Counting heptagons formed using vertices of $n$-sided polygon having no sides in common with the polygon 
Number of Different Heptagons ($7$-sided polygons) which can be formed by joining the vertices of a polygon having $n$ sides, If none of the side of the polygon is the side of heptagon, is

My Try:


*

*Let us take $n$ sided regular polygon has vertices as
$A_{1},A_{2},A_{3},\dotsc,A_{n-1},A_{n}$

*Now here we have to form a Heptagon none of whose side are the side
of $\bf{Polygon}$

*Now If we take $A_{1}$ as one vertices, then we can not take $A_{2}$
and $A_{n}$ So here we have to take

*$6$ vertices from $n-3$ vertices such that no two vertices are
consecutive.

*So the total number of ways equals $n$ times the number of ways  in
which no two vertices are consecutive.


So I did not understand how I can calculate that part, please help me.
Thanks.
 A: I take it that the heptagons are convex.
Let us denote the $7$ chosen vertices as $\Large\circ$ and the rest as $\Large\bullet$
Looking clockwise make blocks of used-unused vertices $\boxed{\Large\circ\Large\bullet},$ treating each as one object.
This ensures that used vertices can never be adjacent,
and hence no side of the polygon is used in forming the heptagon.
There are now $7 + n -14 = n-7$ objects, and the  blocks can be placed in $\binom{n-7}{7}$ ways,
but we are giving the blocks only $(n-7)$ starting points instead of $n$,
so we need to use a multiplication factor of $\dfrac{n}{n-7}$,
giving the answer $\dfrac{n}{n-7}\times\dbinom{n-7}{7}$

Another way
Between the $7$ blocks, there are $7$ "compartments" in which the remaining $(n-14)$ can be placed, in $\binom{n-14+7-1}{7-1} = \binom{n-8}{6}$ ways.
But due to the circular type of arrangement, each pattern will repeat $7$ times,
thus ans = $\dfrac{n}{7}\dbinom{n-8}{6}$ ways
Check that the two answers are equivalent !
A: Let us count the number of such heptagons where one of the vertices is called the leader.
A priori we can choose the leader in $n$ ways. After the leader has been chosen an admissible heptagon can be encoded as a word 
$$L\ \underline{\quad}\ 1\ \underline{\quad}\ 1\ \underline{\quad}\ 1\ \underline{\quad}\ 1\ \underline{\quad}\ 1\ \underline{\quad}\ 1\ \underline{\quad}\ L$$
with $6$ ones and at least one zero in each of the $7$ slots. This leaves $n-14$ zeros to distribute freely in the $7$ slots, which can be done in ${n-14+6\choose 6}$ ways. It follows that there are
$$n\cdot{n-8\choose 6}$$
admissible heptagons with a leading vertex. 
In this way we have counted each "blank" heptagon $7$ times, so that the number of "blank" heptagons comes to
$${n\over 7}\cdot{n-8\choose 6}\ ,$$
as obtained also by true blue anil.
