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Question Number 84261 by 698148290 last updated on 10/Mar/20

Answered by Rio Michael last updated on 10/Mar/20

^n C_r = ((n!)/((n−r)!r!)) ^n C_(12) =^n C_8 ⇒ ((n!)/((n−12)!12!)) = ((n!)/((n−8)!8!)) ⇒ (1/((n−12)!12!)) = (1/((n−8)!8!)) ⇒ (1/((n−12)!(12)(11)(10)(9))) = (1/((n−8)(n−9)(n−10)(n−11)(n−12)!)) ⇒ (1/(12(11)(10)(9))) = (1/((n−8)(n−9)(n−10)(n−11))) 12(11)(10)(9) = (n−8)(n−9)(n−10)(n−11) ⇒ 12 = n−8 ⇔ n = 20

$$\:^{{n}} {C}_{{r}} =\:\frac{{n}!}{\left({n}−{r}\right)!{r}!} \\ $$$$\:^{{n}} {C}_{\mathrm{12}} \:=\:^{{n}} {C}_{\mathrm{8}} \:\Rightarrow\:\:\frac{{n}!}{\left({n}−\mathrm{12}\right)!\mathrm{12}!}\:=\:\frac{{n}!}{\left({n}−\mathrm{8}\right)!\mathrm{8}!} \\ $$$$\:\Rightarrow\:\frac{\mathrm{1}}{\left({n}−\mathrm{12}\right)!\mathrm{12}!}\:=\:\frac{\mathrm{1}}{\left({n}−\mathrm{8}\right)!\mathrm{8}!} \\ $$$$\Rightarrow\:\:\frac{\mathrm{1}}{\left({n}−\mathrm{12}\right)!\left(\mathrm{12}\right)\left(\mathrm{11}\right)\left(\mathrm{10}\right)\left(\mathrm{9}\right)}\:=\:\frac{\mathrm{1}}{\left({n}−\mathrm{8}\right)\left({n}−\mathrm{9}\right)\left({n}−\mathrm{10}\right)\left({n}−\mathrm{11}\right)\left({n}−\mathrm{12}\right)!} \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{\mathrm{12}\left(\mathrm{11}\right)\left(\mathrm{10}\right)\left(\mathrm{9}\right)}\:=\:\frac{\mathrm{1}}{\left({n}−\mathrm{8}\right)\left({n}−\mathrm{9}\right)\left({n}−\mathrm{10}\right)\left({n}−\mathrm{11}\right)} \\ $$$$\mathrm{12}\left(\mathrm{11}\right)\left(\mathrm{10}\right)\left(\mathrm{9}\right)\:=\:\left({n}−\mathrm{8}\right)\left({n}−\mathrm{9}\right)\left({n}−\mathrm{10}\right)\left({n}−\mathrm{11}\right) \\ $$$$\Rightarrow\:\mathrm{12}\:=\:{n}−\mathrm{8}\:\Leftrightarrow\:{n}\:=\:\mathrm{20}\: \\ $$$$ \\ $$

Commented by mr W last updated on 11/Mar/20

C_8 ^n =C_(12) ^n ⇒n=8+12=20 note: C_r ^n =C_(n−r) ^n

$${C}_{\mathrm{8}} ^{{n}} ={C}_{\mathrm{12}} ^{{n}} \\ $$$$\Rightarrow{n}=\mathrm{8}+\mathrm{12}=\mathrm{20} \\ $$$$ \\ $$$${note}:\:\boldsymbol{{C}}_{\boldsymbol{{r}}} ^{\boldsymbol{{n}}} =\boldsymbol{{C}}_{\boldsymbol{{n}}−\boldsymbol{{r}}} ^{\boldsymbol{{n}}} \\ $$

Answered by Rio Michael last updated on 10/Mar/20

^(20) C_r =^(20) C_(r+2) ⇒ ((20!)/((20−r)!r!)) = ((20!)/((20−(r+1))!(r+2)!)) ⇒ (20−r)!r! = (19−r)!(r+2)(r+1)r! ⇒ (20−r)! = (19−r)! (r + 2)(r+1) ⇒ (20−r)(19−r)!= (19−r)!(r+2)(r+1) (20−r) = (r + 2)(r + 1) am sure you can take it from here

$$\:^{\mathrm{20}} {C}_{{r}} \:=\:^{\mathrm{20}} {C}_{{r}+\mathrm{2}} \\ $$$$\:\:\Rightarrow\:\:\frac{\mathrm{20}!}{\left(\mathrm{20}−{r}\right)!{r}!}\:=\:\frac{\mathrm{20}!}{\left(\mathrm{20}−\left({r}+\mathrm{1}\right)\right)!\left({r}+\mathrm{2}\right)!} \\ $$$$\Rightarrow\:\left(\mathrm{20}−{r}\right)!{r}!\:=\:\left(\mathrm{19}−{r}\right)!\left({r}+\mathrm{2}\right)\left({r}+\mathrm{1}\right){r}! \\ $$$$\Rightarrow\:\left(\mathrm{20}−{r}\right)!\:=\:\left(\mathrm{19}−{r}\right)!\:\left({r}\:+\:\mathrm{2}\right)\left({r}+\mathrm{1}\right) \\ $$$$\Rightarrow\:\left(\mathrm{20}−{r}\right)\left(\mathrm{19}−{r}\right)!=\:\left(\mathrm{19}−{r}\right)!\left({r}+\mathrm{2}\right)\left({r}+\mathrm{1}\right) \\ $$$$\left(\mathrm{20}−{r}\right)\:=\:\left({r}\:+\:\mathrm{2}\right)\left({r}\:+\:\mathrm{1}\right)\: \\ $$$$\mathrm{am}\:\mathrm{sure}\:\mathrm{you}\:\mathrm{can}\:\mathrm{take}\:\mathrm{it}\:\mathrm{from}\:\mathrm{here}\: \\ $$

