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JEE Main & Advanced AIEEE Solved Paper-2010

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    Directions: Questions number 87 are Assertion - Reason type questions. Each of these questions contains two statements: Statement - 1 (Assertion) and Statement - 2 (Reason). Each of these questions also has four alternative choices, only one of which is the correct answer. You have to select the correct choice. Let \[{{S}_{1}}=\sum\limits_{j=1}^{10}{j{{(j-1)}^{10}}}{{C}_{j}},{{S}_{2}}=\sum\limits_{j=1}^{10}{{{j}^{10}}{{C}_{j}}}\]and \[{{S}_{3}}=\sum\limits_{j=1}^{10}{{{j}^{2}}^{10}{{C}_{j}}}\] Statement ? 1: \[{{S}_{3}}=55\times {{2}^{9}}.\] Statement ? 2:\[{{S}_{1}}=90\times {{2}^{8}}\]and\[{{S}_{2}}=10\times {{2}^{8}}.\] Statement ? 1 (Assertion) and Statement ? 2 (Reason). Each of these questions also has four alternative choices, only one of which is the correct answer. You have to select the correct choice.       AIEEE  Solved  Paper-2010

    A) Statement -1 is true, Statement -2 is true; Statement -2 is a correct explanation for Statement -1

    B) Statement -1 is true, Statement -2 is true; Statement -2 is not a correct explanation for Statement -1.

    C) Statement -1 is true, Statement -2 is false.

    D) Statement -1 is false, Statement -2 is true.

    Correct Answer: C

    Solution :

    \[{{S}_{1}}=\sum\limits_{j=1}^{10}{j(j-1)}\frac{10!}{j(j-1)(j-2)!(10-j)!}\] \[=90\sum\limits_{j=2}^{10}{\frac{8!}{(j-2)!(8-(j-2))!}}=90\times {{2}^{8}}\] \[{{S}_{2}}=\sum\limits_{j=1}^{10}{j}\frac{10!}{j(j-1)!(9-(j-1))!}\] \[=10\sum\limits_{j=1}^{10}{\frac{9!}{(j-1)!(9-(j-1))!}}=10\times {{2}^{9}}\] \[{{S}_{3}}=\sum\limits_{j=1}^{10}{[j(j-1)+j]}\frac{10!}{j(10-j)!}\]\[=\sum\limits_{j=1}^{10}{{{(j-1)}^{10}}{{C}_{j}}}\] \[=\sum\limits_{j=1}^{10}{{{(j)}^{10}}{{C}_{j}}}={{90.2}^{8}}+{{10.2}^{9}}={{110.2}^{8}}={{55.2}^{9}}\] Hence statement 1 is true, statement 2 is false


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