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 ________.

• A.
• B.
• C.
• D. None of these

Answer 1

The correct option is A

Since the minimum marks to any question is two, the maximum marks that can be assigned

to any questions is 16 (= 30 - 2 × 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 $x^{16}$${\left(\frac{1-x^{15}}{1-x}\right)}^8$

= the coefficient of $x^{14}$ in $(1-x{)}^{-8}.(1-x^{15}{)}^8$

= the coefficient of $x^{14}$ in

$(1+\frac{8}{1!}x+\frac{8.9}{2!}{x}^2+\frac{\mathrm{8.9.10}}{3!}{x}^3+.............)$($1{-}^8{C}_1{x}^{15}+............)$

= the coefficient $x^{14}$ in {1 + ${}^8{C}_{1}x{+}^9{C}_2{x}^2{+}^{10}{C}_3{x}^3$ + ...........}

since the second bracket has powers of $x^0,x^{15}$ etc.

= ${}^{21}{C}_{14}{=}^{21}{C}_7$