Don't Try Counting one by one

Consider the general case for an $\text{n-gon}$ creating an $\text{r-gon}$ . Essentially, we need to choose $r$ points from the $n$ vertices such that no two vertices are adjacent. First assume the $1^{\text{st}}$ vertex isn't a part of this $\text{r-gon}$ . Then, the $n^{\text{th}}$ vertex can be a part of this $\text{r-gon}$ . Imagine removing exactly $r$ points from the $\text{n-gon}$ and choosing $r$ points from the resulting figure. Notice that there may or may not be adjacent vertices in the produced figure. But if we reinsert these $r$ blocks into the spaces where adjacent blocks are chosen, we get the $\text{n-gon}$ again but such that none of the two points are ever adjacent. So there will be $\binom{n-r}{r}$ to choose such shapes.

Similarly consider the case when the $1^{st}$ vertex is a part of the $\text{r-gon}$ ,then the $n^{\text{th}}$ vertex cannot be a part of this $\text{r-gon}$ . So there will be $\binom{n-r-1}{r-1}$ ways of choosing the points.

If $S_{(n,r)}$ denotes the number of ways to get an $\text{r-gon}$ from an $\text{n-gon}$ such that no two vertices are adjacent from the original $\text{n-gon}$ , then $S_{(n,r)}=\binom{n-r}{r}+\binom{n-r-1}{r-1}$

Substituting $n=28,r=10$ , we get $S_{(28,10)} = 68068$ .

Let's do this question using gap method. For understanding see my solution in this

So Let us fix $18$ points(vertices of 28-gon) in a straight line(see above link for reference) .Now there are total 19 spaces which we can choose as vertices(including first and last). If any 10 of these 19 places are chosen then, no one will be adjacent. This can be done in $^{19}C_{10}$ ways.

But if we choose the first and last position and then join the line back to form the polygon then these two points will adjacent.This case is possible by $^{17}C_{8}$ ways (17 places left and 8 more vertices to be chosen because two have been chosen before i.e first and last).

Hence $^{19}C_{10}$ - $^{17}C_{8}= \boxed{68068}$ .