in-how-many-ways-can-the-number-n-be-written-as-a-sum-of-three-positive-integers-if-representations-differing-in-the-order-of-the-terms-are-considered-to-be-different-

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Question Number 153220 by talminator2856791 last updated on 05/Sep/21

$$$$$$\mathrm{in}\mathrm{how}\mathrm{many}\mathrm{ways}\mathrm{can}\mathrm{the}\mathrm{number}$$$$n\mathrm{be}\mathrm{written}\mathrm{as}\mathrm{a}\mathrm{sum}\mathrm{of}\mathrm{three}\mathrm{positive}$$$$\mathrm{integers}\mathrm{if}\mathrm{representations}\mathrm{differing}$$$$\mathrm{in}\mathrm{the}\mathrm{order}\mathrm{of}\mathrm{the}\mathrm{terms}\mathrm{are}\mathrm{considered}$$$$\mathrm{to}\mathrm{be}\mathrm{different}?$$$$$$

Answered by mr W last updated on 05/Sep/21

$$a+b+c=n$$$$a⩾1,b⩾1,c⩾1$$$${(x+x^2+x^3+\dots )}^3=\frac{x^3}{{(1-x)}^3}=\underset{k=0}{\sum ^{\mathrm{\infty }}}{C}_2^{k+2}{x}^{k+3}$$$$k+3=n\Rightarrow k+1=n-1$$$$coef.of{x}^{n}is{C}_2^{n-1},thatmeansthere$$$$are{C}_2^{n-1}ways.$$$$examplen=5:$$$$C_2^4=6ways$$$$5=1+1+4=1+4+1=1+2+2=2+1+2$$$$=2+2+1=4+1+1$$

Commented by Tawa11 last updated on 06/Sep/21

$$\mathrm{great}\mathrm{sir}.$$

Commented by mr W last updated on 06/Sep/21

$$canyouconfirmthisanswerwithan$$$$othermethodyouknow?$$

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