It has nothing to do with permutations/combinations, but it is a counting problem.

Written in prime factorization form: .

Even though it is not the custom to use 1 as an exponent, I included it to make a point.

Add one to each exponent and multiply, .

So has six factors. We add the one to account for using 0 as an exponent.

Here is another example: .

Thus has factors.