Find number of Quadrilaterals that can be formed in a Decagon such that no side of Quadrilateral is common to side of Decagon.
I tried as follows:
Arbitrarily choose $6$ points on a circle. Then we have $6$ Gaps in between them. Choose any four gaps to place other four vertices in each gap which forms a quadrilateral with no side common. So number of ways =$\binom{6}{4}=15$.
But answer is $25$, where did i went wrong?
On
There are $\binom{10}{6}=210 \,\,$ possible quadrilaters. Among these, we have to exclude:
Those with only a single side in common with the decagon. The number of these quadrilaters can be obtained by noting that, once chosen a decagon side (e.g. $AB \,$), there are $\binom{6}{2}=15\,\,$ ways to choose the other two vertices $C,D \,$ of the quadrilater (note that these two vertices cannot be adjacent to $AB\,$, since we are looking for the quadrilaters with only a single side in common with the decagon). Among these $15$ quadrilaters, $5$ must be excluded because have $C$ and $D$ as adjacent vertices. Since the initial side $AB$ can be chosen in $10$ ways, we obtain $10 (15 - 5)=100$ quadrilaters with only a single side in common with the decagon.
Those with two sides in common with the decagon. These two sides can be adjacent or not. The number of quadrilaters with two adjacent sides in common with the decagon can be obtained by noting that there are $10$ ways to choose two adjacent sides, and $5$ ways to choose the fourth vertex (note that, among the remaining $7$ decagon vertices, only $5$ have to be considered because we are counting the quadrilaters with exactly two adjacent sides). So we get $10 \cdot 5=50 \,$ quadrilaters of this type. On the other hand, the number of quadrilaters with two non-adjacent sides in common with the decagon can be obtained by noting that there are $10$ ways to choose an initial side $AB$, and $5$ ways to choose the other side $CD$ so that $AB$ and $CD$ are not adjacent. We also have to consider that, repeating this calculation for each of the $10$ possible initial sides $AB$, each of these quadrilaters is counted twice (the same quadrilater is obtained by choosing $AB$ as initial side and $CD$ as the second one, or vice versa). So the number of quadrilaters with two non-adjacent sides in common with the decagon is given by $\frac{10 \cdot 5}{2}=25 \,\,$.
Lastly, those with three sides in common with the decagon. It is trivial to show that there are $10$ of these quadrilaters.
Therefore, the number of quadrilaterals with no side in common with the Decagon is
$$210-(100+50+25+10)=25$$
The problem is that you are given some fixed decagon $ABCDEFGHIJ$ (left) but after you choose six points on a circle and pick $4$ gaps, you get a quadrilateral in one of several lopsided decagons (right).
To turn the quadrilateral on the right into a quadrilateral inside the decagon on the left, you need to pick a starting point. One of $\{1,1.5,2,3,3.5,4,4.5,5,5.5,6\}$ becomes $A$, and then the rest of the labels are determined by going around the circle clockwise. This can be done in $10$ ways.
This wouldn't be a problem if there were always $10$ possible orderings of the labels on the lopsided decagon. Then we'd be overcounting by a factor of $10$, and the two $10$s would cancel out.
However, there aren't $10$ possible orderings of the labels on the lopsided decagon. There are only $6$, because the original vertices of the hexagon are distinguishable from the four vertices you put in the gaps. The quadrilateral with vertices $1$, $3$, $4$, $5$ couldn't have been obtained in this way, because $1$, $3$, $4$, and $5$ are the vertices of the starting hexagon.
In other words, there are $10$ ways to label the decagon on the left, but only $6$ ways to label the decagon on the right, so when you take labels into account, you undercount by a factor of $\frac{6}{10}$.