Let $$A = \{1, 3, 4, 6, 9\}$$ and $$B = \{2, 4, 5, 8, 10\}$$. Let $$R$$ be a relation defined on $$A \times B$$ such that $$R = \{(a_1, b_1), (a_2, b_2): a_1 \leq b_2 \text{ and } b_1 \leq a_2\}$$. Then the number of elements in the set $$R$$ is

Let $$A = \{1, 3, 4, 6, 9\}$$ and $$B = \{2, 4, 5, 8, 10\}$$. The relation $$R$$ on $$A \times B$$ is defined as:

$$ R = \{((a_1, b_1), (a_2, b_2)) : a_1 \leq b_2 \text{ and } b_1 \leq a_2\} $$

Elements of $$A \times B$$ are ordered pairs $$(a, b)$$ with $$a \in A, b \in B$$. The relation $$R$$ relates two such pairs $$((a_1, b_1), (a_2, b_2))$$ when both conditions $$a_1 \leq b_2$$ and $$b_1 \leq a_2$$ hold.

Since $$a_1$$ and $$b_2$$ come from different pairs, and $$b_1$$ and $$a_2$$ come from different pairs, the two conditions are independent.

Count pairs $$(a_1, b_2)$$ with $$a_1 \in A, b_2 \in B, a_1 \leq b_2$$:

$$a_1 = 1$$: $$b_2 \in \{2,4,5,8,10\}$$ → 5 pairs

$$a_1 = 3$$: $$b_2 \in \{4,5,8,10\}$$ → 4 pairs

$$a_1 = 4$$: $$b_2 \in \{4,5,8,10\}$$ → 4 pairs

$$a_1 = 6$$: $$b_2 \in \{8,10\}$$ → 2 pairs

$$a_1 = 9$$: $$b_2 \in \{10\}$$ → 1 pair

Total: $$5 + 4 + 4 + 2 + 1 = 16$$

Count pairs $$(b_1, a_2)$$ with $$b_1 \in B, a_2 \in A, b_1 \leq a_2$$:

$$b_1 = 2$$: $$a_2 \in \{3,4,6,9\}$$ → 4 pairs

$$b_1 = 4$$: $$a_2 \in \{4,6,9\}$$ → 3 pairs

$$b_1 = 5$$: $$a_2 \in \{6,9\}$$ → 2 pairs

$$b_1 = 8$$: $$a_2 \in \{9\}$$ → 1 pair

$$b_1 = 10$$: $$a_2 \in \{\}$$ → 0 pairs

Total: $$4 + 3 + 2 + 1 + 0 = 10$$

$$ |R| = 16 \times 10 = 160 $$

The number of elements in $$R$$ is Option A: 160.