1. Key Concepts and Formulas

• Set Definition: Given a set $A = \{1,2,3,4\}$, the set $A \times A$ consists of all ordered pairs $(x,y)$ where $x \in A$ and $y \in A$.
• Relation Definition: A relation $R$ on a set $S$ is a subset of $S \times S$. In this problem, $S = A \times A$. Thus, elements of $R$ are of the form $((a,b), (c,d))$ where $(a,b) \in A \times A$ and $(c,d) \in A \times A$.
• Counting Elements: To find the number of elements in a relation defined by an equality between two expressions, we systematically evaluate the possible values of each expression and count how many pairs of inputs yield each value. The total number of relation elements is the sum of the products of the counts for common values.

2. Step-by-Step Solution

The problem defines a relation $R$ on the set $A \times A$, where $A = \{1,2,3,4\}$. The relation is given by $R=\{((a, b),(c, d)): 2 a+3 b=4 c+5 d\}$. We need to find the number of elements in $R$. This means we need to find the number of pairs $((a,b), (c,d))$ such that $a, b, c, d \in \{1,2,3,4\}$ and $2a+3b = 4c+5d$.

Step 1: Determine the possible values and frequencies for the expression $2a+3b$.

We list all possible pairs $(a,b)$ from $A \times A$ and compute the value of $2a+3b$.

Let $V_1 = \{2a+3b \mid a,b \in A\}$. The set of unique values is $V_1 = \{5, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 20\}$. Let $N_1(k)$ be the number of pairs $(a,b)$ such that $2a+3b=k$. From the table: $N_1(5)=1, N_1(7)=1, N_1(8)=1, N_1(9)=1, N_1(10)=1, N_1(11)=2$ (from (1,3) and (4,1)), $N_1(12)=1, N_1(13)=1, N_1(14)=2$ (from (1,4) and (4,2)), $N_1(15)=1, N_1(16)=1, N_1(17)=1, N_1(18)=1, N_1(20)=1$.

Step 2: Determine the possible values and frequencies for the expression $4c+5d$.

We list all possible pairs $(c,d)$ from $A \times A$ and compute the value of $4c+5d$.

Let $V_2 = \{4c+5d \mid c,d \in A\}$. The set of unique values is $V_2 = \{9, 13, 14, 17, 18, 19, 21, 22, 23, 24, 26, 27, 28, 31, 32, 36\}$. Let $N_2(k)$ be the number of pairs $(c,d)$ such that $4c+5d=k$. Since $\gcd(4,5)=1$, for any given sum $k$, there can be at most one pair $(c,d)$ that produces it within a certain range. In this case, each value in $V_2$ is unique for the given domain of $c$ and $d$. Thus, $N_2(k)=1$ for all $k \in V_2$.

Step 3: Find the common values in $V_1$ and $V_2$.

The condition for an element to be in $R$ is $2a+3b = 4c+5d$. We need to find the values $k$ that are present in both $V_1$ and $V_2$. $V_1 = \{5, 7, 8, \mathbf{9}, 10, 11, 12, \mathbf{13}, \mathbf{14}, 15, 16, \mathbf{17}, \mathbf{18}, 20\}$ $V_2 = \{\mathbf{9}, \mathbf{13}, \mathbf{14}, \mathbf{17}, \mathbf{18}, 19, 21, 22, 23, 24, 26, 27, 28, 31, 32, 36\}$

The common values are $K = V_1 \cap V_2 = \{9, 13, 14, 17, 18\}$.

Step 4: Calculate the number of elements in $R$ for each common value and sum them up.

For each common value $k$, the number of elements in $R$ corresponding to this value is $N_1(k) \times N_2(k)$.

• For $k=9$: $N_1(9) = 1$ (pair is (3,1)), $N_2(9) = 1$ (pair is (1,1)). Contribution = $1 \times 1 = 1$. The element is $((3,1),(1,1))$.
• For $k=13$: $N_1(13) = 1$ (pair is (2,3)), $N_2(13) = 1$ (pair is (2,1)). Contribution = $1 \times 1 = 1$. The element is $((2,3),(2,1))$.
• For $k=14$: $N_1(14) = 2$ (pairs are (1,4) and (4,2)), $N_2(14) = 1$ (pair is (1,2)). Contribution = $2 \times 1 = 2$. The elements are $((1,4),(1,2))$ and $((4,2),(1,2))$.
• For $k=17$: $N_1(17) = 1$ (pair is (4,3)), $N_2(17) = 1$ (pair is (3,1)). Contribution = $1 \times 1 = 1$. The element is $((4,3),(3,1))$.
• For $k=18$: $N_1(18) = 1$ (pair is (3,4)), $N_2(18) = 1$ (pair is (2,2)). Contribution = $1 \times 1 = 1$. The element is $((3,4),(2,2))$.

The total number of elements in $R$ is the sum of these contributions: $|R| = 1 + 1 + 2 + 1 + 1 = 6$.

3. Common Mistakes & Tips

• Scope of Variables: Ensure you correctly understand that $a, b, c, d$ are all from the set $A=\{1,2,3,4\}$.
• Counting Frequencies: Carefully count how many different pairs $(a,b)$ can result in the same value for $2a+3b$, and similarly for $4c+5d$. This is crucial for the multiplication principle.
• Intersection of Values: The relation requires the values of $2a+3b$ and $4c+5d$ to be equal, so you must find the common values in the sets of possible outcomes for each expression.

4. Summary

To determine the number of elements in the relation $R=\{((a, b),(c, d)): 2 a+3 b=4 c+5 d\}$ on $A \times A$ where $A=\{1,2,3,4\}$, we systematically calculated all possible values of $2a+3b$ and $4c+5d$ along with their frequencies. We then identified the common values between these two sets of results. For each common value, we multiplied the number of ways to obtain it from $2a+3b$ by the number of ways to obtain it from $4c+5d$. Summing these products gave the total number of elements in $R$. The common values were found to be $\{9, 13, 14, 17, 18\}$, leading to a total count of 6 elements in the relation.

The final answer is $\boxed{6}$.