Permutations And Combinations Question 30
Question: 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
Options:
A) $ ^{21}C_7 $
B) $ ^{30}C _{16} $
C) $ ^{21}C _{16} $
D) None of these
Correct Answer: A 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 $ (1+x+x^2+…+x^{14})^8 $ = the coefficient of $ x^{14} $ in $ {{(1-x)}^{-8}}\ .\ {{(1-x^{15})}^{8}} $ = the coefficient of $ x^{14} $ in $ { 1+\frac{8}{1\ !}x+\frac{8.9}{2\ !}x^{2}+\frac{8.9.10}{3\ !}x^{3}+…… }(1{{-}^{8}}C_1x^{15}+……) $ = the coefficient $ x^{14} $ in $ { 1{{+}^{8}}C_1x{{+}^{9}}C_2x^{2}{{+}^{10}}C_3x^{3}+…. } $ since the second bracket has powers of $ i.e. $ etc. = $ 7\ ! $ .Show Answer
Answer:
Solution:
BETA
