Key Concepts and Formulas

• Cartesian Product: For sets $A$ and $B$, the Cartesian product $A \times B$ is the set of all ordered pairs $(a, b)$ where $a \in A$ and $b \in B$.
• Relation on a Set: A relation $R$ on a set $S$ is a subset of $S \times S$. In this problem, the relation is defined on $A \times B$, meaning $R \subseteq (A \times B) \times (A \times B)$.
• Multiplication Principle: If an event can occur in $m$ ways and a second independent event can occur in $n$ ways, then the number of ways both events can occur is $m \times n$.

Step-by-Step Solution

Step 1: Understand the structure of the relation $R$. The relation $R$ is defined on the set $A \times B$. An element of $R$ is an ordered pair of elements from $A \times B$. Let an element of $R$ be denoted by $((a_1, b_1), (a_2, b_2))$, where $(a_1, b_1) \in A \times B$ and $(a_2, b_2) \in A \times B$. The conditions for an element to be in $R$ are $a_1 \in A$, $b_1 \in B$, $a_2 \in A$, $b_2 \in B$, and the inequalities $a_1 \leq b_2$ and $b_1 \leq a_2$ must hold simultaneously. We are given $A = \{1, 3, 4, 6, 9\}$ and $B = \{2, 4, 5, 8, 10\}$. The number of elements in $R$, denoted by $|R|$, is the number of such pairs $((a_1, b_1), (a_2, b_2))$ that satisfy the given conditions.

Step 2: Decompose the problem using the Multiplication Principle. The conditions for an element to be in $R$ are $a_1 \leq b_2$ and $b_1 \leq a_2$. Notice that the first condition, $a_1 \leq b_2$, involves variables $a_1$ from set $A$ and $b_2$ from set $B$. The choice of $a_1$ and $b_2$ is independent of the choice of $b_1$ and $a_2$. The second condition, $b_1 \leq a_2$, involves variables $b_1$ from set $B$ and $a_2$ from set $A$. Since these two sets of choices are independent, we can count the number of pairs satisfying each condition separately and then multiply the results. Let $N_1$ be the number of ordered pairs $(a_1, b_2)$ such that $a_1 \in A$, $b_2 \in B$, and $a_1 \leq b_2$. Let $N_2$ be the number of ordered pairs $(b_1, a_2)$ such that $b_1 \in B$, $a_2 \in A$, and $b_1 \leq a_2$. Then, $|R| = N_1 \times N_2$.

Step 3: Calculate $N_1$, the number of pairs $(a_1, b_2)$ with $a_1 \in A, b_2 \in B$ and $a_1 \leq b_2$. We list the elements of $A$ and count how many elements in $B$ are greater than or equal to them.

• If $a_1 = 1$: $b_2 \in \{2, 4, 5, 8, 10\}$. All 5 elements satisfy $1 \leq b_2$. (5 pairs)
• If $a_1 = 3$: $b_2 \in \{4, 5, 8, 10\}$. 4 elements satisfy $3 \leq b_2$. (4 pairs)
• If $a_1 = 4$: $b_2 \in \{4, 5, 8, 10\}$. 4 elements satisfy $4 \leq b_2$. (4 pairs)
• If $a_1 = 6$: $b_2 \in \{8, 10\}$. 2 elements satisfy $6 \leq b_2$. (2 pairs)
• If $a_1 = 9$: $b_2 \in \{10\}$. 1 element satisfies $9 \leq b_2$. (1 pair) Summing these counts: $N_1 = 5 + 4 + 4 + 2 + 1 = 16$.

Step 4: Calculate $N_2$, the number of pairs $(b_1, a_2)$ with $b_1 \in B, a_2 \in A$ and $b_1 \leq a_2$. We list the elements of $B$ and count how many elements in $A$ are greater than or equal to them.

• If $b_1 = 2$: $a_2 \in \{3, 4, 6, 9\}$. 4 elements satisfy $2 \leq a_2$. (4 pairs)
• If $b_1 = 4$: $a_2 \in \{4, 6, 9\}$. 3 elements satisfy $4 \leq a_2$. (3 pairs)
• If $b_1 = 5$: $a_2 \in \{6, 9\}$. 2 elements satisfy $5 \leq a_2$. (2 pairs)
• If $b_1 = 8$: $a_2 \in \{9\}$. 1 element satisfies $8 \leq a_2$. (1 pair)
• If $b_1 = 10$: There are no elements in $A$ such that $10 \leq a_2$. (0 pairs) Summing these counts: $N_2 = 4 + 3 + 2 + 1 + 0 = 10$.

Step 5: Calculate the total number of elements in $R$. Using the Multiplication Principle, $|R| = N_1 \times N_2$. $|R| = 16 \times 10 = 160$

Common Mistakes & Tips

• Confusing the roles of $a_i$ and $b_i$: Ensure that $a_1, a_2$ are correctly identified as elements of set $A$ and $b_1, b_2$ are correctly identified as elements of set $B$.
• Incorrectly applying the Multiplication Principle: This principle can only be applied when the conditions are independent. In this case, the choice of $(a_1, b_2)$ pairs is independent of the choice of $(b_1, a_2)$ pairs.
• Errors in counting inequalities: Carefully check each inequality for every possible combination of elements from the respective sets.

Summary The problem requires us to find the number of pairs of pairs $((a_1, b_1), (a_2, b_2))$ from $A \times B$ such that $a_1 \leq b_2$ and $b_1 \leq a_2$. We identified that the condition $a_1 \leq b_2$ involves independent choices from sets $A$ and $B$, and the condition $b_1 \leq a_2$ involves independent choices from sets $B$ and $A$. By applying the Multiplication Principle, we calculated the number of pairs satisfying the first condition ($N_1 = 16$) and the number of pairs satisfying the second condition ($N_2 = 10$). The total number of elements in the relation $R$ is the product of these two numbers, $16 \times 10 = 160$.

The final answer is \boxed{160}. This corresponds to option (A).