jmc

The prime factorization of $2310$ is $2310=2\cdot 3\cdot 5\cdot 7\cdot 11.$ Therefore, we have the equation $$abc=2310=2\cdot 3\cdot 5\cdot 7\cdot 11,$$where $a,b,c$ must be distinct positive integers and order does not matter. There are $3$ ways to assign each prime number on the right-hand side to one of the variables $a,b,c,$ which gives $3^5=243$ solutions for $(a,b,c).$ However, three of these solutions have two $1$s and one $2310,$ which contradicts the fact that $a,b,c$ must be distinct. Because each prime factor appears only once, all other solutions have $a,b,c$ distinct. Correcting for this, we get $243-3=240$ ordered triples $(a,b,c)$ where $a,b,c$ are all distinct.

Finally, since order does not matter, we must divide by $3!,$ the number of ways to order $a,b,c.$ This gives the final answer, $$\frac{240}{3!}=\frac{240}{6}=\boxed{40}.$$