number of ways of choosing $r$ points from $n$ points arranged in a circle such that no consecutive points is taken. 
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 $$\frac{n\cdot^{n-r-1}C_{r-1}}{r}$$

Consider $n$ things in a circle namely $P_1,P_2 \dots P_n$. Without loss of generality let us assume that $P_1,P_2 \dots P_n$ are all arranged in a circle in a clockwise manner. Now let us start the question by counting the number of favourable cases always involving $P_1$. Clearly $P_2$ and $P_n$ cannot be selected now. as they are just adjacent to $P_1$. Therfore we can open the circle at $P_2$ and $P_n$ thus forming a linear arrangement with its end at $P_3$ and $P_{n-1}$. Now we can simply choose $r-1$ things out of these $n-3$ things so that no two are adjacent. By using the formula we will get $C^{n-r-1}_{r-1}$.
Now this was the case in which $P_1$ was definitely included. Similarly there will be cases where in $P_2$ is definitely included and so on up to the case where $P_n$ is definitely included. Thus there are $n$ such cases. Therefore we will be multiplying the answer we got by $n$. Therefore $n \cdot C^{n-r-1}_{r-1}$.
But careful!!! We have counted in excess. I will tell you how. 
Let's consider a special case where in $n=9$ and $r=3$. Suppose I choose $P_1,P_3,P_5$ and so on cases where $P_1$ is definitely included. Now in the case where $P_3$ is definitely included also this case will appear. Also it will appear in the case where $P_5$ is definitely included. And so if you care to see the pattern will be repeated $3$ times.
Therefore in selection of $r$ things the pattern repeats $r$ times. Therefore to get the correct answer we will have to divide it by $r$.
Hence, our final answer will be $\frac{n}{r} C^{n-r-1}_{r-1}$.
Also please note that the formula that I have used for the linear part in between has been already proved by you in the question itself.
Just have to plug in $n$ as $n-3$ and $r$ as $r-1$ and you have your answer....
Good Luck.
Best wishes.
