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-

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} + \dots .)}^{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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