Now, each possible factor of $81000$ is of the form $2^x\cdot 3^y\cdot 5^z$. When we multiply the three factors we want, we need that the three exponents of the $2$s and the $5$s add up to $3$ and that the exponents of the $3$s add up to $4$. That is, we need to find how many non-negative solutions are to $a+b+c=3$, and how many non-negative solutions are to $a+b+c=4$, which we can do with stars and bars. Thus the number of possible ways of factoring $81000$ as three factors will be $\binom{3+3-1}{3}\cdot\binom{4+3-1}{4}\cdot\binom{3+3-1}{3}=1500$.

(The handout should now be updated on Edurila, thank you for letting us know!)