21.2.10 ** Divide a circle with a circumference of 24 into 24 equal segments. Choose 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?
Let the 24 points be $1, 2, \cdots, 24$. Arrange these 24 numbers in the following table:
\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 any two adjacent numbers in the same row is 3 (considering the first and last numbers in a row as adjacent), and the arc length between any two adjacent numbers in the same column is 8 (considering the first and last numbers in a column as adjacent). Therefore, at most one number can be selected from each column, and from 8 columns, at most 8 numbers can be selected. Since we need to select 8 numbers in total, exactly one number must be selected from each column, and the numbers selected from adjacent columns must be in different rows (considering the first and last columns as adjacent).

According to problem 21.2.8, if each column is considered as a sector, and the first, second, and third rows in each column are considered as 3 different colors, then this problem is equivalent to the case of $n=8, m=3$ in problem 21.2.8. Therefore, by the conclusion of problem 21.2.8, the number of different 8-point groups is $a_{8}(3)=2^{8}+(-1)^{8} \cdot 2=258$.