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Question Number 72441 by kouidri last updated on 28/Oct/19

$$\mathrm{If}{}^n{C}_{12}={}^n{C}_8,\mathrm{then}n=$$

Commented by mathmax by abdo last updated on 29/Oct/19

$$C_n^{12}=C_n^8\iff \frac{n!}{\left(12\right)!(n-12)!}=\frac{n!}{8!(n-8)!}\Rightarrow 8!(n-8)!=12!(n-12)!\Rightarrow$$$$8!(n-8)(n-9)(n-10)(n-11)(n-12)!=12.11.10.9.8!(n-12)!\Rightarrow$$$$\Rightarrow (n-8)(n-9)(n-10)(n-11)=12.11.10.9\Rightarrow n=20$$

Answered by JDamian last updated on 28/Oct/19

$$n=20$$

Answered by $@ty@m123 last updated on 29/Oct/19

$${}^n{C}_r={}^n{C}_{n-r}$$$$ATQ$$$$\Rightarrow r=n-r+4$$$$\Rightarrow 2r=n+4$$$$Here,r=12$$$$\therefore n+4=24$$$$\Rightarrow n=20$$

Commented by JDamian last updated on 29/Oct/19

$$Really?{}^{12}{C}_{12}\ne {}^{12}{C}_8$$

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