Catherine rolls a standard 6-sided die six times. If the product of her rolls is \(2500\) then how many different sequences of rolls could there have been? (The order of the rolls matters.)

Guest Jul 6, 2023

#2**0 **

Explanation: The prime factorization of \(2500\) is \(2^2*5^4\), so Catherine will roll 2 twos and 4 fives. Now we just need to deal with the amount of ways to reorganize those rolls. If we have 6 slots:

[] [] [] [] [] [], and place the 2 twos in them, the other 4 will be automatically placed.

And there are \(\frac{6*5}{2}\) 15 ways of doing this (you can also use the combination formula for this). That gives us \(\boxed{15}\) ways of doing so.

SoulKingBrook Jul 6, 2023

#3**0 **

2500 = 2^2 * 5^4

There are only 2 combinations as follows:

145555 ==6! /4! ==30 permutations

225555==6!/4!2!= 15 permutations

**Total ==30 + 15 ==45 different sequences of rolls.**

Guest Jul 6, 2023

#4