If P(11, r) = P(12, r − 1) find r.

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We have, P(11, r) = P(12, r − 1). $\Rightarrow \frac{11!}{(11-\text{r})!}= \frac{12!}{\big[12-(\text{r}-1)\big]!}\ \Big[\because^\text{n}\text{p}_\text{r}=\frac{\text{n!}}{(\text{n}-\text{r})!}\Big]$ $\Rightarrow \frac{11!}{(11-\text{r})!}=\frac{12\times11!}{[12-\text{r}+1]!}$ $\Rightarrow \frac{1}{(11-\text{r})!}=\frac{12}{[13-\text{r}]!}$ $\Rightarrow \frac{1}{(11-\text{r})!}=\frac{12}{[13-\text{r}]!}$ $\Rightarrow \frac{1}{(11-\text{r})!}= \frac{12}{(13-\text{r})\times(13-\text{r}-1)(13-\text{r}-2)!}$ $\Rightarrow \frac{1}{(11-\text{r})!}=\frac{12}{(13-\text{r})\times(12-\text{r})(11-\text{r})!}$ $\Rightarrow \frac{(13-\text{r})(12-\text{r}){(11-\text{r})}}{(11-\text{r})}=12$ $\Rightarrow (13-\text{r})(12-\text{r})=12$ $\Rightarrow 156-13\text{r}-12\text{r}+\text{r}^2=12$ $\Rightarrow \text{r}^2-25\text{r}+156-12=0$ $\Rightarrow \text{r}^2-25\text{r}+144=0$ $\Rightarrow \text{r}^2-16\text{r}-9\text{r}+144=0$ $\Rightarrow \text{r}(\text{r}-16)-9(\text{r}-16)=0$ $\Rightarrow (\text{r}-9)(\text{r}-16)=0$ $\Rightarrow \text{r}-9=0\begin{bmatrix}\ \because\text{r}\ \leq\ 11 \\ \therefore \ \neq\ 16 \end{bmatrix}$ $\Rightarrow \text{r}=9$

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