How many unique outcomes can be made from the 12 river tiles in Carcassonne? I have been pondering this question for some time now and it's a bit out of my scope to solve it. I am curious to know how many arrangements are possible for the river 1 expansion for Carcassonne.
The rules for playing the river are as follows: The source tile is played first, The lake tile is played last, and If two river bends are drawn in sequence they must have opposing orientation.
I understand the following: The first and last tile played will not factor into the counting so we just look at the $10$ tiles in between. Each tile has $2$ orientations and there are $8$ unique tiles, plus $1$ repeated corner, and one repeated straight. We must also exclude possibilities that become unplayable when the river curves in onto itself. 
Here's an image of the $12$ river tiles:

My preliminary guesswork at counting is
$$\frac{(2^8)10!}{2!2!}$$
My rationale 
$10!/2!2!$ Because order of selection matters and repetition tiles are excluded by dividing by $2!$
$2^8$ because each tile can be placed in $2$ ways (the $2$ straight tiles are not included because they are not unique)
I know this is wrong. It's just my first guess at it, and it has not excluded possibilities where the river curves into itself and creates an unplayable game. Any assistance would be much appreciated, thanks! 
 A: The main problem is counting the different selfavoiding layouts that can be produced using the four tiles of the following figure:

An admissible layout begins with $A$, ends with $B$, and contains $4$ $L/R$ tiles. We may assume that the first such tile is an $L$ (and multiply by $2$ at the end). There are $8$ $L/R$ words of length $4$ beginning with $L$, namely
$$LLLL, LLLR, LLRL, LLRR,LRLL, LRLR,LRRL,LRRRR\ .$$
These words have to be decorated with $A$, $B$ and six letters $S$, whereby certain $S$ are compulsory. We then obtain the following $8$ patterns, whereby at the dots  optional letters  $S$ may be filled in:
$$\eqalign{p_1:\quad&A\cdot LS\cdot LS\cdot LS\cdot L\cdot B \cr
p_2:\quad&A\cdot LS\cdot LS\cdot L\cdot R\cdot B \cr
p_3:\quad&A\cdot LS\cdot L\cdot R\cdot L\cdot B \cr
p_4:\quad&A\cdot LS\cdot L\cdot RS\cdot R\cdot B \cr
p_5:\quad&A\cdot L\cdot R\cdot LS\cdot L\cdot B \cr
p_6:\quad&A\cdot L\cdot R\cdot L\cdot R\cdot B \cr
p_7:\quad&A\cdot L\cdot RS\cdot R\cdot L\cdot B \cr
p_8:\quad&A\cdot L\cdot RS\cdot RS\cdot R\cdot B \cr}$$
These $8$ patterns fall into three types:
(i) The pattern $p_1$ contains four consecutive equal turns, and has to be treated using a case analysis in order to ensure selfavoiding.
(ii) The two patterns $p_2$ and $p_7$ contain three consecutive equal turns, and have to be treated using a case analysis in order to ensure selfavoiding. (Note that $p_7$ is equivalent to $p_2$.)
(iii) The remaining patterns are selfavoiding however we fill in the optional letters $S$. The number of ways to do this is a stars and bars problem for each $p_k$, depending on the number of compulsory letters $S$ in $p_k$. 
Assume that the total number $N$ of selfavoiding layouts has been determined. We then have to distribute the picture tiles on these layouts. The number $M$ of possibilities is the same for all layouts. Concerning the two pictures occurring twice assume that they are "secretly" different, and divide by $2\cdot2$ at the end. In this way we obtain
$$M={6!\> 2^6\>4!\over2\cdot2}=276\,480\ .$$
