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Question Number 50861 by majidedalt last updated on 21/Dec/18

$$\frac{\mathrm{n}!}{(\mathrm{n}-5)!}=20\frac{\mathrm{n}!}{(\mathrm{n}-3)!}$$$$$$$$\mathrm{n}=?$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}$$$$\mathrm{please}\mathrm{give}\mathrm{me}\mathrm{simple}\mathrm{solve}.$$$$\mathrm{thanks}$$

Answered by afachri last updated on 21/Dec/18

$$\frac{n(n-1)(n-2)(n-3)(n-4)(n-5)!}{(n-5)!}=\frac{20.n(n-1)(n-2)(n-3)!}{(n-3)!}$$$$n(n-1)(n-2)(n-3)(n-4)=20n(n-1)(n-2)$$$$(n-3)(n-4)=20$$$$n^2-7n+12-20=0$$$$n^2-7n-8=0$$$$(n-8)(n+1)=0$$$$n_1=8\mathrm{and}{n}_2=-1$$$$\mathrm{since}\mathrm{factorial}\mathrm{is}\mathrm{possitive}\mathrm{function},\mathrm{the}$$$$\mathrm{solution}\mathrm{is}\mathrm{only}n=8.$$$$$$

Commented by majidedalt last updated on 21/Dec/18

$$\mathrm{thanks}\mathrm{dear}$$

Answered by afachri last updated on 21/Dec/18

$$\frac{(n-3)!}{(n-5)!}=\frac{20n!}{n!}$$$$\frac{(n-3)(n-4)(n-5)!}{(n-5)!}=20$$$$(n-3)(n-4)=20$$$$n^2-7n+12-20=0$$$$n^2-7n-8=0$$$$(n-8)(n+1)=0\Rightarrow {\mathit{n}}_1=8;{\mathit{n}}_2=-1$$$$\mathbf{the}\mathbf{solution}\mathbf{is}\mathbf{only}n=8.\mathbf{because}\mathbf{factorial}$$$$\mathbf{of}\mathbf{negative}\mathbf{number}{\mathbf{doesn}}^{\prime }\mathbf{t}\mathbf{work}.$$

Commented by majidedalt last updated on 21/Dec/18

$$\mathrm{very}\mathrm{thanks}$$

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