Example 13 In a live military exercise, the Red side has set up 20 posts on a straight line defense. To test 5 different new weapons, it plans to equip 5 posts with these new weapons, with the requirement that the first and last posts are not equipped with new weapons, and at least one post among every 5 adjacent posts is equipped with new weapons, 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 Zhejiang Provincial Preliminary Exam Question)
Solution 1: Let the 20 positions be ordered as $1,2, \cdots, 20$, and let the sequence number of the $k$-th new type of weapon be $a_{k}(k=1,2,3,4,5)$. Define $x_{1}=a_{1}, x_{2}=a_{2}-a_{1}, x_{3}=a_{3}-a_{2}, x_{4}=a_{4}-a_{3}, x_{5}=a_{5}-a_{4}$, $x_{6}=20-a_{5}$, then we have
$$
\left\{\begin{array}{l}
x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}=20, \\
2 \leqslant x_{k} \leqslant 5(k=1,2,3,4,5), 1 \leqslant x_{6} \leqslant 4 .
\end{array}\right.
$$

Make the substitution $y_{k}=x_{k}-2(k=1,2,3,4,5), y_{6}=x_{6}-1$, then
$$
\left\{\begin{array}{l}
y_{1}+y_{2}+y_{3}+y_{4}+y_{5}+y_{5}=9, \\
0 \leqslant y_{i} \leqslant 3(i=1,2,3,4,5,6)
\end{array}\right.
$$

Let the set of non-negative integer solutions to the equation $y_{1}+y_{2}+y_{3}+y_{4}+y_{5}+y_{6}=9$ be $S$, then $|S|=C_{9+6-1}^{6-1}=$ $C_{14}^{5}$. Let $A_{i}$ be the set of integer solutions in $S$ that satisfy $y_{i} \geqslant 4$, and let $y^{\prime}{ }_{i}=y_{i}-4, y^{\prime}{ }_{k}=y_{k}(k \neq i)$, then $\left|A_{i}\right|$ is equal to the number of non-negative integer solutions to the equation $y^{\prime}{ }_{1}+y^{\prime}{ }_{2}+y^{\prime}{ }_{3}+y^{\prime}{ }_{4}+y^{\prime}{ }_{5}+y^{\prime}{ }_{6}=5$, i.e., $\left|A_{i}\right|=C_{5+6-1}^{6-1}=$ $C_{10}^{5}$. Similarly, $\left|A_{i} \cap A_{j}\right|$ is equal to the number of non-negative integer solutions to the equation $y^{\prime \prime}{ }_{1}+y^{\prime \prime}{ }_{2}+\cdots+y^{\prime \prime}{ }_{6}=1$. i.e., $\left|A_{i} \cap A_{j}\right|=$ $C_{1+6-1}^{6-1}=C_{6}^{5}$. When $3 \leqslant r \leqslant 6$, $\left|A_{i_{1}} \cap A_{i_{2}} \cap \cdots \cap A_{i_{1}}\right|=0$, because it is impossible for $y_{1}+y_{2}+\cdots+y_{6}=9$ to have 3 or more $y_{i} \geqslant 4$. By the principle of inclusion-exclusion, we get
$$
\left|\bigcap_{i=1}^{6} \bar{A}_{i}\right|=|S|-\sum_{i=1}^{6}\left|A_{i}\right|+\sum_{1 \leqslant i<j \leqslant 6}\left|A_{i} \cap A_{j}\right|=C_{14}^{5}-C_{6}^{1} C_{10}^{5}+C_{6}^{2} C_{6}^{5}=580 .
$$

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

Solution 2: As in Solution 1, we can derive the integer equation (1), and the number of integer solutions $m$ to (1) is equal to the coefficient of $x^{20}$ in the following polynomial:
$$
\begin{aligned}
f(x) & =\left(x^{2}+x^{3}+x^{4}+x^{5}\right)^{5}\left(x+x^{2}+x^{3}+x^{4}\right) \\
& =x^{11}\left(1+x+x^{2}+x^{3}\right)^{6}=x^{11}\left(\frac{1-x^{4}}{1-x}\right)^{6} \\
& =x^{11}\left(\sum_{i=0}^{6}(-1)^{i} C_{6}^{i} x^{4 i}\right)\left(\sum_{n=0}^{\infty} C_{n+5}^{5} x^{n}\right),
\end{aligned}
$$

i.e.,
$$
m=C_{9+5}^{5}-C_{6}^{1} C_{5+5}^{5}+C_{6}^{2} C_{1+5}^{5}=580 .
$$