I didn't know that till now

Let $n=\Pi_{i=1}^m{p_i}^{r_i}$ , where the $p_i$ are distinct primes and $r_i\in\mathbb{N}$ .

Suppose $n=a\times b$ , where $a$ and $b$ are coprime. Each prime factor $p_i$ may appear in only one of $a$ or $b$ (otherwise they would have a common factor greater than 1). In fact, each element ${p_i}^{r_i}$ must appear in exactly one of $a$ or $b$ .

Therefore, for $i=1,2,3,\dots m$ , decide whether the element ${p_i}^{r_i}$ should be a factor of $a$ or $b$ . There are $2^m$ possibilities for this. Since for every pair $(a,b)$ we have also counted the pair $(b,a)$ , each pair has been counted twice, so our result needs to be halved: $2^{m-1}$ .

Slightly cheating but, for twelve, the only way to express in coprime positive integers is 1x12 or 3x4. This means since m = 2, the answer is either m or 2^m-1. Taking 60 as an example, if the answer was m, you would predict there to be 3 ways (m = 3, {2^2, 3, 5}) to express in coprime integers, whereas if answer is 2^m-1, you expect there to be 2^2 = 4 ways to express. As it turns out, 60 can be expressed as 1x60, 3x20, 4x15, or 5x12. Therefore, answer is 2^m-1.