Five, 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 and 8. How many different ways are there to select such 8-point groups? Explain your reasoning.
（Supplied by Li Chengzhang）
Solution 1: Number the 24 points as $1,2, \cdots, 24$, and arrange them in a $3 \times 8$ number table according to their "bad relationship":
$1,4,7,10,13,16,19,22$,
$9,12,15,18,21,24,3, \quad 6$,
$17,20,23, \quad 2, \quad 5, \quad 8, \quad 11, \quad 14$.
It is easy to see that the arc length between two points represented by adjacent numbers in each row is 3, and the arc length between two points represented by adjacent numbers in each column is 8 (the first and last numbers are also considered adjacent). Thus, the requirement for the 8 points in the problem is transformed into the requirement that the numbers of the 8 points selected in the table are not adjacent. Therefore, exactly one number is taken from each column, and at most 4 non-adjacent numbers are taken from each row. Thus, the number of numbers taken from the 3 rows can only be 4 different cases:
$$
\{4,4,0\},\{4,3,1\},\{4,2,2\},\{3,3,2\} \text {. }
$$
(1) $\{4,4,0\}$. Choose one row from the 3 rows to not take any numbers, there are 3 different ways to do this. In the other two rows, the first row takes 4 non-adjacent numbers, there are 2 different ways to do this. The remaining 4 columns are uniquely determined as the columns for the 4 numbers taken from the other row. By the multiplication principle, there are 6 different ways in this case.
(2) $\{4,3,1\}$. The number of numbers taken from the 3 rows are 4, 3, and 1, respectively, there are $3!=6$ different arrangements. In one row, take 4 non-adjacent numbers, there are 2 different ways to do this. In another row and the remaining 4 columns, choose 1 number, there are 4 different ways to do this. Finally, in the third row, choose 1 number from the remaining 3 columns, the selection is uniquely determined. By the multiplication principle, there are $6 \times 2 \times 4=48$ different ways in this case.
(3) $\{4,2,2\}$. Choose one row from the 3 rows to take 4 non-adjacent numbers, there are 3 different ways to choose the row and 2 different ways to take the numbers, a total of 6 different ways. The remaining 4 columns are non-adjacent. From the 4 columns, choose 2 columns for the second row, there are $\mathrm{C}_{4}^{2}=6$ different ways. By the multiplication principle, there are $6 \times 6=36$ different ways in this case.
(4) $\{3,3,2\}$. From the 3 rows, choose one row to take 2 non-adjacent numbers, there are 3 different ways to choose the row and $\mathrm{C}_{7}^{2}-1=20$ different ways to take the numbers (subtract 1 to exclude the case where the two numbers are in the 1st and 8th columns, respectively).
After choosing the two numbers, the remaining 6 columns are divided into two parts. There are 3 different segment scenarios: $\{1,5\},\{2,4\}$, and $\{3,3\}$. The number of different ways for these segments are 8, 8, and 4, respectively. It is easy to see the different ways:
(i) For the $\{1,5\}$ segment, choose 3 non-adjacent columns, there are 8 different ways;
(ii) For the $\{2,4\}$ segment, choose 3 non-adjacent columns, there are 4 different ways;
(iii) For the $\{3,3\}$ segment, choose 3 non-adjacent columns, there are 2 different ways.
Therefore, the number of different ways in this case is
$3 \times\{8 \times 2+8 \times 4+4 \times 2\}=168$.
In summary, the number of different ways that meet the requirements of the problem is
$$
6+48+36+168=258 \text {. }
$$

Solution 2: As in Solution 1, write the 24 numbers in a $3 \times 8$ number table. Thus, when selecting 8 numbers, exactly one number is taken from each column. At this time, taking 1 number from the 1st column, there are 3 different ways. After the 1st column is determined, the number taken from the 2nd column cannot be in the same row as the number taken from the 1st column, so there are only 2 different ways. For each subsequent column, there are 2 different ways, a total of $3 \times 2^{7}$ different ways. But among these, the ways where the number taken from the 1st column and the number taken from the 8th column are in the same row do not meet the requirements.

If we denote the number of different ways to take one number from each column of a $3 \times n$ number table, and any two adjacent columns (including the $n$-th column and the 1st column) do not take numbers from the same row, as $\dot{x}_{n}$, then the above conclusion can be written as
$$
x_{8}+x_{7}=3 \times 2^{7} \text {. }
$$

Similarly, we can get
$$
x_{n}+x_{n-1}=3 \times 2^{n-1} \text {. }
$$

From this recurrence relation, we get
$$
\begin{aligned}
x_{8} & =3 \times 2^{7}-x_{7} \\
& =3 \times 2^{7}-\left(3 \times 2^{6}-x_{6}\right) \\
& =3 \times\left(2^{7}-2^{6}\right)+x_{6} \\
& =\cdots \cdots \\
& =3 \times\left(2^{7}-2^{6}+2^{5}-2^{4}+2^{3}-2^{2}+2\right) \\
& =3 \times 86=258,
\end{aligned}
$$
i.e., the number of different ways that meet the requirements of the problem is 258.