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Re: Lists: how to?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg110044] Re: Lists: how to?
  • From: "J. Clarke" <jclarke.usenet at cox.net>
  • Date: Sun, 30 May 2010 23:45:05 -0400 (EDT)
  • References: <hto90d$kpp$1@smc.vnet.net>

On 5/28/2010 7:21 AM, S. B. Gray wrote:
> Hello:
>
> I have this line
>
> subs3 = Subsets[Range[numbr], {3}] which for numbr=7 gives subs3=
>
> {{1,2,3}, {1,2,4}, {1,2,5}, {1,2,6}, {1,2,7}, {1,3,4}, {1,3,5},
>    {1,3,6}, {1,3,7}, {1,4,5}, {1,4,6}, {1,4,7}, {1,5,6}, {1,5,7},
>    {1,6,7}, {2,3,4}, {2,3,5}, {2,3,6}, {2,3,7}, {2,4,5}, {2,4,6},
>    {2,4,7}, {2,5,6}, {2,5,7}, {2,6,7}, {3,4,5}, {3,4,6}, {3,4,7},
>    {3,5,6}, {3,5,7}, {3,6,7}, {4,5,6}, {4,5,7}, {4,6,7}, {5,6,7}}
>
> Given one of these sublists, say {1,4,5}, I want a nice way to delete
> all members of the subs3 list which have 1,4, or 5 in any one of the
> three positions. The reduced list in this example would be
> {{2,3,6}, {2,3,7}, {2,6,7}, {3,6,7}}.
>
> The way I'm doing it runs DeleteCases nine times. Is there a better way?

I'm just learning Mathematica myself so have no idea if this is a good 
or bad way to do it, but:

Select[subs3, Intersection[{1, 4, 5}, #] == {} &]

will I think do what you're asking.







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