Let A = \ 2, 3, 4, 5, 6\ . Let R be a relation on the set A × A...

Given the set $A = \{2, 3, 4, 5, 6\}$.

The relation $R$ is defined on $A \times A$ such that $(x, y) R (z, w)$ if and only if $x$ divides $z$ and $y \leq w$.

The number of elements in $R$ is the number of ordered pairs $((x, y), (z, w))$ satisfying these conditions. Since the conditions for $(x, z)$ and $(y, w)$ are independent, the total number of elements in $R$ will be the product of the number of pairs $(x, z)$ satisfying $x$ divides $z$ and the number of pairs $(y, w)$ satisfying $y \leq w$.

First, we find the number of pairs $(x, z) \in A \times A$ such that $x$ divides $z$:
For $x = 2$, $z \in \{2, 4, 6\}$ ($3$ pairs)
For $x = 3$, $z \in \{3, 6\}$ ($2$ pairs)
For $x = 4$, $z \in \{4\}$ ($1$ pair)
For $x = 5$, $z \in \{5\}$ ($1$ pair)
For $x = 6$, $z \in \{6\}$ ($1$ pair)
Total number of pairs $(x, z) = 3 + 2 + 1 + 1 + 1 = 8$.

Next, we find the number of pairs $(y, w) \in A \times A$ such that $y \leq w$:
For $y = 2$, $w \in \{2, 3, 4, 5, 6\}$ ($5$ pairs)
For $y = 3$, $w \in \{3, 4, 5, 6\}$ ($4$ pairs)
For $y = 4$, $w \in \{4, 5, 6\}$ ($3$ pairs)
For $y = 5$, $w \in \{5, 6\}$ ($2$ pairs)
For $y = 6$, $w \in \{6\}$ ($1$ pair)
Total number of pairs $(y, w) = 5 + 4 + 3 + 2 + 1 = 15$.

The total number of elements in the relation $R$ is the product of the number of these pairs:
Total elements = $8 \times 15 = 120$.

Answer: $120$