How many ways can the number $2160$ be written as a product of factors which are relatively prime to each other?

Question

How many ways can the number $2160$ be written as a product of factors which are relatively prime to each other?

I was confused by this question because couldn't we just add $1$ into the factorization every time? For example, $2160$ and $2160 \cdot 1$ would count as distinct factorizations. If that is not the case, then I tried this:

Note that if a prime $p$ divides a factor of $2160$, then it must be the largest power of the prime $p$ dividing $2160$. Then we see that $2160 = 2^4 \cdot 3^3 \cdot 5$ and that $2160$ can't be written as the product of four or more relatively prime factors which aren't $1$. For three factors, we just have $2^4 \cdot 3^3 \cdot 5$. For two factors, we have $(2^4 \cdot 3^3) \cdot 5, (3^3 \cdot 5),$ or $(2^4 \cdot 5) \cdot 3^3$. Finally for one factor we just have $2160$. Therefore there are $5$ different ways in which $2160$ can be written as a product of factors which are relatively prime to each other.

Answer 1

You're right that the problem only makes sense if factors of $1$ are not allowed.

Under this interpretation, your analysis seems to make sense: The number of ways to write $n$ as a product of coprime factors $>1$ (not counting the order of these factors) is exactly the number of partitions of the set of prime factors of $n$.

The number of partitions of a $k$-element set is the Bell number $B_k$. This sequence is A000110 in OEIS.

Since $2160$ has $3$ prime factors your answer is $B_3=5$.

Answer 2

If the number is $n=p_1^{\alpha_1}p_2^{\alpha_2}...p_n^{\alpha_n}$, you can forget about the powers of the primes and find the number of partitions of the set $\{p_1,p_1,...,p_n\}$