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

  • question_answer The number of ways in which an examiner can assign 30 marks to 8 questions, awarding not less than 2 marks to any question is

    A) \[^{21}{{C}_{7}}\]

    B) \[^{30}{{C}_{16}}\]

    C) \[^{21}{{C}_{16}}\]

    D) None of these

    Correct Answer: A

    Solution :

    Since the minimum marks to any question is two, the maximum marks that can be assigned to any  questions is \[16(=30-2\times 7),\ {{n}_{1}}+{{n}_{2}}+........+{{n}_{8}}=30\]. If \[{{n}_{i}}\] are the marks assigned to \[{{i}^{th}}\]questions, then \[{{n}_{1}}+{{n}_{2}}+...........+{{n}_{8}}=30\] with \[2\le {{n}_{i}}\le 16\] for\[i=1,\ 2,........,8\].  Thus the required number of ways = the coefficient of \[{{x}^{30}}\] in \[{{({{x}^{2}}+{{x}^{3}}+.......+{{x}^{16}})}^{8}}\] = the coefficient of \[{{x}^{30}}\] in \[{{x}^{16}}{{(1+x+..........{{x}^{14}})}^{8}}\] = the coefficient of \[{{x}^{30}}\] in \[2P\] = the coefficient of \[{{x}^{14}}\] in \[{{(1-x)}^{-8}}\ .\ {{(1-{{x}^{15}})}^{8}}\] =  the coefficient of \[{{x}^{14}}\] in    \[\left\{ 1+\frac{8}{1\ !}x+\frac{8.9}{2\ !}{{x}^{2}}+\frac{8.9.10}{3\ !}{{x}^{3}}+...... \right\}(1{{-}^{8}}{{C}_{1}}{{x}^{15}}+......)\]    = the coefficient\[{{x}^{14}}\]in \[\left\{ 1{{+}^{8}}{{C}_{1}}x{{+}^{9}}{{C}_{2}}{{x}^{2}}{{+}^{10}}{{C}_{3}}{{x}^{3}}+.... \right\}\] since the second bracket has powers of \[i.e.\] etc. =\[7\ !\].


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