From: Pablo Bianciotto <lists@...>
Date: 2013-04-13T00:34:53+09:00
Subject: Re: Looking for more elegant solutions.

Try the following:

1.upto(n).to_a.permutation(n).to_a.delete_if do |perm|
  perm.any? { |letter| letter == perm[letter -1] }
end

Bye

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