How many unordered pairs of positive integers $(a,b)$ are there such that $\operatorname{lcm}(a,b) = 126000$? 
How many unordered pairs of positive integers $(a,b)$ are there such that $\operatorname{lcm}(a,b) = 126000$?

Attempt:
Let $h= \gcd(A,B)$ so $A=hr$ and $B=hp$, and $$phr=\operatorname{lcm}(A,B)=3^2\cdot 7\cdot 5^3 \cdot 2^4\,.$$ Let $p = 3^a5^b7^c2^d$ and $r = 3^e 5^f 7^g 2^s$. Notice, that given $p$ and $r$, $h$ is determined, so we can count $p$ and $r$. Multiplying $p$ and $r$ we get $$pr = 3^{(a+e)} 5^{(b+f)} 7^{(c+g)} 2^{(d+s)}\,,$$ and so $a+e = 0,1,2$.
For the first case we have $0+1 = 1$ possibility, similarly $2$ and $3$ for the other cases, so the total number is $6$. For $b+f$ we have $b + f = 0,1,2,3$ giving $10$ options. Similarly for $c + g$ we have $3$ choices and for $d + h$ we have $$1+2+3+4+5 = 15$$ choices. Multiplying these together, we get $$15\cdot 3\cdot 6\cdot 10 = 60\cdot 45 = 2700\,,$$ which is not equal to the given answer of $473$.
Edit: sorry for the weird variables. I think I've fixed everything, if not, please do point it out

I can't really follow what's going on in your attempt - it doesn't help that you use some letters with multiple inconsistent meanings and seem to start off with HCF/LCM swapped.
However, writing $A=2^a3^b5^c7^d$ and $B=2^e3^f5^g7^h$ you have $\max(a,e)=4$, etc. This means there are $2\times 5-1=9$ choices for $(a,e)$: choose one to be $4$; the other is one of $5$ options, and subtract $1$ because you just counted $4,4$ twice.
So the number of ordered pairs is $9\times7\times 5\times 3$. This is $1$ less than twice the number of unordered pairs, since every pair can be ordered in two ways except for $(126000,126000)$.
