Assume I have $8$ dice that have different colors. I pick one at random and roll it, and then do it for each of the remaining. What are the total combinations?
For $8$ dice, it would be $8^6$, but because the order matters, I assume we must do permutations.
I thought it was $(8 \cdot 6)^6 + (7 \cdot 6)^6 \ldots$, but this does not seem right.
Example: colors are Red, Yellow, Orange, Green, Blue, Violet, Pink, and White Rolled $8$ dice and they all came up 1, and the colors picked in order were RYOGBVPW. This is a different result than if I did it again and each dice came up as a $1$ again but the colors were in the order RYOGBVWP.
For eight D6s of different colours you'll need to do some multiplication.
Certainly there are $8!$ ways to choose the dice. Each die has six faces. So, the number of total rolls, ignoring colour, is $6^8$, while the total number of ways to arrange the colours is $8!$, so the total number of rolls is $\boxed{8! \cdot 6^8}$.