Question: Find the number of quadrilaterals formed by joining the vertices of a decagon,that share no common side with the decagon.

Approach:

* Select one vertex first:-This can be done in $10\choose1$=$10$ ways.
* Now, we consider the vertices **except** the chosen one, and the two adjacent to it: i.e $7$ vertices. We need to chose three of them, which can be done in $7\choose3$ ways. However, these $7\choose3$ cases include:
* Cases where $2$ vertices are adjacent: $6$ ways to chose 2 adjacent vertices, and $5$ ways to chose the 3rd vertex: so $30$ cases.
* Cases where all $3$ vertices are adjacent: $5$ ways to do so.

Using the IEP, the cases we need to consider for chosing 3 vertices are: $7\choose3$$-30+5=10.$
The Final answer then should be $10\choose1$ $\*10\*1/4$. We divide by 4 since each case gets counted 4 times: for example if the quadrilateral having vertex number:($1,3,7,9$) is counted when we chose $1$ as the 1st vertex, and also when we chose $3$ as the first vertex, and so on.

The final answer matches the one given in my book: although my book writes the answer as :
$$1/4\*{10\choose1}\*{5\choose3}$$

**Whats the interpretation of the ${5\choose3}$ term?**

The answer to this might result in a better approach than what I did:and also could be used for $k$ sided polygons inside $n$ sided polygons, since it's hard to form a closed form based on what I did , for a general case.