number of ways of choosing $r$ points from $n$ ponts arranged in a circle such that no consecutive points is taken.
(I have seen some question on SE related to this. But I am trying to solve it using my own method).
Let the points be taken in a straight line $\{A_1,A_2,…,A_n\}$. I have to choose $r$ points such that no two points are consecutive. Let the chosen points be represented by $\{x_1,x_2,…,x_r\}$. The number of ways of choosing such points is same as the number of ways of placing these $r$ points among $n-r$ objects such that none of the points $\{x_1,x_2,…,x_r\}$ come together. Since I am only talking about "choosing" points, I will assume all the points to be identical.
This can be calculated by assuming gaps between the $n-r$ initial objects. There will be $n-r$ gaps since the leftmost gap and right most gap are same for a circular arrangement. First I will place $x_1$. The number of ways of placing the remaining $r-1$ points in $n-r-1$ gaps is $$^{n-r-1}C_{r-1}$$
(the points are assumed to be identical).
Similarly, If I start with $x_2$, the number of ways will be $$^{n-r-1}C_{r-1}$$
So the total will be $$n\cdot^{n-r-1}C_{r-1}$$
But the answer given is $$\frac{n\cdot^{n-r-1}C_{r-1}}{r}$$
Best Answer
Choose an arbitrary point on the circle and "break" the circle in that point. The resulting configuration can be seen as a string of length $n$ over the alphabet $\Sigma=\{0,1\}$, with the properties:
If both the first character and the last one are "$0$", we are just expressing $n-r$ as a sum of $r+1$ positive integers, that represent the lengths of the blocks of zeroes. By stars and bars, there are $\binom{n-r-1}{r}$ strings of that kind. On the other hand, if the first character is zero while the last one is one, or the opposite, we are expressing $n-r$ as a sum of $r$ positive integers. It follows that the total number of our arrangements is given by:
$$\begin{eqnarray*} \binom{n-r-1}{r}+2\binom{n-r-1}{r-1} &=& \binom{n-r-1}{r-1}+\binom{n-r}{r}\\[0.2cm]&=&\color{red}{\frac{n}{n-r}\binom{n-r}{r}}\\[0.2cm]&=&\color{blue}{\frac{n}{r}\binom{n-r-1}{r-1}}.\end{eqnarray*}$$