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How many hexagons can be constructed by joining the vertices of a 15 sided polygon if none of the sides of the hexagon is also the side of the 15-gon.

My attempt

First calculate in how many ways we can select 6 points from 15 points then subtract which have 1 side same as the 15-gon then 2,3....5 sides common.

But that is very long approach .Any better method.

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Write down $9$ $\times$'s, like this $$ \times\qquad \times\qquad \times\qquad \times\qquad \times\qquad \times\qquad \times\qquad \times\qquad \times$$

These determine $8$ gaps, plus $2$ "endgaps." We either choose $6$ gaps to put a $\circ$ into, or choose one of the two endgaps, plus $5$ real gaps to put a $\circ$ into. This can be done in $$\binom{8}{6}+2\binom{8}{5}$$ ways.

To make a real hexagon inscribed in our $15$-gon out of the pattern of $\circ$ and $\times$, take a fixed vertex of the given $15$-gon, put the leftmost of our $15$ "letters" into it, and the rest counterclockwise as we travel to the right. The $\circ$ will be the vertices of the hexagon.

The method generalizes to $k$-gons in $n$-gons.

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More generally drawing $n$-gons in a $k$-gon but not using any of the original sides,

suppose you know one of the vertices: then you need to divide the $k$ gaps into $n$ parts with each part at least $2$, which is the same as dividing $k-n$ into $n$ parts with each part at least $1$, and using stars and bars, this can be done in $\displaystyle{k-n-1 \choose n-1}$ ways.

But we need to adjust our answer for the fact that we could have had any of $k$ original vertices, but by adjusting for this we get each $n$-gon $n$ times. So the answer is $$\frac{k}{n}{k-n-1 \choose n-1}$$ or here $\frac{15}{6}{8 \choose 5}$. This is the same as André Nicolas's ${k-n-1 \choose n}+ 2 {k-n-1 \choose n-1}$.

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  • $\begingroup$ hen you need to divide the k gaps into n parts with each part at least 2, which is the same as dividing k−n into n parts with each part at least 1,...??What does this means $\endgroup$ Aug 31 '13 at 6:51
  • $\begingroup$ @maths lover: Suppose you had to divide 7 into 3 parts at least 2: you could do this as 3+2+2 or 2+3+2 or 2+2+3. It is the same question as dividing 7-3=4 into 3 parts at least 1: 2+1+1, 1+2+1 or 1+1+2 $\endgroup$
    – Henry
    Aug 31 '13 at 10:19

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