If N is number of pairs of two divisors of 75600 such that the product of these divisors is 60480, then the value of N is

$x=75600=2^4{3}^3{5}^2{7}^1$

Let divisor is $2^{a_1}{.3}^{b_1}{.5}^{c_1}{.7}^{d_1}\text{\hspace{0.17em}and\hspace{0.17em}}{2}^{a_2}{.3}^{b_2}{.5}^{c_2}{.7}^{d_2}$

Product of divisors = $2^6{.3}^3{.5}^1{.7}^1$

$\Rightarrow a_1+a_2=6\Rightarrow 3ways$

$\Rightarrow b_1+b_2=3\Rightarrow 4ways$

$\Rightarrow c_1+c_2=1\Rightarrow 2ways$

$\Rightarrow d_1+d_2=1\Rightarrow 2ways$

Total way $N=\frac{3\times 4\times 2\times 2}{2}=24$