1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Probability - factorial algebraic manipulation

  1. Aug 28, 2012 #1
    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.

    [tex] \frac{n(n-1)...(n-r+1)}{r!} = \frac {n!}{(n-r)!r!} [/tex]

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

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

    From here I tried expanding the factorials then cancelling stuff but I'm still not seeing the pattern. Thanks for any help.
  2. jcsd
  3. Aug 28, 2012 #2
    Look at the left hand of your last equation. We have:
    [tex]n(n-1)\cdots (n-r +1)[/tex]
    [tex](n-r)! = (n-r)(n-r-1)\cdots (2)(1)[/tex]
    Multiplying these together you get
    [tex]n(n-1)\cdots (n-r +1)(n-r)\cdots (2)(1) = n![/tex]
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook