Rearranging Equations with Factorials? Hullo! This is my first time using this site.  I have just begun tutoring a Math 12 student  and I'm a little rusty on factorials. 
The stuff I'm stuck on is why : n!/(n-2)! = n(n-1). 
I seem to be able to come up with the solutions to the following (solving for n) but I can't figure out what to do with the n! in questions like : 
(1) (n+1)!/n! = 6 (I know the answer is n = 5 but don't know where the n! goes..) 
(n+1)!/n! = 6
I know that n has to be one less than 6 because everything else is going to cancel out : 
   6x5x4x3x2x1/5x4x3x2x1 = 6
(2) (n+2!)/n! = 12 
I know the answer is 2 because I got as far as : 
(n+2)! = 12 x n! 
(n+2)!/2 = 6 x n! 
So I somehow got here by subtracting 2 from both sides but I think I'm missing a step :
n!/2 = 4!/2 = 12
Then I'm stumped on : 
(n+1)!/(n-2!) = 20 (n-1) 
and 
(n+1)! = 6 (n-1)! 
(I realised n = 2 here because on the left, 3! = 6 so n = 2 but I can't show the working..)
Any help would be appreciated...I know I must be missing just one fundamental thing somewhere...
 A: Let's work with concrete numbers first. Let $n=10$. Then
$$n!=(10)(9)(8)(7)(6)(5)(4)(3)(2)(1)=(10)(9)\left[(8)(7)(6)(5)(4)(3)(2)(1)\right]=(10)(9)[8!].$$
In exactly the same way, if $n\ge 3$, indeed if $n\ge 2$, we have
$$n!=(n)(n-1)\left[(n-2)!\right].$$
Divide both sides by $(n-2)!$. We get
$$\frac{n!}{(n-2)!}=n(n-1).$$
Added: You also have a question that asks about the equation 
$(n+1)!/(n-2!) = 20 (n-1)$. It may be  a typo for $(n+1)!/(n-2)! = 20 (n-1)$, but let us first assume it is not. Then since $2!=2$, the equation can be rewritten as $(n+1)!=20(n-1)(n-2)$. Note that the factorial on the left grows very fast, while $20(n-1)(n-2)$ grows quite slowly. We have 
$(n+1)!=(n+1)(n)(n-1)(n-2)(n-3)!$. There are several ways in which we can have 
$$(n+1)(n)(n-1)(n-2)(n-3)!=20(n-1)(n-2).$$
Maybe $n=1$, in which case both sides are $0$. We can have $n=2$, in which case again $n=2$. If $n$ is different from $1$ and $2$, we can cancel $(n-1)(n-2)$ from both sides, obtaining $(n+1)(n)(n-3)!=20$. It is not hard to see that this has $n+1=5$ as the only solution. So the solutions are $n=1$, $2$, and $4$. 
If the question was meant to be $(n+1)!/(n-2)!=20(n-1)$, then use the fact that 
$\dfrac{(n+1)!}{(n-2)!}=(n+1)(n)(n-1)$. So we are looking at the equation 
$$(n+1)(n)(n-1)=20(n-1).$$
We cannot have $n=1$, since then $(n-2)!$ would not make sense. So we can cancel $n-1$ from both sides, obtaining $(n+1)(n)=20$. This has the solution $n=4$.  
I hope that this will help with similar questions. 
