How many triples of positive integers $(a,b,c)$ satisfy $a\le b\le c$ and $abc=1,000,000,000?$ 
How many triples of positive integers $(a,b,c)$ satisfy $a\le b\le c$ and $$abc=1,000,000,000?$$

I find combinatorial questions like this quite difficult.
I expressed $1,000,000,000$ as $2^95^9$.
I let $a=2^p5^s$, $b=2^q5^t$ and $c=2^r5^u$. So I need to find the number of non-negative integer solutions for $p+q+r=9$ and $s+t+u=9$, which is $\binom{11}{2}=55$. 
Now I'm stuck. I would just say $55\times55$ but they're not* ordered pairs.
 A: You need to find the number of cases with $a \lt b \lt c$, $a=b\lt c$, $a \lt b =c$ and $a=b=c$, though you can join the second and third together into a set called "two equal", and we can call the first "none equal" and the last "three equal".
"Three equal" is easy: you want solutions to $3p=9$ and $3s=9$ and there only is $1 \times 1 = 1$ "three equal" solution.  
"Two equal" requires the number of solutions to $2p+r=9$ and $2s+u=9$.  This is $5 \times 5 = 25$. These will each appear once either as $(2^p5^s,2^p5^s,2^r5^u)$ or as $(2^r5^u,2^p5^s,2^p5^s)$ so we do not need to adjust for duplication. But $1$ is the "three equal" solution, leaving $25-1=24$ "two equal" solutions.
"None equal" is a little harder.  You have the $55\times 55$ you found.  But you need to subtract three times the $24$ "two equal" solutions [e.g. if you have a solution $(a,a,c)$ then it will appear also as $(a,c,a)$ and $(c,a,a)$] and one times the "three equal" solution.  Even then five-sixths of these "none equal"  solutions are in the wrong order.  So you want $\frac{55\times 55-3\times 24 -1 \times 1}{6}=492$ "none equal" solutions.
And then your answer is $1+24+492=517$ ordered solutions as lnwvr has found. 
