Question

Let \( A = \{2, 3, 4, 5, 6\} \). Let \( R \) be a relation on the set \( A \times A \) given by \( (x, y) R (z, w) \) if and only if \( x \) divides \( z \) and \( y \leq w \). Then the number of elements in \( R \) is _______.

Answer 1

Correct Answer: 15

Step 1: Understanding the Concept:
The relation \(R\) is defined on the Cartesian product \(A \times A\).
An element in \(R\) is a pair of ordered pairs \(((x, y), (z, w))\) such that the specified conditions are met.
The conditions are independent: \(x\) must divide \(z\), and \(y\) must be less than or equal to \(w\).
Since they are independent, we can find the number of valid pairs \((x, z)\) and the number of valid pairs \((y, w)\) separately, and then multiply them.
Step 2: Key Formula or Approach:
Total elements in \(R\) = (Number of pairs \((x, z)\) such that \(x | z\)) \(\times\) (Number of pairs \((y, w)\) such that \(y \le w\)).
We evaluate each condition for the set \(A = \{2, 3, 4, 5, 6\}\).
Step 3: Detailed Explanation:
First, find the number of pairs \((x, z)\) where \(x\) divides \(z\):
If \(x = 2\), \(z \in \{2, 4, 6\}\) (3 pairs).
If \(x = 3\), \(z \in \{3, 6\}\) (2 pairs).
If \(x = 4\), \(z \in \{4\}\) (1 pair).
If \(x = 5\), \(z \in \{5\}\) (1 pair).
If \(x = 6\), \(z \in \{6\}\) (1 pair).
Total valid \((x, z)\) pairs = \(3 + 2 + 1 + 1 + 1 = 8\).
Next, find the number of pairs \((y, w)\) where \(y \le w\):
If \(y = 2\), \(w \in \{2, 3, 4, 5, 6\}\) (5 pairs).
If \(y = 3\), \(w \in \{3, 4, 5, 6\}\) (4 pairs).
If \(y = 4\), \(w \in \{4, 5, 6\}\) (3 pairs).
If \(y = 5\), \(w \in \{5, 6\}\) (2 pairs).
If \(y = 6\), \(w \in \{6\}\) (1 pair).
Total valid \((y, w)\) pairs = \(5 + 4 + 3 + 2 + 1 = 15\).
Finally, the total number of elements in \(R\) is the product of these independent choices.
\[ \text{Total elements} = 8 \times 15 = 120 \] Step 4: Final Answer:
The number of elements in \(R\) is \(120\).