Find n if 55.nP6 = 2 (n+2)P7
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$$\begin{gathered} \mathrm{We} \mathrm{have}: \\ 55{.}^{\mathrm{n}}{\mathrm{P}}_6=2{.}^{\mathrm{n}+2}{\mathrm{P}}_7 \\ \mathrm{As}{ }^{\mathrm{n}}{\mathrm{P}}_{\mathrm{r}}=\frac{\mathrm{n}!}{\left(\mathrm{n}-\mathrm{r}\right)!}, \mathrm{hence} \mathrm{we} \mathrm{get}: \\ 55.\frac{\mathrm{n}!}{\left(\mathrm{n}-6\right)!}=2.\frac{\left(\mathrm{n}+2\right)!}{\left(\mathrm{n}+2-7\right)!} \\ \Rightarrow 55.\frac{\mathrm{n}!}{\left(\mathrm{n}-6\right)!}=2.\frac{\left(\mathrm{n}+2\right)!}{\left(\mathrm{n}-5\right)!} \\ \Rightarrow 55.\frac{\mathrm{n}!}{\left(\mathrm{n}-6\right)!}=2.\frac{\left(\mathrm{n}+2\right)\left(\mathrm{n}+1\right)\mathrm{n}!}{\left(\mathrm{n}-5\right)\left(\mathrm{n}-6\right)!} \\ \Rightarrow 55=2.\frac{{\mathrm{n}}^2+3\mathrm{n}+2}{\mathrm{n}-5} \\ \Rightarrow 2{\mathrm{n}}^2+6\mathrm{n}+4=55\mathrm{n}-275 \\ \Rightarrow 2{\mathrm{n}}^2-49\mathrm{n}+279=0 \\ \Rightarrow 2{\mathrm{n}}^2-18\mathrm{n}-31\mathrm{n}+279=0 \\ \Rightarrow 2\mathrm{n}\left(\mathrm{n}-9\right)-31\left(\mathrm{n}-9\right)=0 \\ \Rightarrow \left(2\mathrm{n}-31\right)\left(\mathrm{n}-9\right)=0 \\ \mathrm{As} \mathrm{n} \mathrm{should} \mathrm{be} \mathrm{a} \mathrm{natural} \mathrm{number}, \mathrm{hence} \\ \mathbf{n}\mathbf{=}\mathbf{9} \end{gathered}$$

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