5. If from the numbers $1,2, \cdots, 14$, we select $a_{1}, a_{2}, a_{3}$ in increasing order, such that
$$
a_{2}-a_{1} \geqslant 3 \text { and } a_{3}-a_{2} \geqslant 3 \text {, }
$$

then the total number of different ways to select them is $\qquad$ kinds.
5. Let $S=\{1,2, \cdots, 14\}, S^{\prime}=\{1,2, \cdots, 10\}$;
$$
\begin{array}{c}
T=\left\{\left(a_{1}, a_{2}, a_{3}\right)\left\{a_{1}, a_{2}, a_{3} \in S, a_{2}-a_{1} \geqslant 3, a_{3}-a_{2} \geqslant 3\right\},\right. \\
T^{\prime}=\left\{\left(a^{\prime}{ }_{1}, a^{\prime}{ }_{2}, a^{\prime}{ }_{3} \in s^{\prime}\right) \mid a^{\prime}{ }_{1}, a^{\prime}{ }_{2}, a^{\prime}{ }_{3} \in S^{\prime}, a^{\prime}{ }_{1}<a^{\prime}{ }_{2}<a^{\prime}{ }_{3}\right\}, \\
a^{\prime}{ }_{1}=a_{1}, a^{\prime}{ }_{2}=a_{2}-2, a_{3}{ }_{3}=a_{3}-4,\left(a_{1}, a_{2}, a_{3}\right) \in T .
\end{array}
$$

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 three different numbers from $S^{\prime}$, which is 120.