Let $A=\{1,2,3,4\}$ and $\mathrm{R}$ be a relation on the set $A \times A$ defined by

$R=\{((a, b),(c, d)): 2 a+3 b=4 c+5 d\}$. Then the number of elements in $\mathrm{R}$ is ____________.

$2a + 3b = 4c + 5d$ <br/><br/> Given A = {1, 2, 3, 4}, the maximum value of $2a + 3b$ is 20, when (a, b) = (4, 4), and the minimum value of $4c + 5d$ is 9, when (c, d) = (1, 1). Therefore, the possible values for $2a + 3b = 4c + 5d$ are 9, 13, 14, 17, 18, and 19. <br/><br/> Now, let's find the combinations of (a, b), (c, d) that satisfy the given equation: <br/><br/> 1. $2a + 3b = 9 \Rightarrow (a, b) = (3, 1) \Rightarrow (c, d) = (1, 1)$<br/><br/> 2. $2a + 3b = 13 \Rightarrow (a, b) = (2, 3) \Rightarrow (c, d) = (2, 1)$<br/><br/> 3. $2a + 3b = 14 \Rightarrow (a, b) = (4, 2) \Rightarrow (c, d) = (1, 2)$<br/><br/> 4. $2a + 3b = 14 \Rightarrow (a, b) = (1, 4) \Rightarrow (c, d) = (1, 2)$<br/><br/> 5. $2a + 3b = 17 \Rightarrow (a, b) = (4, 3) \Rightarrow (c, d) = (3, 1)$<br/><br/> 6. $2a + 3b = 18 \Rightarrow (a, b) = (3, 4) \Rightarrow (c, d) = (2, 2)$<br/><br/> There are a total of 6 elements in the relation R for the given equation with the specified values of a, b, c, and d.