Combinatorial Probability-Rolling 12 fair dice My text says, regarding combinatorial probability, "The number of outcomes associated with any problem involving the rolling of n six-sided dice is $6^n$."
I know that in combinatorial probability $P(A)=m/n$ where $m$ is the number of ways $A$ can happen and $n$ is the number of ways to perform the operation in question. But since we must be consistent in numerator and denominator with respect to order, wouldn't the number in the denominator depend on the whether the example I am in respects order? I.e, in any problem involving rolling $n$ dice, would I expect $P=X/6^n$ regardless of context?
 A: In any problem involving rolling $n$ fair dice once, yes.  By that I mean that the canonical set of outcomes of size $6^n$ is appropriate, and one can expect that probability will have shape $X/6^n$.
For certain problems, a much smaller set of outcomes may make calculations shorter. The great advantage of the canonical set of outcomes is that all the outcomes are equally likely. 
A: A problem in which the order of elements is unimportant can be changed into one where the order is important by counting each arrangement of $k$ elements whose order is unimportant $k!$ times. This can be pictured as numbering all the items.
A: That is usually very good advice, at least until you simplify.  
So for example, to find the probability that rolling twelve dice gives three prime numbers (not necessarily distinct) is $  \frac {3^{12}}{6^{12}}$ but that simplifies to  $\frac {1}{2^{12}}$. 
But it is possible to devise a problem where this advice does not work.  For example, roll a die; then roll a second and keep rolling the second until it is distinct from the first; and then roll a third until it is distinct from each of the first two.  What is the probability the three dice show three prime numbers? And if you do this four times to use all $12$ dice? It is  $\left(\frac 36 \times \frac 25 \times \frac 14\right)^4 =  \frac{1}{20^4}$ and the advice turns out to be wrong in this artificial case.   
My guess is that given the advice in the text, you have a reasonable expectation that all questions will be of the first sort.  But you should remain aware that the second kind is not impossible. 
