How many convex quadrilaterals can be formed between two parallel lines where 5 points lies in one line and 7 in the other?

Question

My approach to solve this problem has been first choose all the possible combinations of 4 points from the 12 available. Which is,

$^{12}C_4=\frac{12!}{4!8!}=495$

Then, since they have to be convexed I get rid of all of the four collinear points. Which are,

$^5C_4=\frac{5!}{4!}=5$ and $^7C_4=\frac{7!}{4!}=210$

So I have that, $^{12}C_4-(^5C_4+^7C_4)=495-(5+210)=495-215=280$

So there are 280 convexed quadrilaterals, but, should I get rid of the 3 collinear points too? If I do it I got,

$280-(^5C_3+^7C_3)=280-(10+35)=280-45=235$

Thank you in advanced for any help on the matter.

Answer 1

In order to have a strictly convex quadrilateral we have to select $2$ points on one line and $2$ points on the other line: $$\binom{5}{2}\cdot \binom{7}{2}=210.$$ Following your approach you should have: $$\binom{12}{4}-\Big(\underbrace{\binom{5}{1}\binom{7}{3}+\binom{7}{1}\binom{5}{3}}_{\text{$3$ on one line and $1$ on the other}}+\underbrace{\binom{7}{4}+\binom{5}{4}}_{\text{$4$ on a line}}\Big)=210.$$

Answer 2

So there are 280 convexed quadrilaterals, but, should I get rid of the 3 collinear points too?

The person who set the question may disagree with me, but I would consider a "quadrilateral" with three collinear points to be a triangle and not a quadrilateral.

If I do it I got,

$280-(^5C_3+^7C_3)=280-(10+35)=280-45=235$

No. Given three collinear points, you need to count all sets of four points which contain them.