Question Number 170833 by LEKOUMA last updated on 01/Jun/22
$${Solve}\: \\ $$$$\begin{cases}{{C}_{{x}} ^{{y}} ={C}_{{x}} ^{{y}+\mathrm{1}} }\\{\mathrm{4}{C}_{{x}} ^{{y}} =\mathrm{5}{C}_{{x}} ^{{y}−\mathrm{1}} }\end{cases}\:{or}\:{C}_{{n}} ^{{k}} =\frac{{n}!}{{k}!\left({n}−{k}\right)!} \\ $$
Answered by aleks041103 last updated on 04/Jun/22
$${C}_{{x}} ^{{y}} ={C}_{{x}} ^{{y}+\mathrm{1}} \\ $$$$\frac{{x}!}{{y}!\left({x}−{y}\right)!}=\frac{{x}!}{\left({y}+\mathrm{1}\right)!\left({x}−{y}−\mathrm{1}\right)!} \\ $$$$\frac{\left({y}+\mathrm{1}\right)!}{{y}!}=\frac{\left({x}−{y}\right)!}{\left({x}−{y}−\mathrm{1}\right)!} \\ $$$${y}+\mathrm{1}={x}−{y} \\ $$$${x}−\mathrm{2}{y}=\mathrm{1}\Rightarrow{x}=\mathrm{1}+\mathrm{2}{y} \\ $$$$\mathrm{4}{C}_{{x}} ^{{y}} =\mathrm{5}{C}_{{x}} ^{{y}−\mathrm{1}} \\ $$$$\mathrm{4}\frac{{x}!}{{y}!\left({x}−{y}\right)!}=\mathrm{5}\frac{{x}!}{\left({y}−\mathrm{1}\right)!\left({x}−{y}+\mathrm{1}\right)!} \\ $$$$\mathrm{4}\frac{\left({x}−{y}+\mathrm{1}\right)!}{\left({x}−{y}\right)!}=\mathrm{5}\frac{{y}!}{\left({y}−\mathrm{1}\right)!} \\ $$$$\mathrm{4}\left({x}−{y}+\mathrm{1}\right)=\mathrm{5}{y} \\ $$$$\mathrm{9}{y}−\mathrm{4}{x}=\mathrm{1} \\ $$$$ \\ $$$$\Rightarrow\mathrm{9}{y}−\mathrm{4}\left(\mathrm{1}+\mathrm{2}{y}\right)=\mathrm{1} \\ $$$$\mathrm{9}{y}−\mathrm{4}−\mathrm{8}{y}=\mathrm{1} \\ $$$${y}=\mathrm{5}\Rightarrow{x}=\mathrm{11} \\ $$