bblast wrote:

why not pick 2 points out of 6 ?

=6C2

now since every 2 point will form a triangle with the 4 remaining points.

hence answer = 6C2*4 = 60

help me

Consider only 3 non-collinear points:

A B C

As per your formula:

Number of triangles = \(C^{3}_{2}*1=3\). But, in reality it is 3/3 = 1.

Because your formula counts same triangle thrice. Thus, at the end you will have to divide by 3.

How so:

A B C are 3 points.

It will choose one line segment using two points:

AB and connect AB to C to form a triangle.

Then,

BC and connect BC to A to form a triangle.

AC and connect AC to B to form a triangle.

You see three counts for the same triangle ABC.

Divide your result by 3 and you will get the answer.

Or simply; select 3 points out of n points to know the number.

If there are n non-collinear points, where n>=3, we can make

\(C^{n}_{3}\) distinct triangles.

_________________

~fluke

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