8 points are lying in a plane if 4 points out of these 8 points are collinear (lying on a line) then how many distinct quadrilaterals can be made using these 8 points??? I am not able to understand anything here can anyone please explain this questions answer in detail.
2026-03-10 06:11:54.1773123114
A problem of Permutations and combinations
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Abridged and revised answer considering only convex quadrilaterals:
The disposition of the 4 collinear points and the 4 "solo" points must be such as to allow only convex quadrilaterals (no point must be inside the triangle formed by the other 3 chosen points)
No point from collinear group and 4 from solo group: $\binom40\binom44 = 1$ quadrilateral.
1 point from collinear group and 3 from solo group $\binom41\binom43 = 16$ quadrilaterals
2 points from collinear group and 2 from solo group: $\binom42\binom42$ = 36 quadrilaterals
We can't take more than 2 points from the collinear group, thus
Number of convex quadrilaterals possible = 1+16+36 = 53