Probability - factorial algebraic manipulation

  • #1
658
2
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.
 

Answers and Replies

  • #2
430
3
Look at the left hand of your last equation. We have:
[tex]n(n-1)\cdots (n-r +1)[/tex]
times
[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]
 

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