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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

$${}^n{C}_r=\frac{n!}{(n-r)!r!}$$$${}^n{C}_{12}={}^n{C}_8\Rightarrow \frac{n!}{(n-12)!12!}=\frac{n!}{(n-8)!8!}$$$$\Rightarrow \frac{1}{(n-12)!12!}=\frac{1}{(n-8)!8!}$$$$\Rightarrow \frac{1}{(n-12)!\left(12\right)\left(11\right)\left(10\right)\left(9\right)}=\frac{1}{(n-8)(n-9)(n-10)(n-11)(n-12)!}$$$$\Rightarrow \frac{1}{12\left(11\right)\left(10\right)\left(9\right)}=\frac{1}{(n-8)(n-9)(n-10)(n-11)}$$$$12\left(11\right)\left(10\right)\left(9\right)=(n-8)(n-9)(n-10)(n-11)$$$$\Rightarrow 12=n-8\iff n=20$$$$$$

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

$$C_8^n=C_{12}^n$$$$\Rightarrow n=8+12=20$$$$$$$$note:{\mathit{C}}_{\mathit{r}}^{\mathit{n}}={\mathit{C}}_{\mathit{n}-\mathit{r}}^{\mathit{n}}$$

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

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

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

$$haveyouchecked?$$$$(20-r)=(r+2)(r+1)hasnosolution$$$$forr\in N.$$$$$$$$yourcalculationsecondlineiswrong.$$$$pleaserecheck!$$

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

$${}^{20}{C}_r={}^{20}{C}_{r+2}$$$$\Rightarrow r+(r+2)=20\Rightarrow r=9$$

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

$$\mathrm{thanks},{\mathrm{i}}^{\prime }\mathrm{ll}\mathrm{corrct}\mathrm{it}$$

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

$${}^n{P}_4={}^n{\mathrm{C}}_3\times {}^n{C}_3$$$$\mathrm{since}{}^n{P}_r=\frac{n!}{(n-r)!}$$$$\Rightarrow {}^n{P}_4=\frac{n!}{(n-4)!}=\frac{n!}{(n-3)!3!}\times \frac{n!}{(n-3)!3!}$$$$\Rightarrow \frac{1}{(n-4)!}=\frac{n(n-1)(n-2)(n-3)!}{(n-3)(n-4)!}\times \frac{1}{(n-3)!3!}$$$$\Rightarrow 1=\frac{n(n-1)(n-2)}{6(n-3)}$$$$6(n-3)=n(n-1)(n-2)$$$$\mathrm{at}\mathrm{this}\mathrm{point}\mathrm{i}\mathrm{believe}\mathrm{to}\mathrm{get}n\mathrm{will}\mathrm{be}\mathrm{easy}$$$$\mathrm{note}\mathrm{that}\mathrm{n}\in \mathbb{N}$$

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

$$buthaveyouchecked?$$$$6(n-3)=n(n-1)(n-2)hasno$$$$solutionforn\in N!$$$$$$$$thislineiswrong:$$$$\Rightarrow \frac{1}{(n-4)!}=\frac{n(n-1)(n-2)(n-3)!}{(n-3)(n-4)!}\times \frac{1}{(n-3)!3!}$$$$pleaserecheck!$$

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