Dice Combination / Permutation with Repetition 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!
 A: 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.
A: 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.
