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Question Number 197578 by mr W last updated on 22/Sep/23

Commented by mr W last updated on 22/Sep/23

$$\text{You can't use 'macro parameter character \#' in math mode}$$

Answered by mr W last updated on 22/Sep/23

$$1)onedigitnumbers:$$$$8\Rightarrow 1number$$$$$$$$2)2digitnumbers:$$$$17,26,35,44,53,62,71,80\Rightarrow 8numbers$$$$$$$$3)3digitnumbers:$$$$abcwith$$$$a=1,2,3,...9$$$$b,c=0,1,2,3,...9$$$$(x+x^2+x^3+...){(1+x+x^2+x^3+...)}^2$$$$=\frac{x}{{(1-x)}^3}=x\underset{k=0}{\sum ^{\mathrm{\infty }}}{C}_2^{k+2}{x}^k$$$$coef.of{x}^{8}is:$$$$C_2^{7+2}=36\Rightarrow 36numbers$$$$$$$$4)4digitnumbers:$$$$abcdwith$$$$a=1,2,3,4,5$$$$b,c,d=0,1,2,3,...9$$$$(x+x^2+x^3+x^4+x^5){(1+x+x^2+x^3+...)}^3$$$$=\frac{x(1-x^5)}{{(1-x)}^4}=x(1-x^5)\underset{k=0}{\sum ^{\mathrm{\infty }}}{C}_3^{k+3}{x}^k$$$$coef.of{x}^{8}is:$$$$C_3^{7+3}-C_3^{2+3}=110\Rightarrow 110numbers$$$$$$$$totally:$$$$1+8+36+110=155numbers$$$$\Rightarrow answerA$$

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