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  3. CEE Kerala Engineering

CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2007

  • question_answer
    If\[n=5,\]then \[{{{{(}^{n}}{{C}_{0}})}^{2}}+{{{{(}^{n}}{{C}_{1}})}^{2}}+{{{{(}^{n}}{{C}_{2}})}^{2}}+.....\] \[+{{{{(}^{n}}{{C}_{5}})}^{2}}\]is equal to

    A)  250                                       

    B)  254

    C)  245                       

    D)         252

    E)  258

    Correct Answer: D

    Solution :

    \[{{{{(}^{n}}{{C}_{0}})}^{2}}+{{{{(}^{n}}{{C}_{1}})}^{2}}+{{{{(}^{n}}{{C}_{2}})}^{2}}+.....+{{{{(}^{n}}{{C}_{5}})}^{2}}\] \[={{{{(}^{5}}{{C}_{0}})}^{2}}+{{{{(}^{5}}{{C}_{1}})}^{2}}+{{{{(}^{5}}{{C}_{2}})}^{2}}+{{{{(}^{n}}{{C}_{3}})}^{3}}+{{{{(}^{5}}{{C}_{4}})}^{4}}\]                                                                 \[+{{{{(}^{5}}{{C}_{5}})}^{2}}\] \[=1+25+100+100+25+1\] \[=252\]


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