JEE Main & Advanced Mathematics Permutations and Combinations Question Bank Critical Thinking Questions

  • question_answer A person goes in for an examination in which there are four papers with a maximum of \[m\] marks from each paper. The number of ways in which one can get \[2m\] marks is

    A) \[^{2m+3}{{C}_{3}}\]

    B) \[\frac{1}{3}(m+1)(2{{m}^{2}}+4m+1)\]

    C) \[\frac{1}{3}(m+1)(2{{m}^{2}}+4m+3)\]

    D) None of these

    Correct Answer: C

    Solution :

    The required number = coefficient of \[{{x}^{2m}}\] in \[{{({{x}^{0}}+{{x}^{1}}+......+{{x}^{m}})}^{4}}\] = coefficient of \[{{x}^{2m}}\] in \[{{\left( \frac{1-{{x}^{m+1}}}{1-x} \right)}^{4}}\] = coefficient of \[{{x}^{2m}}\] in \[{{(1-{{x}^{m+1}})}^{4}}{{(1-x)}^{-4}}\] = coefficient of \[{{x}^{2m}}\] in \[(1-4{{x}^{m+1}}+6{{x}^{2m+2}}+......)\]                                     \[\left( 1+4x+......+\frac{(r+1)(r+2)(r+3)}{3\ !}{{x}^{r}}+.... \right)\] \[=\frac{(2m+1)(2m+2)(2m+3)}{6}-4m\frac{(m+1)(m+2)}{6}\] \[=\frac{(m+1)(2{{m}^{2}}+4m+3)}{3}\].


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