RG-color blinder recognizes blue. He seeing objects red and green but he only knows that these colors are different (and that none of them is blue).
RBG-color blinder sees red, blue and green objects as different, but cannot decide which is red, which is blue and which is green.
For example:
For RG–color blinder: $ {R, R, G} = {R, G, G} \neq {B, G, G}$.
For RBG-color blinder ${R, G, G} = {B, B, G} \neq {R, B, G}$.
Let's consider coloring the tips of a regular pentagon in red, blue and green, but we identify such coloring that one goes to the other with some isometry of the pentagon.
Calculate how many different colors see RG-color blinder and how many RBG-color blinder.
My try:
$$\begin{array}{|c|c|c|c||c|c|}
\hline
G \in G & \text{how many elements of that type}& \text{cycle index}\\ \hline
\text{id} & 1 & x_{1}^{5} \\ \hline
\text{symmetry about the axis passing through the vertex} & 5 & x_1x_{2}^{2} \\ \hline
\text{clockwise rotation once } & 2 & x_{5}^{1}\\ \hline
\text{2 rotation right time } & 2 & x_{5}^{1}\\ \hline
\end{array}$$
$$f(x_1,x_2,x_3,x_4,x_5)=\frac{1}{10}(x_{1}^{5}+5x_{1}x_{2}^{2}+4x_{5})$$
$$\text{RBG:} \frac{1}{10}(1+5+4)=1$$
$$RG: \frac{1}{10}(2^5+6\cdot2^3+4\cdot2)=\frac{32+40+8}{10}=8$$
However, I don't know if my solution makes sense. Can you check it ?