Example 8 If $1,2,3, \cdots, 14$ are taken in increasing 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. Find the total number of all different ways of selection.
(1989, National High School Competition)
Solution: Assignment
$i=1,2, \cdots, 14$. Then any selection of three numbers from the 14 numbers corresponds 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. Therefore, a 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 total 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 select the pattern 1001001, and then insert the remaining 7 zeros into the 4 gaps formed by the 3 ones, hence there are $C_{7+4-1}^{7}=C_{10}^{3}$ ways, which is the total number of all different selections.