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 \( \mathrm{n}(A \cap B) \).

To solve part (a), we need to understand the intersection of prime numbers and numbers that end with the digit 7. The primes that end with 7 less than or equal to \( t \) are as follows: 7, 17, 27, 37, etc. Each of these must be less than or equal to \( t \). Since \( n(P \cap Q) = 5 \), we need at least five prime numbers that fit the criteria given that they end with 7. Thus, we can conclude that the minimum value of \( t \) must be at least 37. In fact, the required primes that meet the conditions are: 7, 17, 27, 37, and if needed also 47 for a total count of five. So \( t \) must be 37 or greater to have those five primes. For part (b), we know that \( n(A \cup B) = n(A) + n(B) - n(A \cap B) \). Given the information provided, we can calculate \( n(A \cup B) \) using the relation with the complement: \( n(A \cup B)^{\prime} + n(A \cup B) = n(\xi) \). So, \( n(A \cup B) = 50 - 8 = 42 \). Thus, we now have: \[ 42 = 32 + 19 - n(A \cap B) \] From this, we can solve for \( n(A \cap B) \): \[ n(A \cap B) = 32 + 19 - 42 = 9. \] The final answers are: (a) The minimum value of \( t \) is 37. (b) The value of \( n(A \cap B) \) is 9.