# Simplifying Factorials

• June 12th 2013, 05:19 AM
theworstatmath
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:

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

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

But the actual/final answer should be:

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

Can anyone explain how to get this? I understand factorials, but only the basics when variables are involved. Thankyou!
• June 12th 2013, 05:45 AM
MarkFL
Re: Simplifying Factorials
It boils down to recognizing:

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

$(n-(n-3))!=3!=6$
• June 12th 2013, 05:53 AM
theworstatmath
Re: Simplifying Factorials
I think I need to go bash my head for not seeing this sooner. Thanks Mark, this makes perfect sense!
• June 12th 2013, 07:08 AM
HallsofIvy
Re: Simplifying Factorials
You can borrow my two by four!
• June 12th 2013, 11:16 AM
topsquark
Re: Simplifying Factorials
Quote:

Originally Posted by HallsofIvy
You can borrow my two by four!

An infraction for HallsofIvy. A shovel works much better.

-Dan