Question Number 72441 by kouidri last updated on 28/Oct/19
$$\mathrm{If}\:^{{n}} {C}_{\mathrm{12}} =\:^{{n}} {C}_{\mathrm{8}} \:,\:\mathrm{then}\:{n}= \\ $$
Commented by mathmax by abdo last updated on 29/Oct/19
$${C}_{{n}} ^{\mathrm{12}} ={C}_{{n}} ^{\mathrm{8}} \:\Leftrightarrow\frac{{n}!}{\left(\mathrm{12}\right)!\left({n}−\mathrm{12}\right)!}\:=\frac{{n}!}{\mathrm{8}!\left({n}−\mathrm{8}\right)!}\:\Rightarrow\mathrm{8}!\left({n}−\mathrm{8}\right)!=\mathrm{12}!\left({n}−\mathrm{12}\right)!\:\Rightarrow \\ $$$$\mathrm{8}!\left({n}−\mathrm{8}\right)\left({n}−\mathrm{9}\right)\left({n}−\mathrm{10}\right)\left({n}−\mathrm{11}\right)\left({n}−\mathrm{12}\right)!=\mathrm{12}.\mathrm{11}.\mathrm{10}.\mathrm{9}.\mathrm{8}!\left({n}−\mathrm{12}\right)!\Rightarrow \\ $$$$\Rightarrow\left({n}−\mathrm{8}\right)\left({n}−\mathrm{9}\right)\left({n}−\mathrm{10}\right)\left({n}−\mathrm{11}\right)=\mathrm{12}.\mathrm{11}.\mathrm{10}.\mathrm{9}\:\Rightarrow{n}=\mathrm{20} \\ $$
Answered by JDamian last updated on 28/Oct/19
$${n}=\mathrm{20} \\ $$
Answered by $@ty@m123 last updated on 29/Oct/19
$$\:^{{n}} {C}_{{r}} =\:^{{n}} {C}_{{n}−{r}} \\ $$$${ATQ} \\ $$$$\Rightarrow{r}={n}−{r}+\mathrm{4} \\ $$$$\Rightarrow\mathrm{2}{r}={n}+\mathrm{4} \\ $$$${Here},\:{r}=\mathrm{12} \\ $$$$\therefore\:{n}+\mathrm{4}=\mathrm{24} \\ $$$$\Rightarrow{n}=\mathrm{20} \\ $$
Commented by JDamian last updated on 29/Oct/19
$${Really}?\:\:\:\:^{\mathrm{12}} {C}_{\mathrm{12}} \neq\:^{\mathrm{12}} {C}_{\mathrm{8}} \: \\ $$