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?

We solve the problem in general for a cycle of $N$ vertices, and electing $k$ non adjacent vertices.
Assume the vertices are $\{1,2,3\dots N\}$
How many ways are there to select $k$ elements so that no two are consecutive (ignoring the fact that $1$ and $n$ are adjacent)?
There is a bijective correspondance between the subsets of $k$ elements of $\{1,2,3\dots N-k+1\}$ and the subsets of $k$ elements of $\{1,2,3\dots N\}$ that don't have two consecutive elements.
The bijection sends subset $a_1<a_2\dots < a_k$ to $a_1,a_2+1,a_3+2\dots,a_k+k-1$.
There are therefore $\binom{N-k+1}{k}$ sequences, now we just need to subtract the ones that contain $1$ and $N$.
The sequences that contain $1$ and $N$ are those which contain $1$ and $N-k+1$ in our bijection. There are clearly $\binom{N-k-1}{k-2}$.
So the final answer is $\binom{N-k+1}{k}-\binom{N-k-1}{k-2}$

In your particular case this yields $\binom{8}{4}-\binom{6}{2}=55$
