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  3. CEE Kerala Engineering

CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2003

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

    A)  0                                            

    B)  1242     

    C)         7315                     

    D)         6345

    E)  3340

    Correct Answer: A

    Solution :

    \[^{20}{{C}_{4}}+{{2.}^{20}}{{C}_{3}}{{+}^{20}}{{C}_{2}}{{-}^{22}}{{C}_{18}}\] \[{{=}^{21}}{{C}_{4}}{{+}^{20}}{{C}_{3}}{{+}^{20}}{{C}_{2}}{{-}^{22}}{{C}_{18}}\] \[{{=}^{21}}{{C}_{4}}{{+}^{21}}{{C}_{3}}{{-}^{22}}{{C}_{18}}\] \[{{=}^{22}}{{C}_{4}}{{-}^{22}}{{C}_{18}}\] \[{{=}^{22}}{{C}_{18}}{{-}^{22}}{{C}_{18}}\] \[=0\]


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