Multiplication partitioning into k distinct elements Let's say I have a list with the prime factors of a number $n$ and their corresponding exponents. Is there a formula to calculate how many multiplications with $k$ distinct factors of $n$ are possible? For example, $12 = 2^2 \times 3$, so $F(12,2)$ would be:
$[2,6][3,4][1,12] = 3$
Note that the order of each set of factors does not matter. I have a feeling I only need the exponents of each prime factor, but I'm at a loss on how to actually proceed.
 A: Yes! From a number theoretical perspective, this is known as the divisor function. Or, even more generally, $\sigma_k (n)$ where the sigma function denotes the sum of all divisors of an integer to the power $k$. If we set $k=0$ we see that this is exactly the same as counting all possible divisors of this integer.
The general form of the divisor function (sometimes denoted $d(n)$) is as follows:
$$\prod_{p^\alpha || n} (\alpha +1)$$ where $p^\alpha || n$ means that $p^\alpha |n$ but $p^{\alpha +1}$ does not. Also, p denotes a prime divisor. Also, if the notation is getting to you, just thinking of it as factorizing an integer, and then taking the product of 1 more than all the exponents of each divisor. (Do an example!)
A: Any number in prime factorization would look like:
N = X_1^a * X_2^b * X_3^c * ... where X_1, X_2, X_3, ... are its prime factors
Notice that all powers of X_1 will be factors of N. So we can vary a from 0 through 1, 2, 3, ... to a. i.e. a can have a+1 values and for all these, X_1^a will be a factor. Similarly X_2 and X_3 and so on will be factors from 0 to b+1 and from 0 to c+1 etc. Also, we can choose X_1, X_2, X_3 etc. simultaneously. So, the total factors of N should be (a+1)(b+1)(c+1)*....
Since you want only pairs and any one factor of N can pair up with exactly one other factor to give N on multiplication, you simply divide number of factors by 2 and you have
[(a+1)(b+1)(c+1)....]/2
