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

Solve the system. ^x C_(y + 1) = 20, ^(x − 1) C_y = 10

$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{system}. \\ $$$$\:\:\:\:\:\:\overset{\mathrm{x}} {\:}\mathrm{C}_{\mathrm{y}\:+\:\mathrm{1}} \:\:=\:\:\mathrm{20},\:\:\:\:\:\:\:\:\:\:\:\overset{\mathrm{x}\:−\:\mathrm{1}} {\:}\mathrm{C}_{\mathrm{y}} \:\:=\:\:\mathrm{10} \\ $$

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

^(x − 1) C_y = 10 ⇒(((x−1)!)/((x−1−y)!y!))=10 ^x C_(y + 1) = 20 ⇒((x!)/((x−y−1)!(y+1)!))=20 ⇒((x(x−1)!)/((x−1−y)!(y+1)y!))=20 ⇒(x/((y+1)))×10=20 ⇒(x/(y+1))=2 ⇒x=2y+2 ⇒(((2y+1)!)/((y+1)!y!))=10 ⇒(((y+y+1)!)/((y+1)!y!))=10 ⇒(((y+1+1)(y+1+2)...(y+1+y))/(1×2×...y))=10 =(((y+1)/1)+1)(((y+1)/2)+1)...(((y+1)/y)+1)=10 since each term is >2, we may have at most 3 terms to get product 10, i.e. y≤3. with y=3: ⇒x=8 ^(x − 1) C_y = ^7 C_3 =35≠10 with y=2: ⇒x=6 ^(x − 1) C_y = ^5 C_2 =10 ⇒ok ⇒solution is x=6, y=2

$$\:\overset{\mathrm{x}\:−\:\mathrm{1}} {\:}\mathrm{C}_{\mathrm{y}} \:\:=\:\:\mathrm{10} \\ $$$$\Rightarrow\frac{\left({x}−\mathrm{1}\right)!}{\left({x}−\mathrm{1}−{y}\right)!{y}!}=\mathrm{10} \\ $$$$ \\ $$$$\:\overset{\mathrm{x}} {\:}\mathrm{C}_{\mathrm{y}\:+\:\mathrm{1}} \:\:=\:\:\mathrm{20} \\ $$$$\Rightarrow\frac{{x}!}{\left({x}−{y}−\mathrm{1}\right)!\left({y}+\mathrm{1}\right)!}=\mathrm{20} \\ $$$$\Rightarrow\frac{{x}\left({x}−\mathrm{1}\right)!}{\left({x}−\mathrm{1}−{y}\right)!\left({y}+\mathrm{1}\right){y}!}=\mathrm{20} \\ $$$$\Rightarrow\frac{{x}}{\left({y}+\mathrm{1}\right)}×\mathrm{10}=\mathrm{20} \\ $$$$\Rightarrow\frac{{x}}{{y}+\mathrm{1}}=\mathrm{2} \\ $$$$\Rightarrow{x}=\mathrm{2}{y}+\mathrm{2} \\ $$$$ \\ $$$$\Rightarrow\frac{\left(\mathrm{2}{y}+\mathrm{1}\right)!}{\left({y}+\mathrm{1}\right)!{y}!}=\mathrm{10} \\ $$$$\Rightarrow\frac{\left({y}+{y}+\mathrm{1}\right)!}{\left({y}+\mathrm{1}\right)!{y}!}=\mathrm{10} \\ $$$$\Rightarrow\frac{\left({y}+\mathrm{1}+\mathrm{1}\right)\left({y}+\mathrm{1}+\mathrm{2}\right)...\left({y}+\mathrm{1}+{y}\right)}{\mathrm{1}×\mathrm{2}×...{y}}=\mathrm{10} \\ $$$$=\left(\frac{{y}+\mathrm{1}}{\mathrm{1}}+\mathrm{1}\right)\left(\frac{{y}+\mathrm{1}}{\mathrm{2}}+\mathrm{1}\right)...\left(\frac{{y}+\mathrm{1}}{{y}}+\mathrm{1}\right)=\mathrm{10} \\ $$$${since}\:{each}\:{term}\:{is}\:>\mathrm{2},\:{we}\:{may}\:{have} \\ $$$${at}\:{most}\:\mathrm{3}\:{terms}\:{to}\:{get}\:{product}\:\mathrm{10}, \\ $$$${i}.{e}.\:{y}\leqslant\mathrm{3}. \\ $$$${with}\:{y}=\mathrm{3}:\:\Rightarrow{x}=\mathrm{8} \\ $$$$\overset{\mathrm{x}\:−\:\mathrm{1}} {\:}\mathrm{C}_{\mathrm{y}} \:\:=\:\:^{\mathrm{7}} {C}_{\mathrm{3}} =\mathrm{35}\neq\mathrm{10} \\ $$$${with}\:{y}=\mathrm{2}:\:\Rightarrow{x}=\mathrm{6} \\ $$$$\overset{\mathrm{x}\:−\:\mathrm{1}} {\:}\mathrm{C}_{\mathrm{y}} \:\:=\:\:^{\mathrm{5}} {C}_{\mathrm{2}} =\mathrm{10}\:\Rightarrow{ok} \\ $$$$ \\ $$$$\Rightarrow{solution}\:{is}\:{x}=\mathrm{6},\:{y}=\mathrm{2} \\ $$

Commented by Tawa1 last updated on 30/Mar/19

God bless you sir. I appreciate.

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

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

nice work

$${nice}\:{work} \\ $$

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