Commented by mr W last updated on 11/Mar/20

have you checked? (20−r) = (r + 2)(r + 1) has no solution for r∈N. your calculation second line is wrong. please recheck!

$${have}\:{you}\:{checked}? \\ $$$$\left(\mathrm{20}−{r}\right)\:=\:\left({r}\:+\:\mathrm{2}\right)\left({r}\:+\:\mathrm{1}\right)\:{has}\:{no}\:{solution} \\ $$$${for}\:{r}\in{N}. \\ $$$$ \\ $$$${your}\:{calculation}\:{second}\:{line}\:{is}\:{wrong}. \\ $$$${please}\:{recheck}! \\ $$

Commented by mr W last updated on 11/Mar/20

^(20) C_r =^(20) C_(r+2) ⇒r+(r+2)=20 ⇒r=9

$$\:^{\mathrm{20}} {C}_{{r}} \:=\:^{\mathrm{20}} {C}_{{r}+\mathrm{2}} \\ $$$$\Rightarrow{r}+\left({r}+\mathrm{2}\right)=\mathrm{20}\:\Rightarrow{r}=\mathrm{9} \\ $$

Commented by Rio Michael last updated on 11/Mar/20

thanks,i′ll corrct it

$$\mathrm{thanks},\mathrm{i}'\mathrm{ll}\:\mathrm{corrct}\:\mathrm{it} \\ $$

Answered by Rio Michael last updated on 10/Mar/20

^n P_4 =^n C_3 ×^n C_3 since ^n P_r = ((n!)/((n−r)!)) ⇒ ^n P_4 = ((n!)/((n−4)!)) = ((n!)/((n−3)!3!)) × ((n!)/((n−3)!3!)) ⇒ (1/((n−4)!)) = ((n(n−1)(n−2)(n−3)!)/((n−3)(n−4)!)) × (1/((n−3)!3!)) ⇒ 1 = ((n(n−1)(n−2))/(6(n−3))) 6(n−3) = n(n−1)(n−2) at this point i believe to get n will be easy note that n∈ N

$$\:^{{n}} {P}_{\mathrm{4}} \:=\:^{{n}} \mathrm{C}_{\mathrm{3}} \:×\:^{{n}} {C}_{\mathrm{3}} \\ $$$$\:\mathrm{since}\:\:^{{n}} {P}_{{r}} \:=\:\frac{{n}!}{\left({n}−{r}\right)!} \\ $$$$\:\Rightarrow\:\:^{{n}} {P}_{\mathrm{4}} \:=\:\frac{{n}!}{\left({n}−\mathrm{4}\right)!}\:=\:\frac{{n}!}{\left({n}−\mathrm{3}\right)!\mathrm{3}!}\:×\:\frac{{n}!}{\left({n}−\mathrm{3}\right)!\mathrm{3}!} \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{\left({n}−\mathrm{4}\right)!}\:=\:\frac{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)\left({n}−\mathrm{3}\right)!}{\left({n}−\mathrm{3}\right)\left({n}−\mathrm{4}\right)!}\:×\:\frac{\mathrm{1}}{\left({n}−\mathrm{3}\right)!\mathrm{3}!} \\ $$$$\Rightarrow\:\mathrm{1}\:=\:\frac{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)}{\mathrm{6}\left({n}−\mathrm{3}\right)} \\ $$$$\:\:\:\:\:\:\mathrm{6}\left({n}−\mathrm{3}\right)\:=\:{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right) \\ $$$$\mathrm{at}\:\mathrm{this}\:\mathrm{point}\:\mathrm{i}\:\mathrm{believe}\:\mathrm{to}\:\mathrm{get}\:{n}\:\mathrm{will}\:\mathrm{be}\:\mathrm{easy} \\ $$$$\mathrm{note}\:\mathrm{that}\:\mathrm{n}\in\:\mathbb{N} \\ $$

Commented by mr W last updated on 11/Mar/20

but have you checked? 6(n−3) = n(n−1)(n−2) has no solution for n∈N! this line is wrong: ⇒ (1/((n−4)!)) = ((n(n−1)(n−2)(n−3)!)/((n−3)(n−4)!)) × (1/((n−3)!3!)) please recheck!

$${but}\:{have}\:{you}\:{checked}? \\ $$$$\mathrm{6}\left({n}−\mathrm{3}\right)\:=\:{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)\:{has}\:{no} \\ $$$${solution}\:{for}\:{n}\in{N}! \\ $$$$ \\ $$$${this}\:{line}\:{is}\:{wrong}: \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{\left({n}−\mathrm{4}\right)!}\:=\:\frac{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)\left({n}−\mathrm{3}\right)!}{\left({n}−\mathrm{3}\right)\left({n}−\mathrm{4}\right)!}\:×\:\frac{\mathrm{1}}{\left({n}−\mathrm{3}\right)!\mathrm{3}!} \\ $$$${please}\:{recheck}! \\ $$

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