Background
For reference, the board for this game is a large regular hexagon consisting of 19 regular hexagonal tiles in a grid like:
\begin{array}{ccccccccc}
& & \text{a} & & \text{b} & & \text{c}\\
& \text{d} & & \text{e} & & \text{f} & & \text{g}\\
\text{h} & & \text{i} & & \text{j} & & \text{k} & & \text{l}\\
& \text{m} & & \text{n} & & \text{o} & & \text{p}\\
& & \text{q} & & \text{r} & & \text{s}
\end{array}
The 19 small hexagons come from a selection as follows: 4 sheep tiles, 4 wood tiles, 4 wheat tiles, 3 brick tiles, 3 ore tiles, and 1 desert tile. In the game, while different sheep tiles (resp. wood tiles, etc.) may look different and have non-symmetric designs on them, they are all functionally equivalent (in any rotation), which is what matters here.
Easy version
I feel there are three versions of this problem, based on what counts as "the same board". Firstly, if rotating the board or a mirror image of the board would count a different arrangement, then counting is relatively simple: There are 19 tiles, and they go in 19 spots ("abcdefghijklmnopqrs" above), and there are some equivalent tiles, so this is similar in spirit to "how many ways can you rearrange the letters of MISSISSIPPI?". The answer to this question is just $\dfrac{19!}{4!4!4!3!3!1!}=244\,432\,188\,000$.
However, the question mentions
for a circle with n elements there are (n-1)! possibles ways to order it [sic]
so I assume that certain boards are intended to be equivalent to other boards, at least if you can get to one from the other by a rotation.
Medium version
Now let's assume that rotations count as the same board (so that if you walk around a board it counts as the same setup). First, note that there are six ways you can rotate a hexagon and have it end up taking up the same space: you can rotate it clockwise by $0^\circ$, $60^\circ$, $120^\circ$, $180^\circ$, $240^\circ$, or $300^\circ$. These would send spot "a" above to spot "a", "c", "l", "s", "q", or "h".
We would like to say "the easy answer overcounts by a factor of 6", but we need to be careful. If some configurations end up the same after a certain rotation (not counting the $0^\circ$ "rotation"), then we would be overcounting by less than a clean factor of 6, so we have to check whether or not that can happen.
Rotation by $60^\circ$ or $300^\circ$ can't leave us with the same setup, because we would need the same type of tile to be in the 6 spots: a,c,l,s,q,h (after $60^\circ$, a would end up on c, the old c would end up on l, etc.), but no type of tile is repeated 6 times.
Rotation by $120^\circ$ or $240^\circ$ can't leave us with the same setup, because we would have nowhere to place the four sheep tiles and the desert tile: even if we put three sheep tiles on a triangle ("bpm", "drg", "alq", "csh", "foi", or "enk") then the last sheep tile and the desert tile would both need to go to a place that doesn't move when you rotate the board, but the center "j" is the only space allowed.
Rotation by $180^\circ$ can't leave us with the same setup, because we would have nowhere to place the three ore tiles and the desert tile. Even if we put two ore tiles in a pair ("as", "br", "cq", "gm", "lh", "pd", eo", "fn", or "ki"), both the third ore tile and the desert tile would need to go in the center "j" spot.
Since a rotation can't leave the setup the same, the easy answer really does overcount by a factor of 6, and the answer to this question is $\dfrac{19!}{4!4!4!3!3!1!}\div6=\boxed{40\,738\,698\,000}$.
Why is your answer wrong?
Your answer comes out to be $\frac1{12}$ of the correct answer to this version of the problem. The issue is that your $(n-1)!$ formula accounts for rotations in a circle, but there are two problems with that:
- A rotation of $30^\circ$ wouldn't keep the hexagon spots the same, so treating the outer ring as a circle is not valid.
- You can't separate the rotations of the inner and outer rings. The rings are connected.
Using $5!$ instead of $6!$ for the inner ring accounts for all 6 rotations of the whole board, so you should just use $12!$ instead of $11!$ for the outer ring.
Hard version
All that said, there's another convention for these sorts of problems that's very common. Often, we count reflections as "the same board". It doesn't really change anything about strategy to play Settlers of Catan by looking in a mirror, after all. There are six reflections of a hexagon:
Unfortunately, unlike the rotations, sometimes a reflection can keep a setup the same, so it's not going to be as simple as "the easy answer overcounts by a factor of 12". How do we deal with the overcounting then? Luckily, there's a great theorem about counting, often called Burnside's Lemma, which says "when you want to count things with symmetry, the answer the the average (across all the operations: in this case rotations and reflections) of the numbers of things (ignoring symmetry) that don't change when you do the operation.
- How many setups don't change when you rotate by $0^\circ$? All $244\,432\,188\,000$ of them!
- How many setups don't change when you rotate by $60^\circ$? We already saw the answer is $0$.
- $120^\circ$? $0$.
- $180^\circ$? $0$.
- $240^\circ$? $0$.
- $300^\circ$? $0$.
- How many setups don't change when you reflect over a vertical line (the line "bjr")? Well, we have a bunch of 8 pairs of spots that have to have the same type of tile ("ac", "dg", "ef", "hl", "ik", "mp", "no", and "qs"), and then a few spots left over ("b","j","r"). Since brick, ore, and desert are the only types of tile with an odd number, one of each of those needs to go in the "bjr" spots ($3!$ possibilities), and then we can arrange two pairs of sheep tiles, two pairs of wood tiles, two pairs of wheat tiles, one pair of brick tiles, and one pair of ore tiles among the 8 pairs, for $\dfrac{8!}{2!2!2!1!1!}$ possibilities. All together, we have $\dfrac{8!}{2!2!2!1!1!}\times3!=30\,240$ setups that don't change.
- How many setups don't change when you reflect over "gjm"? This is just like "bjr", but the picture is rotated. The answer is still $30\,240$.
- "djp"? $30\,240$.
- How many setups don't change when you reflect over a horizontal line: "hijkl"? Now there are only 7 pairs that must match up, so in addition to one brick, one ore, and the desert tile, we can also have two other tiles on the line of reflection ("hijkl"). But since the tiles off this line must match up in pairs, these two other tiles must be of the same type. If all three brick are on the line of reflection, then there are $\dfrac{5!}{3!1!1!}$ possibilities for that line, and $\dfrac{7!}{2!2!2!1!}$ for the other 7 pairs, for a product of $12\,600$. Similarly it's $12\,600$ if all three ore are on the line of reflection. If instead, both brick and ore get to use one of the 7 pairs each, then the extra spots on the line of reflection come from sheep or wood or wheat. We have three cases of $\dfrac{5!}{2!1!1!1!}\times \dfrac{7!}{2!2!1!1!1!}=75\,600$. Putting it all together, we have $2\times 12\,600+3\times75\,600=252\,000$ setups that don't change under this reflection.
- "aejos"? Just like "hijkl": $252\,000$.
- "cfjnq"? $252\,000$.
By Burnside's Lemma, the answer is then $$\dfrac{244\,432\,188\,000+0\times5+30\,240+30\,240+30\,240+252\,000+252\,000+252\,000}{12}$$
$$=\dfrac{244\,433\,034\,720}{12}=\boxed{20\,369\,419\,560}$$
Just dividing by 12 would have undercounted by $70\,560$ possibilities.
