I am trying to find the positive integer solutions to the following permutation equation: $$\phantom{.}^{n+1}P_{3}=4\phantom{.}^{n}P_{2}.$$
I'm really lost I don't know what I'm doing. I've try doing what I read off from other problem but I'm feel like I'm doing it wrong
As $\displaystyle^mP_r=\frac{m!}{(m-r)!}$
$$^{n+1}P_3=4\cdot^nP_2\iff\frac{(n+1)!}{(n+1-3)!}=4\frac{n!}{(n-2)!}$$
$$\iff n+1=4$$
A permutation expression with $n$ in the upper index is just a fancy way to write a polynomial. In particular, we have $$\phantom{.}^{n+1}P_{3}=(n+1)\cdot n \cdot (n-1) \quad \textrm{and} \quad \phantom{.}^{n}P_{2}=n\cdot (n-1).$$
Substitutions these expressions into your equation, we get: \begin{eqnarray} \phantom{.}^{n+1}P_{3} & = & 4\phantom{.}^{n}P_{2} \\ (n+1) n (n-1) & = & 4n(n-1) \end{eqnarray} so we have now reduced the problem to a simple polynomial problem.
The expressions only make sense for $n\geq 2$ (since otherwise $\phantom{.}^{n+1}P_{3}$ would be undefined), so we may safely divide both sides by $n(n-1)$ without losing a solution: $$(n+1) = 4. $$
(This is a slightly subtle point. The polynomials that we found are defined for all $n$, but the underlying permutation equation is only defined for $n\geq 2$. By converting to a polynomial, we have actually created solutions: $n=0$ and $n=1$. But since these are not solutions to the permutation equation, we can, and must disregard them.)
