Distinguishable telephone poles being painted Each of n (distinguishable) telephone poles is painted red, white, blue or yellow.  An odd number are painted blue and an even number yellow.  In how many ways can this be done?
Can some give me a hint how to approach this problem?

Proposition$^{[1]}$ Fix $k\in\mathbb{P}$ and functions $f_1,f_2,\dots,f_k:\mathbb{N}\to K$.  Define a new function $h:\mathbb{N}\to K$ by
$$h(\#S)=\sum f_1(\#T_1)\ f_2(\#T_2)\dots f_k(\#T_k),$$
where $(T_1,\dots,T_k)$ ranges over all weak ordered partitions of $S$ into $k$ blocks, i.e., $T_1,\dots,T_k$ are subsets of $S$ satisfying: 
(i) $T_i\cap T_j=\emptyset\ \text{if}\ i\neq j$, and 
(ii) $T_1\cup \dots\cup t_k=S$
Then:  $$E_h(x)=\prod^k_{i=1}e_{f_i}(x)$$
Solution Using this proposition, now we let $h(n)$ be the desired number.
\begin{align}
E_h(x)&=(\sum_{n\geq0}\frac{x^n}{n!})^2(\sum_{n\geq0}\frac{x^{2n+1}}{(2n+1)!})(\sum_{n\geq0}\frac{x^{2n}}{(2n)!})
\\ \\
&=e^{2x}(\frac{e^x+e^{-x}}{2})(\frac{e^x-e^{-x}}{2})
\\ \\
&=\frac{1}{4}(e^{4x}-1)
\\ \\
&=\sum_{n\geq 1}4^{n-1}\frac{x^n}{n!}
\end{align}
Thus, $\boxed{h(n)=4^{n-1}}$.
$[1]:\ \ \ \ \ $Stanley, Richard. Enumerative Combinatorics: Volume 2. 1st ed. Cambridge, United Kingdom: Cambridge University Press, 2011. 3. Print.
