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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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