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I saw this problem in an algebra 1 book and I can't reconcile the answer in the book with the knowledge I already have (which wasn't presented in the book). Here is the question:

In your turn of a certain game, you roll five dice at the same time. Do the outcomes of rolling the five dice represent a permutation or a combination? How many outcomes are possible?

My answers: It's a combination (I only left this part of the Q in because of how it affects the following).

There should be 252 combinations to my mind. There are 5 dice 🎲 and they have to be divided up among 6 sides. So n = 5, k = 6. Thus (n + k - 1)! / (n! (k - 1)!) = 10!/(5!(5!)) = 252 outcomes. Note that I assume rolling 1, 4, 4, 4, 6 = 6, 4, 4, 4, 1, and as such it's just a combination with repetition. HOWEVER:

The book solution says there's 7776 outcomes. I realize this is just (6)(6)(6)(6)(6)(6). Perhaps they used the word outcomes rather than combinations to suggest a different solution? I can get to this answer via common sense (each dice could have one of six different sides) but I thought maybe they said OUTCOMES to not give away part A of the question (that it's a combination, not permutation). Note that the book only covers simple permutations and combinations and does not cover those cases with repetition.

Am I wrong? Or is the book not correct? Or are we both somehow correct depending on interpretation? Thanks!

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  • $\begingroup$ I think that the book is not considering differently ordered results. I mean that, i.e., $(122334)=(433221)$ $\endgroup$ Jun 19 '17 at 14:30
  • $\begingroup$ Yeah you're right. Seems like a bad way to specify a question since it's not specific! 😜 $\endgroup$
    – Hanzy
    Jun 19 '17 at 14:54
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The thing with dice games is that we often want to know the odds of getting a particular event ('I have a pair of sixes').

Your result is correct for calculating the number of possible combinations, with interchangeable dices. However, if we want to find a result about, say, the sum of the dices, this way of counting cases is not particularly useful since there are less chances to get the combination (6,6,6,6,6) than the combination (1,2,3,4,5).

Your book calculates the number of 'outcomes' as if each dice is of a different colour (hence, not interchangeable) and getting (1,2,3,4,5) is not the same as getting (2,3,4,5,1). The main point of this approach is to have outcomes that all have the same odds of occuring (namely, $1/6^5$). Then you can solve a lot of probability questions (odds of at least one 6, odds of sum>20, etc) just by counting the favourable cases.

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    $\begingroup$ Thank you, this makes a lot of sense. I'm reviewing all my math from Algebra 1 - Calculus to prepare for entering another degree program, quantitatively focused this time. I notice when going backwards like this sometimes more advanced concepts I've learned aren't meant to have been learned by the reader of the book so I confuse myself trying to impose that framework on a simpler question. $\endgroup$
    – Hanzy
    Jun 19 '17 at 17:42
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Clearly it all depends on if we consider the dices to be distinct or interchangeable. From my reading of the question, you would be correct, and in general when considering rolling dices we dont distinguish them.

I guess one might understand the book's solution to be valid, if it had been explained that the dices have different colors for example.

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  • $\begingroup$ Thanks for the reply; can you show me one step further how to apply the idea of a permutation with repetition to this and arrive at 7776? To my mind we have 30 sides taken 5 at a time, but there are 6 copies of each side. Plugging those numbers into my equation doesn't give me the solution. I can tell I'm missing the fact that we aren't arranging all 30 sides into a permutation in that line of thinking though. $\endgroup$
    – Hanzy
    Jun 19 '17 at 14:39
  • $\begingroup$ Duh, after I posted I realized we don't have any duplicate sides on the same dice. 🙄. Still can you take me a step further? $\endgroup$
    – Hanzy
    Jun 19 '17 at 14:46
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    $\begingroup$ Ok, I got it by just thinking simple and not going overly complex. Each dice is just nPr (6,1). Multiply that by 5 dice... that's the 7776. $\endgroup$
    – Hanzy
    Jun 19 '17 at 14:49

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