Q59 / 160 · TS EAMCET 2023 (14 May Shift 1) · Mathematics

There are 10 points in a plane, of which no three points are collinear except 4 . Then, the number of distinct triangles that can be formed by joining any three points of these ten points, such that at least one of the vertices of every triangle formed is from the given 4 collinear points is

  • A
    $80$
  • B
    $100$
  • Correct answer
    C
    $96$
  • D
    $116$

Solution

Number of required triangles: (Select 1 point from 4 collinear and 2 from remaining 6 ) OR (select 2 from collinear points \& 1 from remaining 6 ) $$ \begin{aligned} & =\left({ }^4 C_1 \times{ }^6 C_2\right)+\left({ }^4 C_2 \times{ }^6 C_1\right) \\ & =(4 \times 15)+(6 \times 6)=96 \end{aligned} $$

Hence, the correct answer is option: 'C'.

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