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.)
+40 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.
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.
+40 Oh you're right @guest, completely forgot about that part lol. Good job!
