Combinatorial Counting: Painting Streetlight Poles with Color Restrictions

Forget for the time being the condition that an even number are to be painted green. If you have $n$ poles and $c$ many different colours then there $n^c$ many ways to paint the poles (for each pole you must make $c$ choices, since there are no conditons the total number of choices to be made is the product over the number of choices for each pole). Now I count $6$ different colours in your list. (red white blue yellow purple green)

The amount of ways to colour $n$ poles with an even number being green can be calculated by calculating the number of ways to colour $(n-k)$ poles a colour that is not green and $k$ green - then sum over all even $k≤n$. To calculate this number for a given $k$ (with $k≤n$) you may do the following combinatorics:

choose $k$ poles to be coloured green, there are ${n\choose k}$ many ways of doing this, colour the remaining $n-k$ poles one of the remaining $5$ colours, there are $5^{n-k}$ ways of doing this. This gives then
$${n\choose k}\cdot 5^{n-k}$$
ways of colouring exactly $k$ poles green (with $k≤n$). Then the amount of ways to colour an even number green is:
$$\sum_{k≤n,\ k \text{ even} }{n\choose k}5^{n-k}$$
one way to simplify the condition that only even $k$ are to be allowed in the sum is to note that $\frac{1+(-1)^k}2$ is $1$ when $k$ is even and $0$ else. So the number we are looking for is:
$$\sum_{k=0}^n{n\choose k}5^{n-k}\frac{1+(-1)^k}{2}=\frac12\left[\sum_{k=0}^n{n\choose k}5^{n-k}1^k+\sum_{k=0}^n{n\choose k}5^{n-k}(-1)^k\right] =\frac12\left[(5+1)^n+(5-1)^n\right]$$
which is $\dfrac{6^n+4^n}{2}$.