Let $A$ be the set of all positive integer divisors of $3^6 5^8 11^{10} 17^{15}$. Define the relation $R$ on $A$ as follows. For $x, y \in A, xRy$ when $x | y$. Determine the number of ordered pairs that are in the relation $R$.
I know answers come in the form $\binom{n+r-1}{r}$, the number of ways to choose $r$ elements from a set of $n$ elements if repetitions are allowed.
The answer is $\binom{7+2-1}{2} \binom{9+2-1}{2} \binom{11+2-1}{2} \binom{16+2-1}{2}$, but I don't quite understand why $r$ is $2$ in this case.
You take two divisors: $x$ and $y$. Hence $r=2$. More precisely, for each prime factor you take two numbers: the exponent for $x$ and the exponent for $y$.