Let's analyze the relation $R$. $R$ is a set of ordered pairs of ordered pairs $((a_1, b_1), (a_2, b_2))$ such that $a_1 \in A$, $b_1 \in B$, $a_2 \in A$, $b_2 \in B$, and the conditions $a_1 \le b_2$ and $b_1 \le a_2$ are satisfied. We need to find the number of such ordered pairs. However, the question is formulated wrongly, since we are looking at a relation of pairs. A proper relation is given by $R = {(a_1, b_1) : a_1 \in A, b_1 \in B}$. The number of elements in the relation is not a relation between ordered pairs, it is simply asking to find pairs $(a, b)$ such that $a \le b$. So let $S = {(a,b): a \in A, b \in B, a \le b}$ and find the number of elements in $S$. Now we enumerate:
- If $a=1$, then $b$ can be 2, 4, 5, 8, 10. (5 options)
- If $a=3$, then $b$ can be 4, 5, 8, 10. (4 options)
- If $a=4$, then $b$ can be 4, 5, 8, 10. (4 options)
- If $a=6$, then $b$ can be 8, 10. (2 options)
- If $a=9$, then $b$ can be 10. (1 option)
So the total number of elements in $S$ is $5+4+4+2+1 = 16$.
However, we are looking for elements that obey BOTH rules. That is:
Number of pairs satisfying BOTH $a_1 \le b_2$ and $b_1 \le a_2$.
Let's re-read the provided solution:
- For each $a_1$ in $A = {1, 3, 4, 6, 9}$, we find the possible $b_2$ values in $B = {2, 4, 5, 8, 10}$ such that $a_1 \le b_2$.
- $a_1 = 1$: $b_2$ can be 2, 4, 5, 8, 10 (5 choices)
- $a_1 = 3$: $b_2$ can be 4, 5, 8, 10 (4 choices)
- $a_1 = 4$: $b_2$ can be 4, 5, 8, 10 (4 choices)
- $a_1 = 6$: $b_2$ can be 8, 10 (2 choices)
- $a_1 = 9$: $b_2$ can be 10 (1 choice)
This gives $5+4+4+2+1 = 16$.

- For each $b_1$ in $B = {2, 4, 5, 8, 10}$, we find the possible $a_2$ values in $A = {1, 3, 4, 6, 9}$ such that $b_1 \le a_2$.
- $b_1 = 2$: $a_2$ can be 3, 4, 6, 9 (4 choices)
- $b_1 = 4$: $a_2$ can be 4, 6, 9 (3 choices)
- $b_1 = 5$: $a_2$ can be 6, 9 (2 choices)
- $b_1 = 8$: $a_2$ can be 9 (1 choice)
- $b_1 = 10$: no $a_2$ works (0 choices)
This gives $4+3+2+1+0 = 10$.
So, $16 \times 10 = 160$. Thus the final number of elements is 160.