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

Guest May 16, 2023

#1**0 **

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 May 16, 2023

#2**0 **

There are 2 combinations as follows:

145555 = 6! / 4! = 30 permutations

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

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

Guest May 16, 2023

#3**+1 **

Get the prime factorization of 2500: 5^{4} * 2^{2} or 5 * 5 * 5 * 5 * 2 * 2. There are six numbers so roll each **5** four times and **2** two times (for the six rolls) or roll **5** four times and roll **4 **once as well as rolling a one. These are only possible combinations because 5 * 5 = **25** (bigger than six), and 5 * 2 = 10 (bigger than six). So 2 * 2 = 4 is the only other way to get another combination.

So there are 2 combinations, *5 5 5 5 2 2* = 6! / (4! * 2!) = 15 permutations, and *5 5 5 5 4 1* = 6! / 4! = 30 permutations.

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

*(Sorry for reposting the same solution Guest, but I thought he should know how you got the 2 different combinations just in case) *

Altee May 16, 2023