Example 1 If numbers $1,2, \cdots, 14$ are taken in ascending order as $a_{1}, a_{2}, a_{3}$, 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 the numbers is $\qquad$ kinds.
Let $S=\{1,2, \cdots, 14\}, S^{\prime}=\{1,2, \cdots, 10\}$.
$T=\left\{\left(a_{1}, a_{2}, a_{3}\right) \mid a_{1}, a_{2}, a_{3} \in S, a_{2}-a_{1} \geqslant 3, a_{3}-a_{2} \geqslant 3\right\}$,
$T^{\prime}=\left\{\left(a_{1}^{\prime}, a_{2}^{\prime}, a_{3}^{\prime}\right) \mid a_{1}^{\prime}, a_{2}^{\prime}, a_{3}^{\prime} \in S^{\prime}, a_{1}^{\prime}<a^{\prime}{ }_{2}<a_{3}^{\prime}\right\}$.
Consider the mapping $\left(a_{1}, a_{2}, a_{3}\right) \rightarrow\left(a_{1}^{\prime}, a_{2}^{\prime}, a_{3}^{\prime}\right)$, where $a_{1}^{\prime}=a_{1}, a^{\prime}{ }_{2}=a_{2}-2, a_{3}^{\prime}=a_{3}-4\left(a_{1}, a_{2}, a_{3} \in T\right)$.
It is easy to verify that this mapping is a one-to-one correspondence or bijection, so $|T|=\left|T^{\prime}\right|$.
Thus, the problem is reduced to finding $\mid T^{\prime} \mid$, which is the number of ways to choose 3 different elements from $S^{\prime}$, i.e., $C_{10}^{3}=120$.
Therefore, selecting a subset of three numbers from a set of 14 numbers represents a selection method, corresponding to a permutation $\left(x_{1}, x_{2}, \cdots, x_{14}\right)$, and conversely, any permutation $\left(x_{1}, x_{2}, \cdots, x_{14}\right)$ must correspond to a selection method. Hence, the correspondence between the set of all selection methods and the set of all permutations $\left(x_{1}, x_{2}, \cdots, x_{14}\right)$ is a one-to-one correspondence.

According to the problem's requirements, the total number of selection methods equals the number of permutations $\left(x_{1}, x_{2}, \cdots, x_{14}\right)$ that contain 3 ones and 11 zeros, with at least two zeros between any two ones. To find the number of such permutations, we can consider the pattern 1001001 and then insert the remaining 7 zeros into the 4 gaps formed by the 3 ones, giving $C_{7+4-1}^{7}=C_{10}^{3}$ ways, which is the total number of different selection methods.

Note: This problem can be generalized to a more general case: finding the number of different ways to select $r$ numbers $a_{1}, a_{2}, \cdots, a_{r}$ from the set $\{1,2, \cdots, n\}$ that satisfy the following conditions: (1) $1 \leqslant a_{1}<a_{2}<\cdots<a_{r} \leqslant n$; (2) $a_{k+1}-a_{k} \geqslant m$, $i=1,2, \cdots, r-1$, where $m \in \mathbf{N}^{+}$.

In fact, let the set of all $r$-tuples that satisfy (1) and (2) be denoted as $A$. From (2), we have $a_{k} \leqslant a_{k+1}-n(k=1,2, \cdots, r-1)$, i.e., $a_{k}<a_{k+1}-(m-1)$.
$$
\begin{array}{l}
\text { Therefore, } a_{1}<a_{2}-(m-1)<a_{3}-2(m-1)<a_{4}-3(m-1) \\
<\cdots<a_{r}-(r-1)(m-1) . \\
\text { Let } b_{i}=a_{i}-(i-1)(m-1), i=1,2, \cdots r, \text { then } \\
1 \leqslant b_{1}<b_{2}<\cdots<b_{r} \leqslant n-(r-1)(m-1) .
\end{array}
$$

Let the set of all (non-repeating) $r$-tuples from the set $\{1,2, \cdots, n-(r-1)(m-1)\}$ be denoted as $B$. Then there exists a one-to-one correspondence, i.e., a bijection, between $A$ and $B$.
Thus, $|A|=|B|=C_{n-(r-1)(m-1)}^{r}$.