please help check if this would the correct way to solve this:
$^nP_2 = ^{n+1}C_3$. I want to solve for $n$.
theoretically, I was thinking that:
$^nP_k = k!\times ^nC_k $ hence:
$\dfrac{n!}{(n-2)!} =3!\times\dfrac{(n+1)!}{(n+1-3)!\times3!} $
But eliminating 3! I get:
$\dfrac{n!}{(n-2)!} =\dfrac{(n+1)!}{(n-2)!} $
But again $if \dfrac{A}{B} =\dfrac{A+1}{B}$ then $A=A+1 $ hence
from the equation above $ n! =(n+1)! $
divide both sides by $n!$:
$ 1=\dfrac{(n+1)!}{n!} = \dfrac{(n+1)\times[(n+1)-1]\times[(n+1)-2]!}{n!}$ comes to:
$1=\dfrac{(n+1)\times(n-0)!}{n!} =n+1 $
$\therefore n+1=1 $ hence $ n=0 $
We are presumably to take $n\ge 3$.
The left-hand side is $n(n-1)$. The right-hand side is $\frac{(n+1)(n)(n-1)}{3!}$.
Since $n$ is equal to neither $0$ or $1$, we can divide both sides by $n(n-1)$.
Remark: Sometimes, if $m$ and $k$ are non-negative integers, and $m\lt k$, ${}^mP_k$ and ${}^mC_k$ are defined to be $0$. If so, we get the additional solutions $n=0$ and $n=1$.