Burnside's Lemma on octagon, using one of two colors on each side I am trying to find why the number of colorings of a regular octagon there are such that each side is colored either red or blue is 10. As well as this condition, each color must be used once. Essentially, out of the 8 edges, 4 are red and 4 are blue. If an octagon can be rotated to become another octagon, then they are considered the same. 
My work on this is such: there are ${8 \choose 4} = 70$ possible combinations, forgetting about rotation. If a possible octagon is rotated clockwise once, there is no way it can be considered the same. If it is rotated twice, there are two ways it can be considered the same - a RBRBRBRB or BRBRBRBR color scheme. Similarly, if it is rotated $3$, $5$, or $7$ times, there are $0$ "fixed" combinations. If it is rotated $4$ or $6$ times, there are $2$ fixed points. The answer should then be $\frac{70+0+2+0+2+0+2+0}{8} = 9.5$. However, this is obviously not right, being a non-integer. The correct answer is $10$, found through writing a quick computer program.
Can anyone help me find where I messed up?

We may apply PET here since we require the cycle index $Z(C_8)$ of the
cyclic group $C_8$ anyway in order to apply Burnside. We have
$$Z(C_n) = \frac{1}{n} \sum_{d|n} \varphi(d) a_d^{n/d}$$
With $n=8$ this works out to
$$Z(C_8) = \frac{1}{8} a_1^8 + \frac{1}{8} a_2^4
+ \frac{1}{4} a_4^2 + \frac{1}{2} a_8.$$
We get  
$$[R^4 B^4] Z(C_8; R+B) \\ =
\frac{1}{8} [R^4 B^4] (R+B)^8
+ \frac{1}{8} [R^4 B^4] (R^2+B^2)^4
+ \frac{1}{4} [R^4 B^4] (R^4+B^4)^2
\\ + \frac{1}{2} [R^4 B^4] (R^8+B^8)
\\ = \frac{1}{8} {8\choose 4}
+ \frac{1}{8} [R^2 B^2] (R+B)^4
+ \frac{1}{4} [R B] (R+B)^2
\\ = \frac{1}{8} {8\choose 4} + \frac{1}{8} {4\choose 2}
+ \frac{1}{4} {2\choose 1}
= \frac{35}{4} + \frac{3}{4} + \frac{1}{2} = 10.$$
