Re: Re: obtaining an ordered subset of k elements

*To*: mathgroup at smc.vnet.net*Subject*: [mg73519] Re: [mg73498] Re: obtaining an ordered subset of k elements*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Wed, 21 Feb 2007 01:40:43 -0500 (EST)*References*: <cFfCh.12521$aF5.12165@tornado.fastwebnet.it> <200702201118.GAA07759@smc.vnet.net>

A very much faster way, without the need for the Combinatorica package, is: Flatten[Permutations /@ Subsets[Range[6], {3}], 1] Note that the order of elements returned is not the same. Andrzej Kozlowski On 20 Feb 2007, at 12:18, digrpat wrote: > an example with 3 elements chosen from a set of 6 > > << DiscreteMath`Combinatorica` > > Sort[Flatten[Map[LexicographicPermutations, KSubsets[{1, 2, 3, 4, > 5, 6}, > 3]], 1]] > > > > "GiulioC" <giulio.castorinaNOSPAM at fastwebnet.it> wrote in message > news:cFfCh.12521$aF5.12165 at tornado.fastwebnet.it... >> Hello All, >> >> I am in trouble as I cannot find no way to obtain an ordered >> subsets of k >> elements with mathematica 5.2. I didnt find any >> solutions...Mathematica >> seems hadle just Permutations (n=r) or unordered subsets of k >> elements >> (KSubsets). What I want is what is explained here >> (http://mathworld.wolfram.com/Permutation.html) at the 8th text row >> ...but no Mathematica function is showed :( >> Please note I dont want just the "number" of total rearrangement >> but I >> want a named list of all subsets of k elements >> >> Any solutions? I hope yes... >> >> Cheers >> >> Giulio >> >> >> >> > > > > Posted Via Usenet.com Premium Usenet Newsgroup Services > ---------------------------------------------------------- > ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY ** > ---------------------------------------------------------- > http://www.usenet.com >

**References**:**Re: obtaining an ordered subset of k elements***From:*"digrpat" <digrpat@AOL.com>