Let A = \ 1, 2, 3, 4\ and R be a relation on the set A × A defined...

Given,

Set $A = \left\{1, 2, 3, 4\right\}$

And relation on $A \times A$ as $2a + 3b = 4c + 5d$

Now, Maximum value of $2a + 3b = 20$ at \left(4, 4\right)$</p><p>And Minimum value of$4c + 5d = 9$at \left(1, 1\right)$

So, $4c + 5d$ can be equal to $9, 13, 14, 17, 18, 19$

Now $2a + 3b$ can be $9$ if \left(a, b\right) = \left(3, 1\right) ⇒ \left(c, d\right) = \left(1, 1\right)$</p><p>Similarly,$2a + 3b$can be$13$, if$\left(a, b\right) = \left(2, 3\right) ⇒ \left(c, d\right) = \left(2, 1\right)$</p><p>And$2a + 3b$can be$14$, if$\left(a, b\right) = \left(4, 2\right)\text{or}\left(1, 4\right)$</p><p>$⇒ \left(c, d\right) = \left(1, 2\right)$</p><p>Similarly,$2a + 3b$can be$17$, if$\left(a, b\right) = \left(4, 3\right) ⇒$\left(c, d\right) = \left(3, 1\right)$

And $2a + 3b$ can be $18$ if $\left(a, b\right) = \left(3, 4\right) ⇒$

\left(c, d\right) = \left(2, 2\right)$</p><p>Hence, there are total$6 elements which satisfy the given relation.