• # 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

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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