There's probably a better way to solve this (likely with actual combinatorics), but I'd solve it as follows:

The trick relies on the fact that we're dealing with multiplication (and positive integers), so we can just break up 100 (1*1*100) into factors. We don't have to worry about order as all that matters is that a<b<c; and as long as all of the numbers are different, we can satisfy that inequality.

Then we just go through the possible solutions and eliminate those in which there are at least two equal factors.

Wolfram alpha concurs with my answer:

https://www.wolframalpha.com/input?i=0%3Ca%3Cb%3Cc%3B+a*b*c+%3D+100

Hope this helps!