Step 1: Understand what has to be counted.
The relation \( R \) is defined on \( A \times A \). So, an element of \( R \) is of the form
\[
((x,y),(a,b))
\]
such that:
\[
x \text{ divides } a \quad \text{and} \quad y \leq b
\]
Hence, for each fixed pair \( (a,b) \in A \times A \), we count the number of possible \( x \) and \( y \) satisfying these conditions.
Step 2: Count values of \(x\) such that \(x\mid a\).
Now \( A=\{2,3,4,5,6\} \). For each \( a \in A \), count how many elements of \( A \) divide \( a \):
\[
a=2 \Rightarrow \{2\} \Rightarrow 1
\]
\[
a=3 \Rightarrow \{3\} \Rightarrow 1
\]
\[
a=4 \Rightarrow \{2,4\} \Rightarrow 2
\]
\[
a=5 \Rightarrow \{5\} \Rightarrow 1
\]
\[
a=6 \Rightarrow \{2,3,6\} \Rightarrow 3
\]
So, total divisor-count contribution is:
\[
1+1+2+1+3=8
\]
Step 3: Count values of \(y\) such that \(y \leq b\).
Again \( A=\{2,3,4,5,6\} \). For each \( b \in A \), count how many elements of \( A \) are less than or equal to \( b \):
\[
b=2 \Rightarrow \{2\} \Rightarrow 1
\]
\[
b=3 \Rightarrow \{2,3\} \Rightarrow 2
\]
\[
b=4 \Rightarrow \{2,3,4\} \Rightarrow 3
\]
\[
b=5 \Rightarrow \{2,3,4,5\} \Rightarrow 4
\]
\[
b=6 \Rightarrow \{2,3,4,5,6\} \Rightarrow 5
\]
So, total \(y\)-count contribution is:
\[
1+2+3+4+5=15
\]
Step 4: Find the total number of elements in \(R\).
For each \( (a,b) \), the number of valid pairs \( (x,y) \) is:
\[
(\text{number of }x\text{ dividing }a)\times(\text{number of }y\leq b)
\]
Therefore, total number of elements in \( R \) is:
\[
\left( \sum \text{divisor counts} \right)\left( \sum \text{order counts} \right)
\]
\[
=8 \times 15
\]
\[
=120
\]
Step 5: Conclusion.
Hence, the total number of elements in the relation \( R \) is \(120\).
Final Answer: \(120\)