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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

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

Commented by Tony Lin last updated on 15/Feb/20

first, give 2 marks to each question and there are 8 questions so we still have 14 marks x_1 +x_2 +x_3 +x_4 +x_5 +x_6 +x_7 +x_8 =14 actually, this kind of question we can change to the combination of +&∣ there are 7 + which divide 14∣ into 8 parts so the answer ((21!)/(7!14!))=116280 or we can use H_(14) ^8 =C_(14) ^(8+14−1) =C_7 ^(21) =116280 if the examiner assigned less than 30 marks(including 30) then x_1 +x_2 +x_3 +x_4 +x_5 +x_6 +x_7 +x_8 ≤14 ⇒put one trash can 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

$${first},\:{give}\:\mathrm{2}\:{marks}\:{to}\:{each}\:{question} \\ $$$${and}\:{there}\:{are}\:\mathrm{8}\:{questions} \\ $$$${so}\:{we}\:{still}\:{have}\:\mathrm{14}\:{marks} \\ $$$${x}_{\mathrm{1}} +{x}_{\mathrm{2}} +{x}_{\mathrm{3}} +{x}_{\mathrm{4}} +{x}_{\mathrm{5}} +{x}_{\mathrm{6}} +{x}_{\mathrm{7}} +{x}_{\mathrm{8}} =\mathrm{14} \\ $$$${actually},\:{this}\:{kind}\:{of}\:{question}\:{we} \\ $$$${can}\:{change}\:{to}\:{the}\:{combination}\:{of} \\ $$$$+\&\mid \\ $$$${there}\:{are}\:\mathrm{7}\:+\:{which}\:{divide}\:\mathrm{14}\mid \\ $$$${into}\:\mathrm{8}\:{parts} \\ $$$${so}\:\:{the}\:{answer}\:\frac{\mathrm{21}!}{\mathrm{7}!\mathrm{14}!}=\mathrm{116280} \\ $$$${or}\:{we}\:{can}\:{use}\:{H}_{\mathrm{14}} ^{\mathrm{8}} ={C}_{\mathrm{14}} ^{\mathrm{8}+\mathrm{14}−\mathrm{1}} ={C}_{\mathrm{7}} ^{\mathrm{21}} =\mathrm{116280} \\ $$$${if}\:{the}\:{examiner}\:{assigned}\:{less}\:{than} \\ $$$$\mathrm{30}\:{marks}\left({including}\:\mathrm{30}\right) \\ $$$${then}\:{x}_{\mathrm{1}} +{x}_{\mathrm{2}} +{x}_{\mathrm{3}} +{x}_{\mathrm{4}} +{x}_{\mathrm{5}} +{x}_{\mathrm{6}} +{x}_{\mathrm{7}} +{x}_{\mathrm{8}} \leqslant\mathrm{14} \\ $$$$\Rightarrow{put}\:{one}\:{trash}\:{can} \\ $$$${x}_{\mathrm{1}} +{x}_{\mathrm{2}} +{x}_{\mathrm{3}} +{x}_{\mathrm{4}} +{x}_{\mathrm{5}} +{x}_{\mathrm{6}} +{x}_{\mathrm{7}} +{x}_{\mathrm{8}} +{t}=\mathrm{14} \\ $$$${H}_{\mathrm{14}} ^{\mathrm{9}} ={C}_{\mathrm{14}} ^{\mathrm{9}+\mathrm{14}−\mathrm{1}} ={C}_{\mathrm{8}} ^{\mathrm{22}} =\mathrm{319770} \\ $$

Commented by mr W last updated on 15/Feb/20

(x^2 +x^3 +x^4 +....)^8 =(x^(16) /((1−x)^8 ))=x^(16) Σ_(k=0) ^∞ C_7 ^(7+k) x^k coefficient of x^(30) =C_7 ^(7+14) =C_7 ^(21) =116280 ⇒there are 116280 ways.

$$\left({x}^{\mathrm{2}} +{x}^{\mathrm{3}} +{x}^{\mathrm{4}} +....\right)^{\mathrm{8}} =\frac{{x}^{\mathrm{16}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{8}} }={x}^{\mathrm{16}} \underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{7}} ^{\mathrm{7}+{k}} {x}^{{k}} \\ $$$${coefficient}\:{of}\:{x}^{\mathrm{30}} ={C}_{\mathrm{7}} ^{\mathrm{7}+\mathrm{14}} ={C}_{\mathrm{7}} ^{\mathrm{21}} =\mathrm{116280} \\ $$$$\Rightarrow{there}\:{are}\:\mathrm{116280}\:{ways}. \\ $$

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