Find the number of ordered pairs that have a product of $10!$ and a least common multiple of $9!$ Problem

How many ordered pairs $(a,b)$ of positive integers have a product of $10!$ and a least common multiple of $9!$?

I was told my answer of $16$ to this problem was wrong. I don't see how since we just have to distribute the max powers of $2,3,5,$ and $7$ and there are $16$ ways to do that. We don't need to divide by $2$ since $(3,4)$ is different from $(4,3)$, for example.

You can rephrase it as: how many ordered pairs $(a,b)$ of positive integers satisfy $ab=10!$ and $\gcd(a,b)=10$, because $ab=\gcd(a,b)\text{lcm}(a,b)$.
I.e. let $a=10a_1, b=10b_1$, and you're asking how many ordered pairs $(a_1,b_1)$ of positive integers satisfy $a_1b_1=\frac{10!}{100}=2^6\cdot 3^4\cdot 7$ and $\gcd(a_1,b_1)=1$, so all the cases are $$a_1\in\{1,2^6,3^4,7,2^6\cdot 3^4,3^4\cdot 7,2^6\cdot 7,2^6\cdot 3^4\cdot 7\},$$
so the answer is $8$ ordered pairs.
