This is different to the other questions asking for the number of factors of a natural number based on the number of prime factors because the group is non-commutative.
If I have a monoid $(M, *)$ with some elements which are "prime" (i.e. $\{n :\neg(\exists a,b \in M, a \neq e, b \neq e: a * b = n) \land n \neq e\}$), where each composite element has a unique prime factorization (i.e. $p_1 * p_2 * p_3\ldots* p_a = n$, where the order of $p_1, p_2,\ldots p_a$ is also included in the definition of uniqueness) then given the prime factorization of an element, how many factors does that element have?
I feel like this should be a fairly simple combinatorics problem, but I just can't figure out the answer.
Let $A$ be the set of prime elements. Your condition on $M$ implies that $M$ is the free monoid $A^*$. It follows that the number of (nonempty) factors of an element $a_1 \dotsm a_n$ is $n+1 \choose 2$.