There are 10 points in a plane, no three of which are in the same straight line,excepting 4 points ,which are collinear.Find the (i)number of straight lines obtained from the pairs of these points (ii)number of triangles that can be formed with the vertices as these points..Explain with diagram

the number of straight lines from the given points are
C210-C24+1  [since 4 points are collinear and line formed by these collinear points is 1=10*92-4*32+1=45-6+1=40
the number of triangles formed with non collinear points are C36
the number of triangles formed with one point taken from 4 collinear points and 2 points taken from non collinear points = C14.C26when two points taken from 4 collinear points and 1 point from non collinear point=C24.C16thus the total number of required triangles areC36+C14C26+C24.C16=6*5*43*2+4*6*52+4*32*6=20+60+36=116

hope this helps you

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