Number of $3$-colourings of this chain of $3$ triangles and $2$ trapeziums 
Each of the nine dots in this figure is to be colored red, white or blue. No two dots connected by a segment (with no other dots between) may be the same color. How many ways are there to color the dots of this figure?


Ok, so for the left thre dots, there are $3\cdot2\cdot1$ ways to connect them. Thus, our solution is $3\cdot2\cdot1\cdot3=18.$ I think I am missing something here, thank you in advance!
 A: I've made a labeled diagram.

To make thing show up better I'm using red, green and blue as the colors.  Black vertices haven't been colored yet.  If we color $\triangle ABC$ as shown, then E must be colored green or blue and $D$ must me colored blue or red. Because of this, the color assigned to $F$ uniquely determines the coloring of $\triangle DEF$.  Now the same argument shows that however we color $\triangle DEF$, we have three choices for $\triangle GHI$.  Of course, there are six ways to color $\triangle ABC$, giving $6\cdot3\cdot3=54$ ways in all.
A: This is essentially the same as saulspatz's answer (which I didn't read carefully until after posting), just viewed and presented slightly differently.
Color the three points of the middle triangle Red, White, and Blue, which can be done in any of $6$ ways, then look to color the points of the triangles to its left and right. For each of these, the color of the vertex of the middle triangle furthest away can be used in any of the $3$ positions, but as soon as it is, the colors of the other two vertices are determined by the constraint. As a result, the total number of colorings is
$$6\cdot3^2=54$$
Remark: The only (small) advantage this has over saulspatz's answer is that its symmetry makes it clear from the get-go that the answer will be of the form $6n^2$, where $n$ counts the ways to color the triangles to left and right.
