The number of ordered pairs of integers (a, b) such that a, b are divisors of 720 but ab is not is R, then which is/are CORRECT?

First consider the case a, b $>$  0. We have  $720=2^4{.3}^2.5$ , so the number of divisors of 720 is 5*3*2 = 30.
We consider the number of ways to select an ordered pair (a, b)
such that a, b, ab all divide 720.
Using the balls and urns method on each of the prime factors,
we find the number of ways to distribute the
factors of 2 across a and b is  ${(}_2^6)$ , the factors of 3 is  ${(}_2^4)$, the factors of 5 is  ${(}_2^3)$ .
So the total number of ways to select (a, b) with a, b, ab
all dividing 720 is 15 * 6 * 3 = 270. The number of ways to select any
(a, b) with a and b dividing 720 is 30 * 30 = 900, so there are
900 – 270 = 630 ways to select a and b such that a,b divide 720 but ab doesn’t.
Now, each a, b $>$  0 corresponds to four solutions $(\pm a,\pm b)$ 
giving the final answer of 2520. (Note that  $ab\ne 0$ .)
2520 = 4 * 630 = $2^3{3}^2{5}^1{7}^1$