Example 2.3.4 In a live military exercise, the Red side has set up 20 posts along a straight line defense. To test 5 new types of weapons, they plan to equip 5 posts with 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 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?
Solution: Let the 20 positions be ordered as $1,2, \cdots, 20$, and let the positions of the new weapons be $a_{i}(i=1,2,3,4,5)$ in ascending order. Let $x_{1}=a_{1}, x_{2}=a_{2}-a_{1}, x_{3}$ $=a_{3}-a_{2}, x_{4}=a_{4}-a_{3}, x_{5}=a_{5}-a_{4}, x_{6}=20-a_{5}$,

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 $y_{k}=x_{k}-1(k=1,2,3,4,5), y_{6}=x_{6}$, thus we have $y_{1}+y_{2}+y_{3}+y_{4}+y_{5}+y_{6}=15$, where $1 \leqslant y_{k} \leqslant 4(k=1,2,3,4,5,6)$. As in Example 2.3.3, we can find that there are 580 solutions.
The 5 new weapons are all different, and swapping their positions results in different allocation schemes, so the total number of allocation schemes for the new weapons is $580 \times 5!=69600$.