The figure below shows a ring made of six small sections which you are to paint on a wall. You have four paint colors available and you will paint each of the six sections a solid color. Find the number of ways you can choose to paint the sections if no two adjacent sections can be painted with the same color.
[asy] draw(Circle((0,0), 4)); draw(Circle((0,0), 3)); draw((0,4)--(0,3)); draw((0,-4)--(0,-3)); draw((-2.598, 1.5)--(-3.4641, 2)); draw((-2.598, -1.5)--(-3.4641, -2)); draw((2.598, -1.5)--(3.4641, -2)); draw((2.598, 1.5)--(3.4641, 2)); [/asy]
Choose a section to start coloring. Assume, WLOG, that this section is color $1$. We proceed coloring clockwise around the ring. Let $f(n,C)$ be the number of ways to color the first $n$ sections (proceeding clockwise) such that the last section has color $C$. In general (except for when we complete the coloring), we see that
\[f(n,C_i)=\sum_{j\ne i} f(n-1,C_j),\]
i.e., $f(n,C_i)$ is equal to the number of colorings of $n-1$ sections that end in any color other than $C_i$. Using this, we can compute the values of $f(n,C)$ in the following table.
$\begin{tabular}{c|c|c|c|c } \multicolumn{1}{c}{}&\multicolumn{4}{c}{\(C\)}\\ \(n\)&1 & 2 & 3& 4 \\ \hline 1& 1 & 0 & 0 & 0\\ 2 & 0 & 1 & 1 & 1 \\ 3& 3 & 2 & 2 & 2 \\ 4 & 6 & 7 & 7 & 7 \\ 5 & 21 & 20 & 20 & 20\\ 6& 0& 61 & 61 & 61\\ \end{tabular}$
Note that $f(6, 1)=0$ because then $2$ adjacent sections are both color $1$. We multiply this by $4$ to account for the fact that the initial section can be any color.  Thus the desired answer is $(61+61+61) \cdot 4 = \boxed{732}$.
Solution by Shaddoll