- (q) ; - (p) ; - (s) ; - (r)

Sol. 8 C 4 + 8 C 3 × 5 C 1 + 8 C 2 × 5 C 2

The number of ways of selecting 3 points out of 12 points is 12 C 3 .

Three points out of 7 collinear points can be selected in 7 C 3 ways.

Hence, the number of triangles formed is 12 C 3 – 7 C 3 = 185.

m C 2 × n C 2

Two circles intersect in 2 points.

∴ Maximum number of points of intersection of two circles = 2 × number of selections of two circles from 8 circles.

= 2 × 8 C 2 = 2 × 28 = 56

∴ Maximum number of points of intersection of two straight line = 1 × number of selections of two straight line from 4 straight line = 4 C 2 = 6

∴ Maximum number of points of intersection of one straight line and one circle = 2 × number of selections of one straight line from 4 straight line and number of selections of one circles from 8 circles

= 4 C 1 . 8 C 1 .2 = 64

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