+0

# Help ASAP!!!

0
30
5

Catherine rolls a standard -sided die six times. If the product of her rolls is 2700 then how many different sequences of rolls could there have been? (The order of the rolls matters.)

Jun 25, 2023

#1
-1

The prime factorization of 2700 is 2^2⋅3^3⋅5^1. This means that Catherine rolled a 2 twice, a 3 three times, and a 5 once. The order of the rolls matters, so there are 6!/2!3!=30 possible sequences of rolls.

Here are the 30 possible sequences:

1. 2 2 3 3 3 5 2. 2 2 3 3 5 3 3. 2 2 3 5 3 3 4. 2 2 5 3 3 3 5. 2 3 2 3 3 5 6. 2 3 2 3 5 3 7. 2 3 2 5 3 3 8. 2 3 5 2 3 3 9. 2 3 5 3 2 3 10. 2 3 5 3 3 2 11. 2 5 2 3 3 3 12. 2 5 3 2 3 3 13. 2 5 3 3 2 3 14. 2 5 3 3 3 2 15. 3 2 2 3 3 5 16. 3 2 2 3 5 3 17. 3 2 2 5 3 3 18. 3 2 3 2 3 5 19. 3 2 3 2 5 3 20. 3 2 3 5 2 3 21. 3 2 3 5 3 2 22. 3 2 5 2 3 3 23. 3 2 5 3 2 3 24. 3 2 5 3 3 2 25. 3 5 2 2 3 3 26. 3 5 2 3 2 3 27. 3 5 2 3 3 2 28. 3 5 3 2 2 3 29. 3 5 3 2 3 2 30. 3 5 3 3 2 2

I hope this helps!

Jun 25, 2023
#5
0

Thanks, that's right!

Guest Jun 26, 2023
#2
-1

2700 = 2^2 * 3^3 * 5^2

There are a total of 6 combinations as follows:

125569==6! / 2! ==360 permutations

134559==6! / 2! ==360 permutations

135566==6! / 2!2! ==180 permutations

223559==6! /2!2! ==180 permutations

233556==6! /2!2! ==180 permutations

333455==6! /3!2! ==60    permutations

Total ==360+360+180+180+180+60==1,320 different sequences of rolls.

Jun 25, 2023
edited by Guest  Jun 25, 2023
#4
+2

You had the right idea, but remember you are tossing 6-sided die 6 times. So you can't have numbers > 6.

135566 ( 180 ) , 233556 ( 180 ) , 333455 ( 60 ) , >>Total combinations = 3

>>Totals in parentheses = 420 permutations (or different sequences of rolls)

Guest Jun 25, 2023
#3
-1

To find the number of different sequences of rolls that could have resulted in a product of 2700, we can consider the prime factorization of 2700.

The prime factorization of 2700 is:

2700 = 2^2 * 3^3 * 5^2

Since we are rolling a standard 6-sided die, the possible outcomes for each roll are numbers 1 to 6. We need to distribute the prime factors among the six rolls.

Let's consider the factor 2. Since 2^2 is one of the prime factors, we need to assign it to two of the rolls. There are 6C2 ways to select two rolls out of six, which can be calculated as:

6C2 = 6! / (2! * (6 - 2)!) = 6! / (2! * 4!) = (6 * 5) / (2 * 1) = 15

Similarly, for the factor 3, we have 3^3. We need to assign three 3's to three of the rolls. There are 6C3 ways to select three rolls out of six, which can be calculated as:

6C3 = 6! / (3! * (6 - 3)!) = 6! / (3! * 3!) = (6 * 5 * 4) / (3 * 2 * 1) = 20

For the factor 5, we have 5^2. We need to assign two 5's to two of the rolls. There are 6C2 ways to select two rolls out of six, which is the same as the selection for the factor 2, so it is also 15.

To find the total number of different sequences, we multiply these counts together:

Total number of sequences = 15 * 20 * 15 = 4500

Therefore, there could have been 4500 different sequences of rolls that would result in a product of 2700.

Jun 25, 2023