【Example 3】 6 people stand in a row, A must not stand at the head of the line.
Since A must not stand at the head of the line, A is a restricted element, and since the head of the line must not be occupied by A, the head of the line is a restricted position.
For Example 3, Solution 1 first arranges the restricted element Jia to stand in one of the back positions, with $A_{5}^{1}$ different ways; then arranges the remaining 5 people to stand in the remaining positions, with $A_{5}^{5}$ different ways. Therefore, there are a total of $A_{5}^{1} \cdot A_{5}^{5}=5 \cdot 5$! different ways to arrange the queue.

For Example 3, Solution 2 first selects 1 person from the 5 people other than Jia to stand at the head of the queue, with $A_{5}^{1}$ different ways; then arranges the remaining 5 people in the remaining 5 positions, with $A_{5}^{5}$ different ways, as there are no further restrictions at this point. Therefore, there are a total of $A_{5}^{1} \cdot A_{5}^{5}=5 \cdot 5$! different ways to arrange the queue.

In the above two solutions, the focus is on prioritizing the restricted element and the restricted object, respectively. For Example 5, two similar solutions can also be provided.

The insertion method, on the other hand, focuses entirely on the opposite. It first arranges the general elements and then inserts the restricted elements into the allowed positions.
For Example 3, Solution 3 first arranges the 5 people other than Jia in a horizontal line, with $A_{5}^{5}$ different ways; then arranges Jia to be inserted into one of the allowed positions, with 5 different ways (since he cannot be inserted at the very front, he can be inserted at the back or between any two people). Therefore, there are a total of $5 A_{5}^{5}=5 \cdot 5$! different ways to arrange the queue.