JCECE Engineering JCECE Engineering Solved Paper-2005

  • question_answer
    \[^{20}{{C}_{4}}+2{{\cdot }^{20}}{{C}_{3}}{{+}^{20}}{{C}_{2}}{{-}^{22}}{{C}_{18}}\]is equal to :

    A) \[0\]                                     

    B) \[1242\]

    C) \[7315\]                              

    D) \[6345\]

    Correct Answer: A

    Solution :

    Key Idea: If \[n\] and \[r\]positive integer and\[0\le r<n\], then                 \[^{n}{{C}_{r}}{{+}^{n}}{{C}_{r-1}}{{=}^{n+1}}{{C}_{r}}\] \[\therefore \]  \[^{20}{{C}_{4}}+{{2.}^{20}}{{C}_{3}}{{+}^{20}}{{C}_{2}}{{-}^{22}}{{C}_{18}}\]                 \[={{(}^{20}}{{C}_{4}}{{+}^{30}}{{C}_{3}})+{{(}^{20}}{{C}_{3}}{{+}^{20}}{{C}_{2}}){{-}^{22}}{{C}_{18}}\]                 \[{{=}^{21}}{{C}_{4}}{{+}^{21}}{{C}_{3}}{{-}^{22}}{{C}_{18}}\]                 \[{{=}^{22}}{{C}_{4}}{{-}^{22}}{{C}_{18}}\]                 \[{{=}^{22}}{{C}_{18}}{{-}^{22}}{{C}_{18}}\]                          \[=0\]


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