The number of possible ways to select the four vertices of eleven vertices A cycle graph is a graph that consists of a single cycle or in the other words some number of vertices connected in a close chain and denoted by $c_n$.
I need the number of possibility ways for to select the four vertices of eleven vertices such that no two vertices are adjacent.
Can you help me?

An admissible selection of $k$ nonadjacent vertices from an $n$-cycle can be realized as follows: Choose an arbitrary first vertex, determining  a train of $n-1$ consecutive vertices in between. In the end there will be $n-k$ unchosen vertices, and $k-1$ more chosen vertices in $k-1$ different  slots between the unchosen vertices. The first choice can be made in $n$ ways, and then the slots can be chosen in ${n-k-1\choose k-1}$ ways. Since we have arbitrarily called one of the $k$ chosen vertices the first we have to divide by $k$ in order to arrive at the end result
$$N={n\over k}{n-k-1\choose k-1}\ .$$
If $n=11$ and $k=4$ we obtain $N={11\over4}{6\choose3}=55$.
