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.)

wiseowl Feb 22, 2021

#1**0 **

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.)

thedudemanguyperson Feb 23, 2021