Probability - factorial algebraic manipulation

[DrummingAtom]

I saw this in my book and I'm having a difficult time figuring out how this formula was manipulated algebraically to equal the other side. I know that it works because I've used it several times but I can't show that it is true.

\frac{n(n-1)...(n-r+1)}{r!} = \frac {n!}{(n-r)!r!}

One of the things I tried was cross multiplied the r! and canceled it then multilpied the (n-r)!to the other side. Which I was left with:

[n(n-1)...(n-r+1)]*[(n-r)!]= n!

From here I tried expanding the factorials then cancelling stuff but I'm still not seeing the pattern. Thanks for any help.

 

[rasmhop]

Look at the left hand of your last equation. We have:
n(n-1)\cdots (n-r +1)
times
(n-r)! = (n-r)(n-r-1)\cdots (2)(1)
Multiplying these together you get
n(n-1)\cdots (n-r +1)(n-r)\cdots (2)(1) = n!