Chapter 11 - 11.6 - Counting Principles - 11.6 Exercises - Page 825: 78

$n=2$

$_{n+2}P_3=6·~_{n+2}P_1$ $\frac{(n+2)!}{(n+2-3)!}=6~\frac{(n+2)!}{(n+2-1)!}$ $\frac{(n+2)!~(n+2-1)!}{(n+2)!~(n+2-3)!}=6$ $\frac{(n+1)!}{(n-1)!}=6$ $\frac{(n+1)n(n-1)!}{(n-1)!}=6$ $n^2+n-6=0$ $n^2+3n-2n-6=0$ $n(n+3)-2(n+3)=0$ $(n-2)(n+3)=0$ $n-2=0$ $n=2$ $n+3=0$ $n=-3~~$ (Invalid solution. $n$ must be greater than 0)