Counting heptagons formed using vertices of $n$-sided polygon having no sides in common with the polygon

Question

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.

Are the sides of the heptagon allowed to intersect each other? E.g. do the blue and green 'polygons' in this image count as heptagons? — Frentos, Feb 20 '16 at 09:03
This http://math.stackexchange.com/questions/490405/choosing-numbers-without-consecutive-numbers might help — Mathematics, Feb 20 '16 at 13:23

true blue anil · Accepted Answer · 2016-02-20T15:22:14.860

3

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 !

edited Feb 20 '16 at 15:22

answered Feb 20 '16 at 13:46

true blue anil

41,295

Could you elaborate on your ${n\over n-7}$ argument? (Anyway, it leads to the correct result.) – Christian Blatter Feb 20 '16 at 15:09
@ChristianBlatter: The first used vertex (say) should be allowed to be any of the $n$, but by forming "blocks", it has only $(n-7)$ choices. – true blue anil Feb 20 '16 at 15:34
@ChristianBlatter: I also simultaneously worked out a similar solution to yours which you have presented in a better way ! – true blue anil Feb 20 '16 at 15:38

score 3 · Answer 2 · answered Feb 20 '16 at 15:05

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.

Counting heptagons formed using vertices of $n$-sided polygon having no sides in common with the polygon

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