Example 36 (1989 National High School Mathematics League Question) If from $1,2,3, \cdots, 14$, 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. Find the total number of all different ways to select them.
Solve the assignment
$$
x_{i}=\left\{\begin{array}{l}
1, \text { if number } i \text { is selected, } \\
0, \text { if number } i \text { is not selected. }
\end{array}\right.
$$

Then, any selection of three numbers from 14 numbers corresponds to a permutation $\left(x_{1}, x_{2}, \cdots, x_{14}\right)$. Conversely, any permutation $\left(x_{1}, x_{2}, \cdots, x_{14}\right)$ must correspond to a selection. Therefore, each selection $\rightarrow\left(x_{1}, x_{2}, \cdots, x_{14}\right)$ is a one-to-one mapping.

According to the requirements of the problem, the number of selections equals the number of permutations $\left(x_{1}, x_{2}, \cdots, x_{14}\right)$ that have 3 ones and 11 zeros, and where each pair of ones is separated by at least two zeros. To find the number of such permutations, we first arrange the pattern 1001001, and then insert the remaining 7 zeros into the 4 gaps formed by the 3 ones. Thus, there are $C_{7+1-1}^{7}=C_{10}^{3}$ ways. This is the total number of all different selections.