If P(n, 5) : P(n, 3) = 2 : 1, find n.

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We have,P(n, 5) : P(n, 3) = 2 : 1
$\Rightarrow \frac{\text{p}(\text{n},5)}{\text{p}(\text{n},3)}=\frac{2}{1}$ $\Rightarrow\frac{\frac{\text{(n)!}}{(\text{(n-5)!})}}{\frac{\text{n}!}{\text{(n-3)!}}}=\frac{2}{1}$ $\Rightarrow \frac{\text{n}!\times\text{(n-3)!}}{\text{(n-5)!}\times\text{n}!}=2$ $\Rightarrow \frac{\text{(n-3)!}}{\text{(n-5)}!}=2$ $\Rightarrow \frac{(\text{n}-3)(\text{n}-4)(\text{n}-5)!}{(\text{n}-5)!}= 2$ $\Rightarrow (\text{n}-3)(\text{n}-4) =2$ $\Rightarrow \text{n}^2-4\text{n}-3\text{n}+12=2$ $\Rightarrow \text{n}^2+7\text{n}+12=2$ $\Rightarrow \text{n}^2+7\text{n}+12-2=0$ $\Rightarrow \text{n}^2+7\text{n}+10=0$ $\Rightarrow \text{n}^2-5\text{n}-2\text{n}+10=0$ $\Rightarrow \text{n}(\text{n}-5)-2(\text{n}-5)= 0$ $\Rightarrow (\text{n}-5)(\text{n}-2) =0$ $\Rightarrow \text{n}=5 \ \begin{bmatrix}\ \because\text{r}\ \geq\ 5 \\ \therefore \ \neq\ 2 \end{bmatrix}$ Hence, $\text{n}=5$

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