BCECE Engineering BCECE Engineering Solved Paper-2001

  • question_answer
    The number of a triangle that can be formed by 5 points in a line and 3 points on a parallel line, is:

    A) \[{{\,}^{8}}{{C}_{3}}\]                                    

    B)  \[{{\,}^{8}}{{C}_{3}}-{{\,}^{5}}{{C}_{3}}\]             

    C)         \[{{\,}^{8}}{{C}_{3}}-{{\,}^{5}}{{C}_{3}}-1\]

    D)         none of these

    Correct Answer: C

    Solution :

    Key Idea: If there are \[n\]different points, the total number of triangles formed by these point are\[{{\,}^{n}}{{C}_{3}}.\] The total number of points are 8. So number of triangles formed by these points are \[{{\,}^{8}}{{C}_{3}}.\] But we have to subtract those triangle in which the point lie on a line. \[\therefore \] Required number of ways\[={{\,}^{8}}{{C}_{3}}-{{\,}^{5}}{{C}_{3}}{{-}^{3}}{{C}_{3}}\]                                                 \[={{\,}^{8}}{{C}_{3}}-{{\,}^{5}}{{C}_{3}}-1\] Note: The triangle will be formed, if not more than two points in a line.


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