Ten Sided Regular Polygon

Method 1:

There are $\dbinom{10}{3} = 120$ possible combinations of $3$ vertices without restrictions.

$10$ of these are composed of $3$ consecutive vertices.

Now look at those that have precisely $2$ consecutive vertices along with one that is not adjacent to either of the others. For each of the $10$ consecutive pairings, there are $6$ vertices that are not adjacent to either of the consecutive vertices. There are then $10*6 = 60$ such combinations.

Thus the number of desired combinations is $120 - 10 - 60 = \boxed{50}.$

Method 2:

Choose a vertex at random, and then look at all those combinations involving this vertex and two other vertices that are pair-wise non-adjacent. Then there are three "gaps" between these vertices going clockwise from the vertex chosen initially that are each composed of at least one vertex. If $a,b,c$ are the number of vertices in these "gaps", then the following equation can be formed:

$a + b + c = 7$ such that $a,b,c \ge 1.$

This is a "stars and bars" problem with solution $\dbinom{6}{2} = 15.$

This will be the case for any of the $10$ vertices that can be chosen initially, which would give us $10*15 = 150$ combinations. But each of these combinations will have been counted three times, one time for each of the vertices of any combination. Thus we need to divide $150$ by $3$ to get the final solution of $\boxed{50}.$

Total number of ways of choosing $2$ consecutive vertices is $10$ : $\begin{cases} (1,2) \\ (2,3) \\ . \\ . \\ . \\ . \\ (9,10) \\ (10,1) \end{cases}$ .

Now,for each of the above cases,we have $6$ choices for the $3^{rd}$ vertex because we are counting the number of ways in which to choose $2$ consecutive vertices but not $3$ (we will count that later).Thus,total $=10\times 6=60$ .

Now,the number of ways of choosing $3$ consecutive vertices is also $10$ : $\begin{cases} (1,2,3) \\ (2,3,4) \\ (3,4,5) \\ . \\ . \\ .\\ (9,10,1) \\ (10,1,2) \end{cases}$ .

Thus,total number of unfavourable cases is $60+10=70$ ,and total number of ways $=\dbinom{10}{3}=120$ .Thus,total number of favourable cases $=120-70=50$ .

Done!

requd no. of ways = no.of ways we can select any three vertices - no. of ways we can select three consecutive vertices - no. of ways we can select 2 consecutive vertices = C(10,3) - 10 - 10*C(6,1)