There are 20 points on a plane, no 3 are collinear except 4. Find no. of lines formed, and total no. of triangles that can be formed.


No of lines will be $$\binom{20}{2}-\binom{4}{2}+1 = 185$$

No of Triangles using only given points will be : $$\binom{20}{3}-\binom{4}{3} = 1140 - 4 = 1136$$

But the question asks for total no of triangles. so we would also consider triangles with some vertices that are not one of the given twenty points.

So how would we calculate them?

I attempted like so:

A triangle would form if three lines intersect. But these three lines should not intersect at same point. SO we subtract 3-line combinations that meet at same point from total no of 3-line combinations.

so it should be $$\binom{185}{3}-\left[16\times\binom{19}{3}+4\times\binom{17}{3}\right]$$

$$= 1038220-18224 = \boxed{1019996}$$

But i seriously doubt if this is correct. Such a small number can't produce an astronomical no. of Triangles! Please help. Thank you.

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As pointed by joriki, the answer would depend on position of points, and nothing may be said until that is known. But will my answer hold if no 2 lines are parallel?


The question can't be meant like that, since the result would depend on the positions of the points: You can make a pair of lines parallel to remove one of those triangles without adding another. So it's probably just a sloppy way of asking for the number of triangles formed by the vertices.

  • $\begingroup$ Thank you! i get that point that two lines can be parallel and so no triangle would form using the third line (transversal). But if we add a condition that no line is parallel, then will my answer hold or will it still be wrong? Thank you. $\endgroup$ – Max Payne Aug 6 '15 at 13:09
  • $\begingroup$ In that case I agree, the alternate interpretation makes no sense because there is no way to know this with the current information. $\endgroup$ – Morgan Rodgers Aug 6 '15 at 13:24
  • $\begingroup$ @TimKrul: Yes, under that assumption, your answer seems correct. Well counted. $\endgroup$ – joriki Aug 6 '15 at 13:26
  • $\begingroup$ Thank you! Actually question was framed like "find maximum no of triangles that can be formed." $\endgroup$ – Max Payne Aug 7 '15 at 0:43
  • $\begingroup$ @joriki No i meant in the book it was written maximum. I asked a different question. I didn't mean to ask maximum. $\endgroup$ – Max Payne Aug 7 '15 at 10:28

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