Question

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

Answer 1

We have,
${n+5}_{P_{n+1}}=\frac{11(n-1)}{2}{n+3}_{P_n}\Rightarrow \frac{(n+5)!}{[n+5-(n+1\left)\right]!}=\frac{11(n-1)}{2}\times \frac{(n+3)!}{[n+3-n]!}\Rightarrow \frac{(n+5)!}{[n+5-n-1]!}=\frac{11(n-1)}{2}\times \frac{(n+3)!}{3!}\Rightarrow \frac{(n+5)!}{4!}=\frac{11(n-1)}{2}\times \frac{(n+3)!}{3!}\Rightarrow \frac{(n+5)(n+4)(n+3)!}{4!}=\frac{11(n-1)}{2}\times \frac{(n+3)!}{3!}\Rightarrow \frac{(n+5)(n+4)}{4\times 3!}=\frac{11(n-1)}{2\times 3!}\Rightarrow (n+5)(n+4)=\frac{11(n-1)\times 4}{2}\Rightarrow (n+5)(n+4)=22(n-1)\Rightarrow n^2+4n+5n+20=22n-22\Rightarrow n^2+9n-22n+20+22=0\Rightarrow n^2-13n+42=0\Rightarrow n^2-6n-7n+42=0\Rightarrow n(n-6)-7(n-6)=0\Rightarrow n-6or,n=7$
Hence, n=6 or, 7