Example 7 In a live military exercise, the Red side has set up 20 posts along a straight line. To test 5 new types of weapons, it is planned to equip 5 posts with these new weapons, with the requirement that the first and last posts are not equipped with new weapons, and among every 5 adjacent posts, at least one post is equipped with a new weapon, and no two adjacent posts are equipped with new weapons at the same time. How many ways are there to equip the new weapons?
(2005, National High School League Zhejiang Province Preliminary Competition)
Let the 20 positions be ordered as $1,2, \cdots, 20$, and let the sequence number of the $k$-th new weapon be $a_{k}$,
$$
k=1,2,3,4,5 \text {. }
$$
$$
\begin{array}{l}
\text { Let } x_{1}=a_{1}, x_{2}=a_{2}-a_{1}, x_{3}=a_{3}-a_{2}, \\
x_{4}=a_{4}-a_{3}, x_{5}=a_{5}-a_{4}, x_{6}=20-a_{5} .
\end{array}
$$

Then we have
$$
x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}=20,
$$

where $2 \leqslant x_{k} \leqslant 5(k=1,2,3,4,5), 1 \leqslant x_{6} \leqslant 4$.
Make the substitution
$$
y_{k}=x_{k}-1(k=1,2,3,4,5), y_{6}=x_{6} \text {. }
$$

Then $y_{1}+y_{2}+y_{3}+y_{4}+y_{5}+y_{6}=15$,
where $1 \leqslant y_{k} \leqslant 4(k=1,2,3,4,5,6)$.
Let $I$ be the set of all positive integer solutions to equation (1), and $A_{k}$ be the set of solutions in $I$ where $y_{k} > 4$. Then
$$
\begin{array}{l}
\left|\bigcap_{k=1}^{6} \overline{A_{k}}\right|=|I|-\left|\bigcap_{k=1}^{6} A_{k}\right| \\
=|I|-\sum_{k=1}^{6}\left|A_{k}\right|+\sum_{j4 、 y_{j}>4 、 y_{k}>4$ of the positive integer solutions to $\sum_{k=1}^{6} y_{k}=15$, so,
$$
I \bigcap_{k=1}^{6} \widehat{A_{k}} \mid=\mathrm{C}_{14}^{5}-6 \mathrm{C}_{10}^{5}+\mathrm{C}_{6}^{2} \mathrm{C}_{6}^{5}=580 .
$$

Since the 5 new weapons are all different, swapping their positions results in different arrangements, thus the number of ways to equip the new weapons is $580 \times 5!=69600$.