5. If from the numbers $1,2, \cdots, 14$, we select $a_{1}, a_{2}, a_{3}$ in ascending order, such that both $a_{2}-a_{1} \geqslant 3$ and $a_{3}$ $-a_{2} \geqslant 3$ are satisfied, then the total number of different ways to select them is $\qquad$ kinds.
120
5.【Analysis and Solution】Let $S=\{1,2, \cdots, 14\}, S^{\prime}=\{1,2, \cdots, 10\}$;
$$
\begin{array}{l}
T=\left\{\left(a_{1}, a_{2}, a_{3}\right) \mid a_{1}, a_{2}, a_{3} \in S, a_{2}-a_{1} \geqslant 3, a_{3}-a_{2} \geqslant 3\right\}, \\
T^{\prime}=\left\{\left(a_{1}^{\prime}, a_{2}^{\prime}, a_{3}^{\prime} \in \mathbf{S}^{\prime}\right) \mid a_{1}^{\prime}, a_{2}^{\prime}, a_{3}^{\prime} \in \mathbf{S}^{\prime}, a_{1}^{\prime}<a_{2}^{\prime}<a_{3}^{\prime}\right\} . \\
a_{1}^{\prime}=a_{1}, a_{2}^{\prime}=a_{2}-2, a_{3}^{\prime}=a_{3}-4,\left(a_{1}, a_{2}, a_{3}\right) \in T .
\end{array}
$$

It is easy to prove that $f$ is a one-to-one correspondence between $T$ and $T^{\prime}$, so the number of ways to choose, is exactly equal to the number of ways to choose any three different numbers from $S^{\prime}$, which is 120 in total.