Question 187: 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 each set of 5 adjacent posts must have at least one post 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?
Question 187: Analysis: Noting that there are five different new weapons, we might as well first assume they are five identical weapons, and then multiply the result obtained by 5! at the end. We consider the sequence numbers of the positions to which the new weapons are allocated, transforming the problem into one of finding the number of positive integer solutions to an indeterminate equation with constraints, thereby solving the problem.

Solution: Let the 20 positions be sequentially numbered as $1, 2, \ldots, 20$, and let the sequence number of the $i$-th new weapon be $a_{i}$ $(i=1, 2, 3, 4, 5)$. Let $x_{1}=a_{1}-1, x_{2}=a_{2}-a_{1}-1, x_{3}=a_{3}-a_{2}-1, x_{4}=a_{4}-a_{3}-1, x_{5}=a_{5}-a_{4}-1, x_{6}=20-a_{5}$, then we have
$$
\mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}+\mathrm{x}_{4}+\mathrm{x}_{5}+\mathrm{x}_{6}=15
$$

where $1 \leq x_{i} \leq 4 (i=1, 2, 3, 4, 5, 6)$.
Now we need to find the number of solutions to the indeterminate equation (1) with constraints.
Let $S$ be the set of positive integer solutions to $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}=15$, and let $A_{1}$ be the subset of $S$ where $x_{1}>4$. Note that $\sum_{i=1}^{6} x_{i}=15$ has no positive integer solutions that simultaneously satisfy $x_{i}>4, x_{j}>4, x_{k}>4$, hence $A_{i} \cap A_{i} \cap A_{k}=\Phi$. By the inclusion-exclusion principle, the number of solutions sought is:
$$
\begin{array}{l}
\left|\mathrm{C}_{S} \mathrm{~A}_{1} \cap \mathrm{C}_{S} \mathrm{~A}_{2} \cap \mathrm{C}_{S} \mathrm{~A}_{3} \cap \mathrm{C}_{S} \mathrm{~A}_{4} \cap \mathrm{C}_{S} \mathrm{~A}_{5} \cap C_{S} \mathrm{~A}_{6}\right| \\
=|\mathrm{S}|-\sum_{i=1}^{6}\left|\mathrm{~A}_{4}\right|+\sum_{1 \leq \mathrm{i}<\leq 6}\left|\mathrm{~A}_{1} \cap \mathrm{A}_{\mathrm{j}}\right| \\
=\mathrm{C}_{14}^{5}-\mathrm{C}_{6}^{1} \cdot \mathrm{C}_{10}^{5}+\mathrm{C}_{6}^{2} \cdot \mathrm{C}_{6}^{5} \\
=580
\end{array}
$$

Since the five new weapons are all different, swapping their positions results in different arrangements, so the number of schemes for allocating the new weapons is $580 \times 5!=69600$.

Note: Generally, to find the number of integer solutions to an indeterminate equation with constraints: $x_{1}+x_{2}+x_{3}+\ldots+x_{m}=n$ (where $0 \leq x_{i} \leq n_{i}$, $i=1, 2, \ldots, m$, and $m, n, n_{i}$ are all positive integers), similar to this problem, the inclusion-exclusion principle can be used to solve it.