1. ## Simplifying Factorials

Hi Guys,

Doing some practise papers for my DM final and have come across something I don't quite understand. I didn't do high school math (last time I did any was 8 years ago) so I find this course very challenging/frustrating but enjoyable!

Question

Write down:

$\displaystyle \left(\begin{array}{cc}n\\n-3\end{array}\right)$

$\displaystyle \frac{n!}{(n-3)!(n-(n-3))!}$

But the actual/final answer should be:

$\displaystyle \frac{n(n-1)(n-2)}{6}$

Can anyone explain how to get this? I understand factorials, but only the basics when variables are involved. Thankyou!

2. ## Re: Simplifying Factorials

It boils down to recognizing:

$\displaystyle n!=n(n-1)(n-2)(n-3)!$

$\displaystyle (n-(n-3))!=3!=6$

3. ## Re: Simplifying Factorials

I think I need to go bash my head for not seeing this sooner. Thanks Mark, this makes perfect sense!

4. ## Re: Simplifying Factorials

You can borrow my two by four!

5. ## Re: Simplifying Factorials

Originally Posted by HallsofIvy
You can borrow my two by four!
An infraction for HallsofIvy. A shovel works much better.

-Dan