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?
 A: 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 $N={11\over4}{6\choose3}=55$.
A: 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 $\binom{8}{4}-\binom{6}{2}=55$
