15. In a live military exercise, the Red side has set up 20 positions along a straight line. To test 5 different new weapons, it is planned to equip 5 positions with these new weapons, with the requirement that the first and last positions are not equipped with new weapons, and that among every 5 consecutive positions, at least one position is equipped with a new weapon, and no two adjacent positions are equipped with new weapons at the same time. How many ways are there to equip the new weapons?

将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
15. Let 20 positions be sequentially numbered as $1,2, \cdots, 20$, and let the sequence number of the $k$-th new weapon be $a_{k}(k=1,2,3,4,5)$.
$$
\begin{array}{l}
\text { Let } x_{1}=a_{1}, x_{2}=a_{2}-a_{1}, x_{3}=a_{3}-a_{2}, \\
x_{4}=a_{4}-a_{3}, x_{5}=a_{5}-a_{4}, x_{6}=20-a_{5} .
\end{array}
$$

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 $y_{k}=x_{k}-1(k=1,2,3,4,5), y_{6}=x_{6}$, thus, we have
$$
y_{1}+y_{2}+y_{3}+y_{4}+y_{5}+y_{6}=15,
$$

where $1 \leqslant y_{k} \leqslant 4(k=1,2,3,4,5,6)$.
Let $I$ be the set of all positive integer solutions to $y_{1}+y_{2}+y_{3}+y_{4}+y_{5}+y_{6}=15$, and $A_{k}$ be the set of solutions in $I$ where $y_{k}$ satisfies $y_{k}>4$. Then
$$
\begin{array}{l}
\left|\bigcap_{k=1}^{6} \overline{A_{k}}\right|=|I|-\left|\bigcup_{k=1}^{6} A_{k}\right| \\
=|I|-\sum_{k=1}^{6}\left|A_{k}\right|+\sum_{j4, y_{j}>4, y_{k}>4$ of $\sum_{k=1}^{6} y_{k}=15$ in positive integers, then
$$
\begin{array}{l}
\left|\bigcap_{k=1}^{6} \overline{A_{k}}\right|=\mathrm{C}_{14}^{5}-6 \mathrm{C}_{10}^{5}+\mathrm{C}_{5}^{2} \mathrm{C}_{6}^{5} \\
=2002-1512+90=580 .
\end{array}
$$

Since the 5 new weapons are all different, swapping their positions results in different arrangements, so the number of ways to equip the new weapons is
$$
580 \times 5!=69600 \text {. }
$$
(Provided by Kong Yangxin)