Question

There are 15 points in a plane out of which only 6 are in a straight line, then 

(a) How many different straight lines can be made? 

(b) How many triangles can be made?

Answer 1

(a) Straight line can be formed by joining 2 points, so number of ways in which we select 2 points from 15 points is = ¹⁵C₂ 

But it also include that in which 6 points are in straight line. From those 6 points only one line can be formed. 

So. We have to subtract ⁶C₂ – 1 from the above result. 

So, total lines are = ¹⁵C₂ – ⁶C₂ + 1 

=\(\frac{15!}{2!\space13!}\) − \(\frac{6!}{2!\space4!}\) + 1 

= \(\frac{15\times14}{2\times1}\) − \(\frac{6\times5}{2\times1}\) + 1 

= 105 − 15 + 1 

= 91 

(b) Noncollinear point 9 and collinear point 6 

Case I we choose 1 Noncollinear and 2 collinear point to make triangle 

⁹C₁ × ⁶C₂ = \(\frac{9!}{8!\space1!}\) × \(\frac{6!}{4!\space2!}\) 

= \(\frac{9\times6\times5}{2}\)

= 135 

Case II we choose 2 Noncollinear point and 1 collinear point to make a triangle 

= ⁹C₂ × ⁶C1 = \(\frac{9!}{3!\space6!}\)

= \(\frac{6 . 8 . 7}{3 . 2}\) = 84 

Total number of triangles = 135 + 216 + 84 

= 435