+1671 Catherine rolls a 6-sided die four times, and the product of her rolls is 300. How many different sequences of rolls could there have been? (The order of the rolls matters.)
+600 The best way is to list here.
Clearly none can be $1$.
If one is $2$, then the rest multiply to $150$, so they can be $5,5,6$ only. This gives $\frac{4!}{2!}=12$ permutations.
If one is $3$, then the rest multiply to $100$, so they can be $5,5,4$ only. This again gives $12$ permutations.
If one is $4$, then the rest multiply to $75$, so they can be $3,5,5$ only. These have already all been counted.
If one is $5$, then the result multiply to $60$, so they can be $5,2,6$ or $5,3,4$. Both have already been counted.
If one is $6$, we find that all have already been counted again.
So $12+12=\boxed{24}$.
I don't like casework... You can use roots of unity filter but it's overkill (the outline is to prime factorize $300$ and then note that they are all of the form $2^a*3^b*5^c$. Use Stars and Bars and then RoUF for permutations.)
