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Question Number 57074 by Tawa1 last updated on 30/Mar/19

$$\mathrm{Solve}\mathrm{the}\mathrm{system}.$$$$\stackrel{\mathrm{x}}{}{\mathrm{C}}_{\mathrm{y}+1}=20,\stackrel{\mathrm{x}-1}{}{\mathrm{C}}_{\mathrm{y}}=10$$

Answered by mr W last updated on 30/Mar/19

$$\stackrel{\mathrm{x}-1}{}{\mathrm{C}}_{\mathrm{y}}=10$$$$\Rightarrow \frac{(x-1)!}{(x-1-y)!y!}=10$$$$$$$$\stackrel{\mathrm{x}}{}{\mathrm{C}}_{\mathrm{y}+1}=20$$$$\Rightarrow \frac{x!}{(x-y-1)!(y+1)!}=20$$$$\Rightarrow \frac{x(x-1)!}{(x-1-y)!(y+1)y!}=20$$$$\Rightarrow \frac{x}{(y+1)}\times 10=20$$$$\Rightarrow \frac{x}{y+1}=2$$$$\Rightarrow x=2y+2$$$$$$$$\Rightarrow \frac{(2y+1)!}{(y+1)!y!}=10$$$$\Rightarrow \frac{(y+y+1)!}{(y+1)!y!}=10$$$$\Rightarrow \frac{(y+1+1)(y+1+2)...(y+1+y)}{1\times 2\times ...y}=10$$$$=(\frac{y+1}{1}+1)(\frac{y+1}{2}+1)...(\frac{y+1}{y}+1)=10$$$$sinceeachtermis>2,wemayhave$$$$atmost3termstogetproduct10,$$$$i.e.y⩽3.$$$$withy=3:\Rightarrow x=8$$$$\stackrel{\mathrm{x}-1}{}{\mathrm{C}}_{\mathrm{y}}={}^7{C}_3=35\ne 10$$$$withy=2:\Rightarrow x=6$$$$\stackrel{\mathrm{x}-1}{}{\mathrm{C}}_{\mathrm{y}}={}^5{C}_2=10\Rightarrow ok$$$$$$$$\Rightarrow solutionisx=6,y=2$$

Commented by Tawa1 last updated on 30/Mar/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{I}\mathrm{appreciate}.$$

Commented by peter frank last updated on 30/Mar/19

$$nicework$$

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