Determining the number of pairs $(a,b)$ where $a\cdot b=n$ and $gcd(a,b)=1$

Question

There is a random number $n$.

Assuming $a$ and $b$ are divisors of $n$, I need to determine the number of pairs $(a,b)$ such that $a\cdot b=n$ and the greatest common divisor of $a$ and $b$ is $1$.

I have no idea how to do this, I've seen some problems involving gcd being solved with Euler's totient function, but that wasn't not part of the class material. Can anyone help me with this problem?

P.S.: The pairs $(a,b)$ and $(b,a)$ are considered equal.

Answer 1

Suppose the distinct prime divisors of $n$ are $p_1,\cdots, p_k$. Then we can build $a$ by picking which primes to include in it. Hence there are $2^k$ ways to choose $a$. By declaring that $(a,b)=(b,a)$ you introduce a symmetry of order $2$, so the answer becomes $2^{k-1}$.