281.

If $8!\left(\frac{1}{3!} + \frac{5}{4!}\right) = {}_9\mathrm{P}_{\mathrm{r}}$, then the value of r is

• 4

• 5

• 3

• 2

$$\begin{gathered} \mathrm{We} \mathrm{have}, 8!\left[\frac{1}{3!} + \frac{5}{4!}\right] = {}_9\mathrm{P}_{\mathrm{r}} \\ \Rightarrow 8!\left[\frac{1}{3!} + \frac{5}{4\left(3!\right)}\right] = {}_9\mathrm{P}_{\mathrm{r}} \\ \Rightarrow \frac{8!}{3!}\left(1 + \frac{5}{4}\right) = {}_9\mathrm{P}_{\mathrm{r}} \\ \Rightarrow \frac{8! 9}{3! 4} = {}_9\mathrm{P}_{\mathrm{r}} \\ \Rightarrow \frac{9!}{4!} = \frac{9!}{\left(9 - \mathrm{r}\right)!} \\ \mathrm{On} \mathrm{comparing}, \mathrm{we} \mathrm{get} \\ 9 - \mathrm{r} = 4 \Rightarrow \mathrm{r} = 5 \end{gathered}$$