How many $6$-sided and $8$-sided standard dice exist? Standard means that the sum of the opposite sides is $7$ for $6$-sided die and $9$ for $8$-sided die.
 A: Another way to do the counting:
Let's start by counting how many ways we can put numbers $1$ through $2n$
on a regular $2n$-sided die if we fix the position of the die in space (do not allow rotating the die in order to change where the numbers appear).
This is just the number of ways we can permute the $2n$ numbers among the $2n$ places where they can be written;
the number of ways is 
$$(2n)! = (2n)(2n-1)(2n-2)\cdots(3)(2)(1).$$
Now count the number of ways we can orient the die (allowing it to rotate now) so that it occupies the same space as before, but not necessarily with the same numbers facing the same directions.
Call this number $|R_{2n}|.$
This is the number of times each possible numbering of the die was repeated when we numbered the faces without considering the possibility of turning the die.
So the total number of ways to number a regular $2n$-sided die, not counting rotations of the same die as different numberings, is
$$\frac{(2n)!}{|R_{2n}|}.$$
But as pointed out in another answer,
the total number of ways to number a regular $2n$-sided die
is exactly $(2n-1)!! = (2n-1)(2n-3)(2n-5)\cdots(5)(3)(1)$
times the number of standard numberings of the die
(if any standard numberings exist).
So the total number of standard numberings is
$$\frac{(2n)!}{(2n-1)!!\cdot|R_{2n}|} = \frac{(2n)!!}{|R_{2n}|},$$
where $(2n)!! = (2n)(2n-2)(2n-4)\cdots(6)(4)(2).$
There are only five regular polyhedra, for which these numbers come out to:
\begin{array}{rrrrrr}
2n & (2n)! & |R_{2n}| & \text{number of dice} & (2n)!! & \text{standard}\\ \hline
 4 &    24 & 12 &    2 &   8 &  - \\
 6 &   720 & 24 &   30 &  48 &  2 \\
 8 & 40320 & 24 & 1680 & 384 & 16 \\
12 &           479001600 & 60 &           7983360 &      46080 &      768 \\
20 & 2432902008176640000 & 60 & 40548366802944000 & 3715891200 & 61931520
\end{array}
There obviously are no "standard" numberings for a $4$-sided die, since no face has an "opposite" face, but the formulas still count how many ways the die can be numbered without that restriction.
A: A fun way to do d8:
First, rotate it so that the 1 is facing you, point up.  The 8 is on the opposite side.
Take the top left edge.  There are 6 possible numbers there.
Going clockwise, there are 4 possibilities for the next one, and then 2 for the last one.
This is 6*4*2, which is 48.
Now, the first choice -- the top left edge -- is symmetric in picking any edge.  In fact, we have enumerated each arrangment 3 times because of this.
So 48/3 = 16.
We can do the same with a 6 sided die.  First, put the 1 facing you.
There are 4 possiblities for the top, leaving 2 for the next clockwise side.  The choice to pick top has 4 equivalent choices to start with, so we overcounted by a factor of 4, leaving 2 arrangements total.
We can then solve the d20 case.  "Clockwise around the 1" doesn't work, but any path of length N/2-1 that doesn't touch opposites work.  In the d20 case, there is a 3 fold-symmetry (like the d8 case).
Thus there are 18*16*14*12*10*8*6*4*2/3, or 61931520, different arrangements of a 20-sided die, and 10*8*6*4*2/5, or 768, of a 12-sided die.
Or, for a dN with K-fold symmetry around a side and each side determining its opposite, there are (N-2)!!/K ways to number the die.
As there are only 4 different platonic solids with opposite sides, this is less than completely useful.
A sanity check on the 12 sided die:
 2 3
4 1 5
  6

5! arrangements of 2-6.  Each one can be swapped for its opposite (2^5).  This double-counts up to 5-fold rotation (/5).  Answer is 768.
Double checking for d20 is trickier.
A: For six-sided there are just two possibilities. This is because the $1$, $2$ and $3$ have to meet at a corner and once you place them, the rest of the die is determined. You can either have $1,2,3$ going clockwise or anticlockwise round the corner, leading to two possibilities which are mirror images of each other.
For eight-sided it is much more complicated, as there are many ways to arrange $1,2,3,4$ without having any of them opposite another. If $1$ and $2$ are adjacent (i.e. on two faces which share an edge), there are $4$ choices for where to place $3$ and for each of those there are $2$ choices for which face is $4$ (then $5,6,7,8$ are determined). This gives $8$ different options, and there are $8$ more obtained by swapping $2$ and $7$ so that $2$ and $1$ are not adjacent. So in total there are $16$ arrangements. 
(edit) Incidentally, the total number of ways to number a die with $2n$ faces is the number of standard ways times $(2n-1)!!$ (i.e. the product of all odd numbers up to $(2n-1)$). This is because there are $(2n-1)!!$ ways to divide the numbers into $n$ pairs, and then the number of ways to put each pair on opposite faces is the same as the original problem. So there are $30$ six-sided dice and $1680$ eight-sided dice.
A: Here is another way of counting the octahedral (8=sided) case.
Bring the triangle with $1$ on to the front. It has three numbers adjacent from the pairs $(2,7);(3,6);(4,5)$. Rotate so that $2$ or $7$ is at the base of the triangle ($2$ possibilities). For each case there are $4$ possibilities for the left side of $1$ and two possibilities remain for the right-hand side. Total $16$.
For the cube. I can always rotate so I have $1$ at the front and $2$ at the base. Then there are two possibilities for placing $3$ and $4$.
