JEE Main & Advanced Mathematics Permutations and Combinations Question Bank Geometrical Problems

  • question_answer
    There are \[m\] points on a straight line \[AB\] and \[n\] points on another line \[AC\], none of them being the point \[A\]. Triangles are formed from these points as vertices when (i) \[A\]is excluded (ii) \[A\] is included. Then the ratio of the number of triangles in the two cases is

    A) \[\frac{m+n-2}{m+n}\]

    B) \[\frac{m+n-2}{2}\]

    C) \[\frac{m+n-2}{m+n+2}\]

    D) None of these

    Correct Answer: A

    Solution :

    Case I: When \[A\] is excluded. Number of triangles = selection of 2 points from \[AB\] and one point from \[AC+\] selection of one point from \[AB\] and two points from \[AC\] \[{{=}^{m}}{{C}_{2}}^{n}{{C}_{1}}{{+}^{m}}{{C}_{1}}^{n}{{C}_{2}}=\frac{1}{2}(m+n-2)mn\] ?..(i) Case II: When \[A\] is included. The triangles with one vertex at \[A=\] selection of  one point from \[AB\] and one point from \[AC=mn\]. \[\therefore \] Number of triangles \[=mn+\frac{1}{2}mn(m+n-2)\]\[=\frac{1}{2}mn(m+n)\] ?..(ii) \[\therefore \]Required ratio\[=\frac{(m+n-2)}{(m+n)}\].


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