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
$$n(n-1)\cdots (n-r +1)(n-r)\cdots (2)(1) = n!$$