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


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Question Number 81724 by zainal tanjung last updated on 14/Feb/20

$The number of ways in which an$ $examiner can assign 30 marks to 8$ $questions, giving not less than 2 marks$ $to any question is$
Commented by Tony Lin last updated on 15/Feb/20

$f i r s t, g i v e 2 m a r k s t o e a c h q u e s t i o n$ $a n d t h e r e a r e 8 q u e s t i o n s$ $s o w e s t i l l h a v e 14 m a r k s$ $x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} + x_{7} + x_{8} = 14$ $a c t u a l l y, t h i s k i n d o f q u e s t i o n w e$ $c a n c h a n g e t o t h e c o m b i n a t i o n o f$ $+ & ∣$ $t h e r e a r e 7 + w h i c h d i v i d e 14 ∣$ $i n t o 8 p a r t s$ $s o t h e a n s w e r \frac{21!}{7! 14!} = 116280$ $o r w e c a n u s e H_{14}^{8} = C_{14}^{8 + 14 - 1} = C_{7}^{21} = 116280$ $i f t h e e x a m i n e r a s s i g n e d l e s s t h a n$ $30 m a r k s (i n c l u d i n g 30)$ $t h e n x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} + x_{7} + x_{8} ⩽ 14$ $\Rightarrow p u t o n e t r a s h c a n$ $x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} + x_{7} + x_{8} + t = 14$ $H_{14}^{9} = C_{14}^{9 + 14 - 1} = C_{8}^{22} = 319770$
Commented by mr W last updated on 15/Feb/20

${(x^{2} + x^{3} + x^{4} + . . . .)}^{8} = \frac{x^{16}}{{(1 - x)}^{8}} = x^{16} \underset{k = 0}{\sum^{\infty}} C_{7}^{7 + k} x^{k}$ $c o e f f i c i e n t o f x^{30} = C_{7}^{7 + 14} = C_{7}^{21} = 116280$ $\Rightarrow t h e r e a r e 116280 w a y s .$
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