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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
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1 Answer

up vote 4 down vote accepted

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 right 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.

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That was really helpful, thank you so much! Just one question - what happens if the numbers are much larger than those where you know the factors of 20? I'm sufficiently familiar with multiplication tables to know the factors of 20 and figure out (n) (n-1) = 20. If the numbers were less friendly (or I suppose less easily predicatble), then is there another solution? I guess what I'm trying to ask is, can you work backwards from these equations to solve for much larger/less tidy numbers? –  Andy Nov 5 '12 at 18:20
    
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
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