JEE 2023 — Mathematics PYQ

We have, $A=\{0,3,4,6,7,8,9,10\}$.

Case I: $x-y$ is odd, if one is even and one is even and x>y.
$\therefore$ Possibilities are $\{(3,0),(4,3),(6,3),(7,6),(7,4)\},(7,0),(8,7),(8,3),(9,8),(9,6),(9,4),(9,0),(10,9),(10,7),(10,3)$.
No. of cases = 15

Case II: $x-y=2$
$\therefore$ Possibilities are $\{(6,4),(8,6),(9,7),(10,8)\}$.
No. of cases = 4

So, minimum ordered pair to be added = 15 + 4 = 19