Re: Help on Partitions, Again!!!

*To*: mathgroup at smc.vnet.net*Subject*: [mg24651] Re: Help on Partitions, Again!!!*From*: "Allan Hayes" <hay at haystack.demon.co.uk>*Date*: Mon, 31 Jul 2000 09:23:21 -0400 (EDT)*References*: <8lt1mt$2m2@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

We can use the Combinatorica function KSubsets (but a custom function should be quicker) << DiscreteMath`Combinatorica` lst = {A, B, C, D, E, F}; KSubsets[lst, 3] {#, Complement[lst, #]} & /@ % Sort /@ % Union[%] {{{A, B, C}, {D, E, F}}, {{A, B, D}, {C, E, F}}, {{A, B, E}, {C, D, F}}, {{A, B, F}, {C, D, E}}, {{A, C, D}, {B, E, F}}, {{A, C, E}, {B, D, F}}, {{A, C, F}, {B, D, E}}, {{A, D, E}, {B, C, F}}, {{A, D, F}, {B, C, E}}, {{A, E, F}, {B, C, D}}} Length[%] 10 -- Allan --------------------- Allan Hayes Mathematica Training and Consulting Leicester UK www.haystack.demon.co.uk hay at haystack.demon.co.uk Voice: +44 (0)116 271 4198 Fax: +44 (0)870 164 0565 "Jose Prado de Melo" <jpmelo at iconet.com.br> wrote in message news:8lt1mt$2m2 at smc.vnet.net... > > > Hello, MathGroup > First of all, thanks for your attention. > To be more specific: > It's not too dificult to calculate the solution of the problem: > How many ways, can the set {A,B,C,D,E,F} be separeted into two parts > with three elements in each? > Answer: x = 6!/(2!.3!.3!) = 10 > I'm looking for a function to generate all the partitions using > Mathematica 3.0 . > I'm not sure, but I think the package Combinatorica doesn't have a > function to do this. > For example, I'm trying to think up a function f like this one: > > In[ ] = f [ {A,B,C,D,E,F},{3,3}] > Out [ ] = { { {A,B,C},{D,E,F} }, { { > A,B,F},{C,D,E}},...................} and so on. > In [ ] = Length[%] > Out [ ] = 10 > > Please, help me. > Thanks! > > >