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Manhattan Review/Combinatorics Question

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by Gat0rbabe » Tue Mar 09, 2010 5:57 pm
"A game is being played with a die. The die is rolled three times and each roll recorded. How many different arrangements are possible? What if previously rolled numbers do not count?" Manhattan Review/Chapter 10 Combinatorics Practice Problem #2

First of all, what is the question asking? Second, I do not believe Manhattan review has answer keys to practice questions. I have all 4 books, study guide and explanation guide.

Thanks in advance for posting an answer!
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by harsh.champ » Fri Mar 12, 2010 6:18 am
Gat0rbabe wrote:"A game is being played with a die. The die is rolled three times and each roll recorded. How many different arrangements are possible? What if previously rolled numbers do not count?" Manhattan Review/Chapter 10 Combinatorics Practice Problem #2

First of all, what is the question asking? Second, I do not believe Manhattan review has answer keys to practice questions. I have all 4 books, study guide and explanation guide.

Thanks in advance for posting an answer!

What if previously rolled numbers do not count? - Thats quite confusing.

6^3.-That would be my answer.
i don't trust this question's source.

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by jpjp » Sat Mar 13, 2010 7:43 pm
Sounds like it's two questions:

How many different combinations can you have?

and

How many different combinations can you have if you dont count rolls where a repeat number comes up (ie if you roll a 6 in the first roll, then u can't roll another 6. If you do, it doesn't count and u roll until u get something different than a 6).


The answer to the first would simply be 6^3 = 216
The answer to the second would be 6*5*4 = 120
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