Question

If ^{n +5}P_{n +1} = $\frac{11(n-1)}{2}$^{n +3}P_n, find n.

Answer 1

^{n +5}P_{n +1} = $\frac{11(n-1)}{2}$^{n +3}P_n
$$\begin{gathered} \Rightarrow \frac{\left(n+5\right)!}{\left(n+5-n-1\right)!}=\frac{11\left(n-1\right)}{2}\times \frac{\left(n+3\right)!}{\left(n+3-n\right)!} \\ \Rightarrow \frac{\left(n+5\right)!}{4!}=\frac{11\left(n-1\right)}{2}\times \frac{\left(n+3\right)!}{3!} \\ \Rightarrow \frac{\left(n+5\right)!}{\left(n+3\right)!}=\frac{11\left(n-1\right)}{2}\times \frac{4!}{3!} \\ \Rightarrow \frac{\left(n+5\right)\left(n+4\right)\left(n+3\right)!}{\left(n+3\right)!}=\frac{11\left(n-1\right)}{2}\times \frac{4\times 3!}{3!} \\ \Rightarrow \left(n+5\right)\left(n+4\right)=22\left(n-1\right) \\ \Rightarrow n^2+9n+20=22n-22 \\ \Rightarrow n^2-13n+42=0 \\ \Rightarrow n=7,6 \end{gathered}$$