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.