# 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...

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$n!=n(n-1)(n-2)!$ – André Nicolas Oct 30 '12 at 3:17
All you are missing is that $n! (n+1) = (n+1)!$. – JavaMan Oct 30 '12 at 3:23
Thank you JavaMan! – Andy Nov 5 '12 at 18:14
Usage of $\LaTeX$ will be appreciated. – Aditya Agarwal Aug 28 '15 at 15:18

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 $n=4$.
So you may be looking for an integer $n$ such that $n(n-1)=a$, where $a$ may be quite large. For "most" $a$ there is no such integer $n$. If there is one, you can find it by rewriting the equation as $n^2-n-a=0$, and then using the Quadratic Formula. So $n=\frac{1+\sqrt{1+4a}}{2}$. For large $a$, it is somewhat simpler to note that if the answer is an integer, then that integer is the smallest integer greater than $\sqrt{a}$. With a calculator, the work is easy. Without, not so much! – André Nicolas Nov 5 '12 at 18:33