In how many ways can we enumerate from 1 to 20 the sides of a icosahedral In how many ways can we enumerate from 1 to 20 the sides of a icosahedral? Consider the sides indistinguishable.
I thought about starting off by fixing 1 to one of the sides and then choosing the number of its opposite side but I can't go much further than that
 A: Suppose we start by ignoring symmetry and count all $20!$ labelings. In doing so, we've counted each distinct labeling in exactly 60 different orientations:

*

*Some (arbitrary) base orientation $x$

*$x$ rotated about an axis that goes through a pair of opposite vertices (6 axes $\times$ 4 nontrivial rotations = 24 rotations)

*$x$ rotated about an axis that goes through the centers of two opposite faces (10 axes $\times$ 2 nontrivial rotations = 20 rotations)

*$x$ rotated about an axis that goes through the centers of two opposite edges (15 axes $\times$ 1 nontrivial rotation = 15 rotations)

So the number of distinct labelings is $20!/60$. This agrees with Mathew's answer.
We can think of this as a simple application of Burnside's Lemma.
A: I am assuming you want to put a unique number on each face. For example, having a 1 on two or more faces is not allowed.
Say you pick up such a icosohedron. You can turn it so that the 1-face is facing you. Once you do this, notice there are three sides touching the 1-face. You can rotate the icosohedron so that the face with the smallest number is directly below the 1-face. The choice for number for the below face must be one of the following: $\{2,3,...20\}$. Lets consider each case separately.
If you pick 2, there are 18 choices for the face to the upper left of the 1-face and 17 for the face to the upper right (because you can't repeat numbers). Once you pick these numbers, we have numbered 4 of the 20 faces, so there are 16 faces left and 16 numbers to put on them. Therefore there are 16! ways to do this. Thus we have $18*17*16!$ total ways in this case.
If you pick 3, there are 17 choices for the face to the upper left of the 1-face and 16 for the face to the upper right. This is because the bottom face had to be the smallest number, so we can't pick 2 for either of the upper faces. Then, just as before, we have 16 faces remaining and 16 numbers for the faces, so we have $16!$ ways to do this. Thus we have $17*16*16!$ total ways in this case.
Hopefully you see the pattern. If we count up all the possible ways we get:
$(18*17+17*16+...2*1)*16!$ total numberings. If you would like a simplified expression, I invite you to try and do so yourself.
A: Label one arbitrary face with number $1$.
Choose three of the remaining $19$ numbers; that can be done in $\binom{19}3$ ways.
Apply the three chosen numbers to the three faces bordering face number $1$; that can be done in $2$ distinguishable ways, clockwise or counterclockwise.
Apply the $16$ remaining numbers to the other $16$ faces; this can be done in $16!$ different ways.
The final answer is $$\binom{19}3\cdot2\cdot16!=\boxed{\frac{19!}3}$$ which agrees with the other answers.
A: Another approach to doing this calculation is to consider the placements of two specific numbers. With two numbers positioned, you have the orientation given uniquely in almost all cases (the exception is when they're opposite each other, in which case a third number is necessary). Note, however, that we must factor in the distance between the two numbers.
For simplicity, we'll use the defining numbers as $1$ and $2$.
Suppose that a specific side is labelled $1$. Now, how many unique placements are there for the $2$? Well, there are three ways that the $2$ can be adjacent to the $1$... but they're equivalent, so we treat all three as identical, and thus only one position.
The $2$ could be in a position that is adjacent to one of those three sides. There are six such faces. You might be tempted to say that they're all identical... but they're not. A close examination reveals that half of them are clockwise from a $1$-adjacent side, while the other half are anti-clockwise.
The same applies for the next two "layers", giving a total of 6 possible placements of the $2$ that aren't opposite each other.
For these positions, we can then place the remaining 18 numbers in any arrangement we choose, so that's a total of $6\times 18!$ - but we haven't considered the case that the $2$ is opposite the $1$.
In that situation, we choose the placement of the $3$, with the $2$ locked into place. This gives us $6\times 17!$ arrangements.
And so, the total number of unique arrangements is $6\times 18!+6\times 17!$, which can be simplified to $6\times 19\times 17!$ - this is the same as the answers given by Mathew and Karl
