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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.)

 Jul 9, 2021

Best Answer 

 #2
avatar+128 
+2

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

 Jul 9, 2021
 #1
avatar+128 
+1

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

 Jul 9, 2021
 #2
avatar+128 
+2
Best Answer

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

HighSchoolDx Jul 9, 2021

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