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Re: Re: obtaining an ordered subset of k elements


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
>>
>>
>>
>>
>
>
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