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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.

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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 !

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  • $\begingroup$ Could you elaborate on your ${n\over n-7}$ argument? (Anyway, it leads to the correct result.) $\endgroup$ – Christian Blatter Feb 20 '16 at 15:09
  • $\begingroup$ @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. $\endgroup$ – true blue anil Feb 20 '16 at 15:34
  • $\begingroup$ @ChristianBlatter: I also simultaneously worked out a similar solution to yours which you have presented in a better way ! $\endgroup$ – true blue anil Feb 20 '16 at 15:38
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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.

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