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If-n-C-12-n-C-8-then-n-




Question Number 72441 by kouidri last updated on 28/Oct/19
If^n C_(12) =^n C_8  , then n=
$$\mathrm{If}\:^{{n}} {C}_{\mathrm{12}} =\:^{{n}} {C}_{\mathrm{8}} \:,\:\mathrm{then}\:{n}= \\ $$
Commented by mathmax by abdo last updated on 29/Oct/19
C_n ^(12) =C_n ^8  ⇔((n!)/((12)!(n−12)!)) =((n!)/(8!(n−8)!)) ⇒8!(n−8)!=12!(n−12)! ⇒  8!(n−8)(n−9)(n−10)(n−11)(n−12)!=12.11.10.9.8!(n−12)!⇒  ⇒(n−8)(n−9)(n−10)(n−11)=12.11.10.9 ⇒n=20
$${C}_{{n}} ^{\mathrm{12}} ={C}_{{n}} ^{\mathrm{8}} \:\Leftrightarrow\frac{{n}!}{\left(\mathrm{12}\right)!\left({n}−\mathrm{12}\right)!}\:=\frac{{n}!}{\mathrm{8}!\left({n}−\mathrm{8}\right)!}\:\Rightarrow\mathrm{8}!\left({n}−\mathrm{8}\right)!=\mathrm{12}!\left({n}−\mathrm{12}\right)!\:\Rightarrow \\ $$$$\mathrm{8}!\left({n}−\mathrm{8}\right)\left({n}−\mathrm{9}\right)\left({n}−\mathrm{10}\right)\left({n}−\mathrm{11}\right)\left({n}−\mathrm{12}\right)!=\mathrm{12}.\mathrm{11}.\mathrm{10}.\mathrm{9}.\mathrm{8}!\left({n}−\mathrm{12}\right)!\Rightarrow \\ $$$$\Rightarrow\left({n}−\mathrm{8}\right)\left({n}−\mathrm{9}\right)\left({n}−\mathrm{10}\right)\left({n}−\mathrm{11}\right)=\mathrm{12}.\mathrm{11}.\mathrm{10}.\mathrm{9}\:\Rightarrow{n}=\mathrm{20} \\ $$
Answered by JDamian last updated on 28/Oct/19
n=20
$${n}=\mathrm{20} \\ $$
Answered by $@ty@m123 last updated on 29/Oct/19
^n C_r =^n C_(n−r)   ATQ  ⇒r=n−r+4  ⇒2r=n+4  Here, r=12  ∴ n+4=24  ⇒n=20
$$\:^{{n}} {C}_{{r}} =\:^{{n}} {C}_{{n}−{r}} \\ $$$${ATQ} \\ $$$$\Rightarrow{r}={n}−{r}+\mathrm{4} \\ $$$$\Rightarrow\mathrm{2}{r}={n}+\mathrm{4} \\ $$$${Here},\:{r}=\mathrm{12} \\ $$$$\therefore\:{n}+\mathrm{4}=\mathrm{24} \\ $$$$\Rightarrow{n}=\mathrm{20} \\ $$
Commented by JDamian last updated on 29/Oct/19
Really?   ^(12) C_(12) ≠^(12) C_8
$${Really}?\:\:\:\:^{\mathrm{12}} {C}_{\mathrm{12}} \neq\:^{\mathrm{12}} {C}_{\mathrm{8}} \: \\ $$

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