# Thread: Combinations - Problem with simplifying Factorials.

1. ## Combinations - Problem with simplifying Factorials.

Hey folks, I'm having a problem with a combination question, hoping to get some ideas of where I'm going wrong from some pros!

Sorry for the bad format, I usually use a program called MathType5 to type out my work, but I can't seem to get the code to translate over here, so I'm gonna do my best with microsoft word and show my work.

Update, so sorry, I just can't get the fractions to translate over.

Solve for "n" in n-1C11 = nC12

I started by using the "n! / (n-r)!r!" formula on both sides, and tried to simplify from there by gathering like terms and then cross multiplying and dividing to get
n!(n-2)!11! = (n-1)!((n-12)!12!)
and then simplifying this into
(n^2-12n)!11n! = (n^2-13n!)(12n-12)!
And it is here that I am stuck, I don't know how to isolate n with factorials being attached to everything.

Again sorry for the bad formatting, any help is appreciated and thanks for taking the time to look at my problem!

2. Hello,

Uh-oh... $a!b! \neq (ab)!$

Here is the beginning (click on the pictures to get the code)

$^{n-1}C_{11}=^n C_{12}$

$\frac{(n-1)!}{11!(n-12)!}=\frac{n!}{12!(n-12)!}$

You can see that (n-12)! cancels on each side :

$\frac{(n-1)!}{11!}=\frac{n!}{12!}$

Now note that $n!=n\times(n-1)\times\dots\times 1=n\times (n-1)!$

Can you finish it ?

3. That did it, I got n = 12.

Thanks so much for the help!