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JEE Main & Advanced Mathematics Permutations and Combinations Question Bank Self Evaluation Test - Permutations and Combinatioins

  • question_answer
    If \[^{n}{{C}_{r-1}}{{+}^{n+1}}{{C}_{r-1}}{{+}^{n+2}}{{C}_{r-1}}+...{{+}^{2n}}{{C}_{r-1}}\] \[{{=}^{2n+1}}{{C}_{{{r}^{2}}-132}}{{-}^{n}}{{C}_{r}},\] Then the value of r and the minimum value of n are

    A) 10

    B) 11

    C) 12

    D) 13

    Correct Answer: C

    Solution :

    [c] Given, \[^{n}{{C}_{r}}{{+}^{n}}{{C}_{r-1}}{{+}^{n+1}}{{C}_{r-1}}{{+}^{n+2}}{{C}_{r-1}}+{{...}^{2n}}{{C}_{r-1}}\] \[{{=}^{2n+1}}{{C}_{{{r}^{2}}-132}}\] \[\Rightarrow \,\,\,\,{{\,}^{n+1}}{{C}_{r}}{{+}^{n+1}}{{C}_{r-1}}+...{{+}^{2n}}{{C}_{r-1}}{{=}^{2n+1}}{{C}_{{{r}^{2}}-132}}\] \[{{\Rightarrow }^{2n}}{{C}_{r}}{{+}^{2n}}{{C}_{r-1}}{{=}^{2n+1}}{{C}_{{{r}^{2}}-132}}\] \[{{\Rightarrow }^{2n+1}}{{C}_{r}}{{=}^{2n+1}}{{C}_{{{r}^{2}}-132}}\] \[\Rightarrow {{r}^{2}}-r-132=0\] \[\Rightarrow (r-12)(r+11)=0\Rightarrow r=12\Rightarrow n\ge 12\] So, minimum value of n = 12.


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