How many positive integers are factors of a given number? I've been trying to find / generate a formula for the following problem: 


Given a number, how many positive integers are factors of this number.


In practice, you could simply build a table as such (lets assume the number is 36):

1 |  36 
2 |  18 
3 |  12 
4 |  9  
6 |  6

Thus there are 9 positive integers that are factors to 36.

This method seems like it would be taxing if the number was large. So I was thinking that if you knew the prime factorization for a number such as  $\,2 \cdot 2 \cdot 3 \cdot 3 = 36, \,$ there should be a formula for the number of unique combinations you can multiply these prime factors together. That's where I am stuck. Thanks in advance!
Hint: if the prime factorization of $\,n\in\Bbb N\,$ is

$$n=\prod_{k=1}^rp_k^{a_k}$$

then the number of divisors of $\,n\,$ is

$$\prod_{k=1}^r(a_k+1)$$
Let me explain this with the help of an example.
Let the number be 630.
Now the prime factorization of $630 = 2^1 × 3^2 × 5^1 × 7^1$.
Any factor is a combination of one or more numbers with their power ranging from 0 to maximum.

For 2 , the power can be 0 or 1 i.e. 2 ways.
For 3 , the power can be 0 or 1 or 2 i.e. 3 ways.
For 5, the power can be 0 or 1 i.e. 2 ways
For 7 , same as 5 i.e. 2 ways.

Now if you know basics of permutation and combination, the number of ways we can select any power of the above prime factorization is 

2×3×2×2 = 24.

Hence 630 has 24 factors.

Now the formula is Let  $N = (A^r)×(B^m)×(C^n)$ where A,B,C are the prime factors of N, then the total number of positive factors of N is given by (r+1)(m+1)(n+1).