Catherine rolls a 6-sided die five times, and the product of her rolls is 1500. How many different sequences of rolls could there have been? (The order of the rolls matters.)
+137 OOPS! I misread your problem and thought the 5 was a 2.
For this problem with 1500, it is still similar. 55562 and 35554 are there, so it is 5! / 3! and 5! / 3! = 20 and 20. 20 + 20 = 40
+137 ABCDE = 1200
1200 prime factorization = 24 * 3 * 52
Rolls: 5, 5, _, _, _
Fit 24 and 3 as product of 3 numbers (6 or less):
4, 4, 3
6, 4, 2
5, 5, 4, 4, 3 and 5, 5, 6, 4, 2
Ways to arrange 55443 = 5! / (2! * 2!) = 30
Ways to arrange 55642 = 5! / 2! = 60
30 + 60 = 90
+137 OOPS! I misread your problem and thought the 5 was a 2.
For this problem with 1500, it is still similar. 55562 and 35554 are there, so it is 5! / 3! and 5! / 3! = 20 and 20. 20 + 20 = 40
