Given 5 dice, how many ways can you roll exactly two 6s?
I just saw the question you posted that had a question 1 and not its context. Yes, that was my attempt at an answer. I thought you were commenting on the question's ambiguity (it was poorly worded). Now that I reread the context I see you weren't looking for an interpretation that was easily answered. I was answering:
You mentioned that it was an extremely advanced question, but because of how poorly it is worded, I offered an easily answered interpretation. Again, I just misread the context. Sorry about that.1.How many possible sums can you roll if you have the choice to use at most 3 dice? You must use at least one die.
comesz
That is a easy question to answer. " how did you calculate your answer?", we understand how to calculate dice outcomes.
A die has six outcomes: $1,~2,~3,~4,~5,~6$. Toss $n$ dice. Then there are $6^n$ different outcomes. We can model this with n-tuples.
If $0\le k\le n$ we can ask how many of those contain exactly $k$ $6's$? ANSWER: $\dbinom{n}{k}\cdot 5^{n-k}$.
We choose $k$ places to put a $6$, then there are $n-k$ places left to put any one of five other numbers.