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magical cook wrote:S is a set of points in the plane. How many distinct triangles can be drawn that have three
of the points in S as vertices?
(1) The number of distinct points in S is 5.
(2) No three of the points in S are collinear.

3 points are needed to draw a triangle .. so if there are N distinct points ( no three of the points out of the N points are collinear) then the number of triangles that can be drawn is Nc3 ..

example:- if we have 3 distnict points and they are not collinear( that is all 3 of them do not fall on the same line), then the number of triangles drawn = 3c3 = 1 ..

Now, from statement 1 .. we know that there are 5 distinct points, but we do not know how many of them are collinear, for all we know all 5 of them could fall on the same line in which case not even a single triangle can be drawn... so the ist statement is insufficient.

From the 2nd statement .. we know that none of the points are collinear, but the second statement doesn't say anything about the number of points .. so, second statement is again insufficient..

Combine the 2 statement and we get that there are 5 points, no three of which are collinear .. so the number of triangles that can be drawn is 5c3 .. Hence the answer is C ..

Yes OA is C - Thanks gabriel!