How many hexagons can be made 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.
Suppose, we take a 9 sided polygon and use Gap Method (Gap and string method in permutations and combinations) to fill other 6 points to be filled in between the present 9 points in order to have a 15 sided polygon
Also, in order to make hexagon according to the question, we just have to choose 6 "gaps" between the present points and put the points in the places. So, by doing all this we will have a 15 sided polygon.
And in this way, no. of ways of making a 15 sided polygon will mean same as selecting 6 points from a 15 sided polygon which aren't adjacent. (which helps in the idea of making a hexagon which will have no side which is also a side of the 15 sided polygon)
[ "×" - present points, "_" - gaps]
                    ×         _            ×
                  _                          _ 
                  
                 ×                             ×
                  _                            _           

                    ×                          ×
                      _                       _
                        ×                   ×
                            _            _
                                   ×  

No. of gaps=9
So, now according to me,
No. of such Hexagons which can be formed = $\binom{9}{6}$
= 84
But, this is not the correct answer. What am I missing?
(Correct Answer- 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.)
 A: The $15$-sided polygon is a fixed object.  You need a reference point.  Say it is vertex $A$.
There are two possibilities: vertex $A$ is a vertex of the hexagon, or it is not.
Suppose $A$ is not a vertex of the hexagon.  Arrange nine blue balls in the form of a circle.  Label one of those balls $A$.  Choose six of the nine gaps in which to place a green ball to ensure that no two green balls are adjacent.  Now label the vertices in alphabetical order as you proceed clockwise around the circle, starting at $A$.  The letters on the green balls are the vertices of the hexagon.  There are $\binom{9}{6}$ ways to do this, which agrees with your answer.
Now suppose $A$ is a vertex of the hexagon.  Color it green.  Arrange nine blue balls around $A$ to form a circle which includes the ball labeled $A$.  Doing so creates $10$ spaces in which to place a green ball, but two of these are adjacent to $A$, leaving eight spaces where a green ball could be placed so that it is not adjacent to $A$.  Choose five of these eight spaces in which to place a green ball to ensure that no two green balls are adjacent.  Now label the balls in alphabetical order as you proceed clockwise around the circle, starting at $A$.  The letters on the green balls are the vertices of the hexagon.  There are $\binom{8}{5}$ ways to do this, which you have not taken into account.
Since the two cases are mutually exclusive and exhaustive, the number of admissible hexagons is
$$\binom{9}{6} + \binom{8}{5}$$
which agrees with the linked answers provided by Andre Nicolas and Henry.
A: What you are missing is a multiplication factor of $\frac{15}{15-6}$
ie the answer should be $\binom96\times \frac{15}{9}$
The multiplication factor is arising because you are necessarily attaching a gap (clockwise, say) immediately after each vertex, which restricts the number of places the vertex-gap "blocks" can occupy.
