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}$$
Whats the interpretation of the ${5\choose3}$ term?
The answer to this might result in a better approach than what I did:and also could be used for $k$ sided polygons inside $n$ sided polygons, since it's hard to form a closed form based on what I did , for a general case.
Best Answer
There are $10$ ways to choose the first vertex. If you move clockwise around the decagon, let $x_1$ be the number of vertices between the first and second vertices, $x_2$ be the number of vertices between the second and third vertices, and $x_3$ be the number of vertices between the third and fourth vertices, and $x_4$ be the number of vertices between the fourth and first vertices. Then $$x_1 + x_2 + x_3 + x_4 = 10 - 4 = 6 \tag{1}$$ which is an equation in the positive integers since there must be at least one vertex between successive vertices of the quadratilateral. A particular solution of equation 1 corresponds to the placement of $4 - 1 = 3$ addition signs in the $6 - 1 = 5$ spaces between successive ones in a row of six ones. $$1 \square 1 \square 1 \square 1 \square 1 \square 1$$ For instance, choosing the second, fourth, and fifth spaces corresponds to the solution $x_1 = 2$, $x_2 = 2$, $x_3 = 1$, and $x_4 = 1$. The number of solutions of equation $1$ is $$\binom{6 - 1}{4 - 1} = \binom{5}{3}$$ However, as you note, if we simply multiply these numbers, we will have counted each quadrilateral four times, once for each way we could have designated one of the vertices of the quadrilateral as the "first" vertex. Hence, there are $$\frac{1}{4}\binom{10}{1}\binom{5}{3}$$ quadrilaterals that can be formed by joining nonadjacent vertices of a decagon.