【Example 6】There are $2 n$ people sitting in a row, and $r$ people need to be called out from them $(r<n)$. It is required that no two of the called-out people sit next to each other. How many different ways are there to call them out? If the original $2 n$ people are sitting in a circle, how does the situation change?
If $2 n$ people originally sit in a row, then the described method of calling people is equivalent to making a limited-distance combination, and $m=2$. Thus, there are $C_{2 n-r+1}^{r}$ different ways of calling.

If $2 n$ people sit in a circle, then the number of ways in which the first and last person are both called out should be subtracted from this number. This number is equivalent to the number of ways to select $r-2$ numbers from $\{3,4, \cdots, 2 n-2\}$ without any two being adjacent, which is $C_{2 n-r-1}^{r-2}$. Therefore, there are
$$
C_{2 n-r+1}^{r}-C_{2 n-r+1}^{r-2}=\frac{2 n}{(2 n-r) C_{2 n-r}^{r}}
$$
ways that meet the requirements.

Alternatively, consider the case where $2 n$ people sit in a circle as follows: First, imagine that $2 n$ people sit in a row. If the person sitting in the first position is not called out, then for the $r$ people called out from the remaining $2 n-1$ people without any two being adjacent, they will still not be adjacent when forming a circle, and there are $C_{2 n-r}^{r}$ ways of doing this. If the person sitting in the first position is called out, then the person sitting in the second position and the person sitting in the $2 n$-th position cannot be called out, so we need to call out $r-1$ people from the remaining $2 n-3$ people without any two being adjacent, which has $C_{2 n-3-(r-2)}^{r-1}=C_{2 n-r-1}^{r-1}$ ways. Therefore, there are a total of $C_{2 n-r}^{r} + C_{2 n-r-1}^{r-1} = \left(2 n \cdot C_{2 n-r}^{r}\right) / (2 n-r)$ ways.