【Example 5】 There are $n$ young people and $r$ elderly people, $n>2r$, to be arranged in a row. It is required that each elderly person has a young person on both sides to support them (but each pair of young people only supports one elderly person). How many different arrangements are there?
Solution 1. First, arrange the $n$ young people in a row, which can be done in $n$! ways. Then, insert the $r$ elderly people into the $n-1$ gaps. However, to ensure that each elderly person is flanked by a young person on either side, the gaps chosen for insertion should not be adjacent. Therefore, we should first select $r$ gaps from the $n-1$ gaps with a minimum distance of $m=2$, and then arrange the elderly people in these gaps. Thus, the total number of different arrangements is
$$
n! \cdot C_{n-1-(r-1)-1}^{r} \cdot r! = \frac{n! \cdot (n-r)!}{(n-2r)!}
$$

Solution 2. First, arrange the $r$ elderly people in a row, which can be done in $r$! ways. Then, assign two young people to each elderly person, which can be done in $C_{n}^{2r} \cdot (2r)!$ ways. The remaining $n-2r$ young people can be inserted into the ends of the $r$ groups of three people or into the $r-1$ gaps between these groups. The gaps can be chosen with repetition, so the number of different ways to do this is
$$
C_{(r+1)+(n-2r)-1}^{n-2r} \cdot (n-2r)! = \frac{(n-r)!}{r!}
$$

Thus, the total number of arrangements is
$$
r! \cdot C_{n}^{2r} \cdot (2r)! \cdot \frac{(n-r)!}{r!} = \frac{n! \cdot (n-r)!}{(n-2r)!}
$$