Question

Let $1440=x\cdot y$, where $x$ and $y$ are coprime to each other, then the possible pairs $(x,y)$ is

Answer 1

$1440=2^5\cdot 3^3\cdot 5$
Here, total different prime factors are $3$
$\therefore$ number of ways in which $1440$ can be resolved as product of two coprime numbers is $2^{3-1}=4$
However, $(x,y)$ can interchange their places.
$\therefore$ total possible pairs $(x,y)$ is $8$