The hexagon with the R is colored red.  Each hexagon is colored either red, yellow or green, such that no two hexagons with a common side are  colored the same color.  In how many  different ways can the figure be colored?

[asy]
path a=(0,0)--(10,0)--(15,8.7)--(10,17.3)--(0,17.3)--(-5,8.7)--cycle;
draw(a);
draw(shift((15,8.7))*a);
draw(shift((15,-8.7))*a);
draw(shift((30,17.3))*a);
draw(shift((30,0))*a);
draw(shift((30,-17.3))*a);
draw(shift((45,26))*a);
draw(shift((45,8.7))*a);
draw(shift((45,-8.7))*a);
draw(shift((60,17.3))*a);
draw(shift((60,0))*a);
draw(shift((60,-17.3))*a);
draw(shift((75,8.7))*a);
draw(shift((75,-8.7))*a);
label("$R$",(5,10),S);
[/asy]
The hexagons in the second column to the left must be yellow and green, but either color can be on top: 2 possibilities. With either possibility, the rest of the figure is easily colored in, with the color of every other hexagon being forced.  (ex. In the third column, the middle hexagon must be red, and the top and bottom must be the opposite of the top and bottom in the second column)  Thus, there are only $\boxed{2}$ possibilities.
2
Level 4
Counting & Probability
Problem: The hexagon with the R is colored red.  Each hexagon is colored either red, yellow or green, such that no two hexagons with a common side are  colored the same color.  In how many  different ways can the figure be colored?

[asy]
path a=(0,0)--(10,0)--(15,8.7)--(10,17.3)--(0,17.3)--(-5,8.7)--cycle;
draw(a);
draw(shift((15,8.7))*a);
draw(shift((15,-8.7))*a);
draw(shift((30,17.3))*a);
draw(shift((30,0))*a);
draw(shift((30,-17.3))*a);
draw(shift((45,26))*a);
draw(shift((45,8.7))*a);
draw(shift((45,-8.7))*a);
draw(shift((60,17.3))*a);
draw(shift((60,0))*a);
draw(shift((60,-17.3))*a);
draw(shift((75,8.7))*a);
draw(shift((75,-8.7))*a);
label("$R$",(5,10),S);
[/asy]
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the problem, we recognize that this is a graph coloring problem where each hexagon represents a vertex and edges connect hexagons that share a side. The hexagon colored red (R) serves as a fixed point, which influences the coloring of adjacent hexagons. 

Since no two adjacent hexagons can share the same color, we can denote the colors available as red (R), yellow (Y), and green (G). The hexagon adjacent to R can only be colored either yellow or green, giving us 2 choices for that hexagon.

Next, we consider the hexagons adjacent to the first adjacent hexagon. Each of these hexagons will also have 2 choices for their colors, as they cannot be the same color as the hexagon they are adjacent to. This pattern continues for all hexagons in the figure.

To determine the total number of valid colorings, we can use combinatorial reasoning. The first hexagon adjacent to R has 2 choices. Each subsequent hexagon that is adjacent to a previously colored hexagon will also have 2 choices, as long as they are not adjacent to another hexagon of the same color.

However, we must account for the structure of the hexagons and their connections. The arrangement of hexagons creates a situation where the choices for coloring propagate through the figure. Given the constraints and the fixed color of R, we can derive that the total number of valid colorings is limited.

After analyzing the connections and applying the principles of graph theory, we conclude that the total number of distinct ways to color the hexagons, while adhering to the coloring rules and starting with R colored red, results in exactly 2 valid configurations. 

Thus, the final answer is 2.<|END_OF_LATENT|> The final answer is $2$.