19. (a) Given that \( \xi=\{1,2,3,4, \ldots, t\} \) where \( t \) is a positive integer, \( P \) and \( Q \) are two subsets of \( \xi \) where \( P=\{x: x \) is a prime number \( \} \) and \( Q=\{x: \) the unit digit of \( x \) is 7\( \} \) If \( n(P \cap Q)=5 \), write down the minimum value of \( t \). (b) Two sets \( A \) and \( B \) are such that \( \mathrm{n}(A)=32 \) and \( \mathrm{n}(B)=19 \). Given \( \mathrm{n}(\xi)=50, B \not \subset A \) and \( \mathrm{n}(A \cup B)^{\prime}=8 \), find the value of \( \mathrm{n}(A \cap B) \).

**(a)** We are given the set \(\xi = \{1, 2, 3, \ldots, t\}\), with \[ P = \{x : x \text{ is a prime number}\} \quad \text{and} \quad Q = \{x : \text{the unit digit of } x \text{ is } 7\}. \] We need to find the minimum value of \(t\) such that \[ n(P \cap Q) = 5. \] The set \(P \cap Q\) consists of prime numbers whose unit digit is 7. Listing them in increasing order: - \(7\) is prime and ends in 7. - \(17\) is prime and ends in 7. - \(37\) is prime and ends in 7. - \(47\) is prime and ends in 7. - \(67\) is prime and ends in 7. These are the first five primes with unit digit 7. To include \(67\) in the set \(\xi\), we must have \[ t \ge 67. \] Thus, the minimum value of \(t\) is \(\boxed{67}\). --- **(b)** We are given two sets \(A\) and \(B\) with \[ n(A) = 32, \quad n(B) = 19, \] and the universal set \(\xi\) has \[ n(\xi) = 50. \] Also, it is given that \(B \not\subset A\) and \[ n((A \cup B)') = 8. \] Since \((A \cup B)'\) is the complement of \(A \cup B\), its number of elements is \[ n((A \cup B)') = n(\xi) - n(A \cup B). \] Thus, we have \[ n(A \cup B) = 50 - 8 = 42. \] Using the principle of inclusion and exclusion, we know that \[ n(A \cup B) = n(A) + n(B) - n(A \cap B). \] Substituting the known values: \[ 42 = 32 + 19 - n(A \cap B). \] Simplify the equation: \[ 42 = 51 - n(A \cap B). \] Solving for \(n(A \cap B)\), we find: \[ n(A \cap B) = 51 - 42 = 9. \] Thus, the number of elements in the intersection is \(\boxed{9}\).