$10$ points lie in a plane, of which $4$ points are collinear. Barring these $4$ points, no $3$ of the $10$ points are collinear. How many distinct quadrilaterals can be drawn?
I have solved it in the following manner:
Considering the first case; $1$ point from the line and rest $3$ out of $6$, which are not collinear I get: $4C1 \cdot 6C3=80$
second case: 2 points from the collinear ones and rest from the 6 I get:
$4C2 \cdot 6C2=90$
third case: using four points from the six non-collinear ones, I get: $6C4=15$
Adding them up I got $185$ ways, but the answer is $209$, which I don't understand how. Please help.
Best Answer
You are correct. To obtain the answer $209$, you would have to consider triangles to be quadrilaterals.
As a check, observe that there are $\binom{10}{4}$ ways to select four of the ten points in the plane to be the vertices of the quadrilateral. If three or more of the vertices are collinear, we obtain fewer than four sides, so we must subtract these cases from the total. There is only one way to select all four points on the line. For each of the four ways we could select three of these four points, there are six ways to pick the additional vertex from the points that are not on the line. Hence, the number of quadrilaterals that can be formed is $$\binom{10}{4} - \binom{4}{4} - \binom{4}{3}\binom{6}{1} = 185$$ as you found directly.