ways to roll a die with probability and permutation how many ways can I roll a die so it is all different numbers every time?
...for instance, what is all the possibilities of rolling all 3 different numbers...
so 1 2 3, 2 3 4, etc... but 3 2 1 also counts ..
I don't understand this concept, I now the total different methods is 6 cubed, but we only need the different combinations here..
I tried 6! 5! 4! but it doesn't make sense it shouldn't be bigger than 216
 A: According to the original post, you want to count the ways to roll an unbiased 6-sided die $k$ times so that each roll has a distinct result, for $k\in\{1,2,3,4,5,6\}$.
So for each $k$ you count the ways to select and permutate $k$ of $6$ results.
$$\begin{align}\text{count} &= \sum_{k=1}^6 {^6\mathbf P_k}
\\ & = \sum_{k=1}^6 {6\choose k}k!
\\ & = 6! \sum_{k=1}^6 \frac{1}{(6-k)!}
\end{align}$$

According to one of the comments, you just want to count the ways to roll an unbiased 6-sided die $3$ times so that each roll has a distinct result.  This is simply:
$$\begin{align}\text{count}_2 &= {^6\mathbf P_3}
\\ & = {6\choose 3}3!
\\ & = \frac{6!}{3!}
\end{align}$$
A: $6 * 5 * 4 = 120$ since there are $6$ possible outcomes of the first die roll, $5$ for the 2nd roll such that it doesn't match the first, and $4$ for the third die rolls such that it doesn't match any previous rolls.  Here, we are considering $1,2,3$ (for example) as different than $3,2,1$.  If you instead don't consider those different rolls (such as you rolled all $3$ dice at once and then sorted the numbers from low to high such that $3,2,1$ would become $1,2,3$, then there are only $6 \choose 3$ = $20$ different ways to roll $3$ different numbers.
