4 \), then \( n(B) = 10 \). Thus, the possible values for \( n(B) \) are \( 6, 7, 8, 9, 10 \). (b) If \( A \cap B = \varnothing \), it means that there is no overlap between sets A and B. In this scenario, since \( n(A \cap B) = 0 \), we can apply the formula directly: \[ n(B) = 6 + 0 = 6 \] Thus, the only possible value for \( n(B) \) when \( A \) and \( B \) are disjoint is \( 6 \). **Final Answers:** (a) The possible value(s) for \( n(B) \) is/are \( 6, 7, 8, 9, 10 \). (b) The only possible value of \( n(B) \) when \( A \cap B=\varnothing \) is \( 6 \).