Example 11 Divide a circle with a circumference of 24 into 24 equal segments, and select 8 points from the 24 points such that the arc length between any two points is not equal to 3 or 8. How many different ways are there to select such a group of 8 points? Explain your reasoning.
(2001 CMO Problem)
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline 1 & 4 & 7 & 10 & 13 & 16 & 19 & 22 \\
\hline 9 & 12 & 15 & 18 & 21 & 24 & 3 & 6 \\
\hline 17 & 20 & 23 & 2 & 5 & 8 & 11 & 14 \\
\hline
\end{tabular}

In the table, the arc length between two adjacent numbers in each row is 3 (considering the first and last numbers in the same row as adjacent), and the arc length between two adjacent numbers in each column is 8 (considering the first and last numbers in the same column as adjacent). Therefore, at most one number can be taken from each column, and from 8 columns, at most 8 numbers can be taken. However, a total of 8 numbers need to be taken, so exactly one number must be taken from each column, and the numbers taken from adjacent columns must be in different rows (considering the first and last columns as adjacent).

According to the recursive method in Example 1, if each column is considered as a sector, and the first, second, and third rows in each column are considered as 3 colors, then this problem is equivalent to the case of $n=8, m=3$ in Example 1. Therefore, by Example 1, the number of different ways to choose 8 points is $a_{8}=(3-1)^{8}+(-1)^{8}(3-1)=258$.