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MathGroup Archive 2000

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Re: Combination Formula, HELP

  • To: mathgroup at smc.vnet.net
  • Subject: [mg21421] Re: Combination Formula, HELP
  • From: Tobias Oed <tobias at physics.odu.edu>
  • Date: Thu, 6 Jan 2000 01:46:37 -0500 (EST)
  • Organization: Old Dominion University
  • References: <84s6uf$q6e@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

::necro:: wrote:
> 
> hi, I don't suppose anyone could help me? I have a bit of a problem with
> some maths revision, I am stuck, the question is as follows:
> 
> e.g.1 has 2 * 2, and in e.g.2 the numbers are 1 * 3, then the 3 is doubled!
you probably mean eq. (equation)

> why? Is it because it is partnered with a 1 and must have a higher value?
> 
> e.g.1
> 4
>   C  =        4!                 4!                 24
>     2     ------------- = ------------- = --------------- = 6
>              (4-2)!2!        2! * 2!           2 * 2
> 
> e.g.2
> 
> 4
>   C  =        4!                 4!                 24
>     3     ------------- = ------------- = --------------- = 4
>              (4-3)!3!        1! * 3!           1 * 6
> 

There are a bunch of ! in your equation. They mean factorial, which is
defined by
n!= n (n-1)!, 0!=1
this is the same thing as saying
n! = n (n-1) (n-2) .... 2 1
therefore,
the denominator of eq.1 is (4-2)! 2! = 2! 2! = (2 1) (2 1) = 2 2 = 4 
the denominator of eq.2 is (4-3)! 3! = 1! 3! = (1) (3 2 1) = 6 = 6
and the numerator is 4! = 4 3 2 1= 24
Is this what you want ?

Tobias.


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