Re: Permutations.

*To*: mathgroup at smc.vnet.net*Subject*: [mg14829] Re: Permutations.*From*: "vdmcc" <w.meeussen.vdmcc at vandemoortele.be>*Date*: Wed, 18 Nov 1998 01:29:33 -0500*Organization*: EUnet Belgium, Leuven, Belgium*References*: <72jefh$3fn@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Alan, I don't see the problem : first you should use the KSubsets[list,k], and then map the Permutations into that : Flatten[ Permutations/@KSubsets[ yourlist, k ] , 1] or am I missing the point? wouter w.meeussen.vdmcc at vandemoortele.be *************************************************** Alan W.Hopper wrote in message <72jefh$3fn at smc.vnet.net>... >Greetings, > > >For the combinations of n objects taken k at a time, (where order counts >and there is no duplication), the function KSubsets is the one to use. >e.g. > In[1]:= <<DiscreteMath`Combinatorica` > > In[2]:= Table[KSubsets[{a,b,c,d}, k], {k, 4}] > > Out[3]= {{{a}, {b}, {c}, {d}}, > {{a,b}, {a,c}, {b,c}, {b,d}, {c,d}}, > {{a,b,c}, {a,b,d}, {a,c,d}, {b,c,d}}, > {{a,b,c,d}}} > >But likewise in wanting to find all the permutation subsets (with no >duplication and order not counting), of a numerical or symbolic list, >there does not seem to be a function anywhere (including the packages), >to achieve this goal. > >(By n!/(n-k)!, there will be be; 4, 12, 24, 24 permutations taken k = >1, 2, 3, 4 at a time, for a 4 element list). > >The built-in function Permutations and also LexicographicPermutations >(from Combinatorica) do not take a second argument, as KSubsets does, >and so only the (n, k=n) permutations (24 in the example ) can be >found. > >I would appreciate some assistance to find a way to generate all of the >permutation subsets, in List, Table or Column form. > > >Alan W.Hopper >awhopper at hermes.net.au >