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. 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.
 A: 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. 
A: 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 ${k-n-1 \choose n}+ 2 {k-n-1 \choose n-1}$.
