If \(n(A) = 2\) and the total number of possible relations from set \(A\) to set \(B\) is 1024, then \(n(B)\) is:
The number of relations from set A to set B is given by \(2^{n(A) \cdot n(B)}\). We are given that \(n(A) = 2\) and the number of relations is 1024. So, we have:
\(2^{2 \cdot n(B)} = 1024\)
Since \(1024 = 2^{10}\), we have:
\(2^{2 \cdot n(B)} = 2^{10}\)
Therefore, \(2 \cdot n(B) = 10\), which means \(n(B) = 5\).
Answer: (D) 5
Let \( S = \{p_1, p_2, \dots, p_{10}\} \) be the set of the first ten prime numbers. Let \( A = S \cup P \), where \( P \) is the set of all possible products of distinct elements of \( S \). Then the number of all ordered pairs \( (x, y) \), where \( x \in S \), \( y \in A \), and \( x \) divides \( y \), is _________.