Difference in implementation of multiset and tuple in dice throws I recently learned about multisets and am told that a simultaneous throw of k-dice is a k-Multiset over the set of {1,2,3,4,5,6}.. mathematically expressed as $\frac{(n+1-k)!}{k!(n-1)!}$. However intuitively, I thought the number of outcomes would be $6^k$ and did a search and came across this other stackexchange question. Am unfortunately still rather confused. (Probability of predicting, then throwing, a particular multiset for 5 dice.)
Question is, what then is the correct method in determining the number of outcomes of a simultaneous cast of k dice? 
As far as I understand, a multiset should be used when we're just interested in finding the number of RESULTS and $6^k$ when we're interested in the number of WAYS to throw every possible result? Am I on the right track?
 A: Yes, you are. You can describe an outcome by a $6$-tuple $\langle d_1,d_2,d_3,d_4,d_5,d_6\rangle$ of non-negative integers, where $d_i$ is the number of dice that landed with face $i$ showing; clearly $\sum_{i=1}^6d_i=k$. The binomial coefficient $$\binom{k+6-1}{6-1}=\binom{k+5}5$$ is the number of such $6$-tuples.
However, most of these outcomes can be achieved in more than one way. If you give the $k$ dice individual identities, say by labelling them from $1$ through $k$, then an outcome (i.e., a specific way of achieving one of the $\binom{k+5}5$ outcomes counted above), is described by a $k$-tuple $\langle d_1,\dots,d_k\rangle$ of integers from the set $\{1,2,3,4,5,6\}$: here $d_i$ is the number showing on die number $i$. There are $6^k$ such $k$-tuples.
If you’re interested in the probability of some set of outcomes, you generally have to consider the more specific notion of outcome in the second paragraph: each of those $6^k$ sequences is equally likely and so occurs with probability $\frac1{6^k}$, while the multiset outcomes of the first paragraph are not equally likely.
