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.
Best Answer
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 !