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Re: Arrangements

  • To: mathgroup at smc.vnet.net
  • Subject: [mg108909] Re: Arrangements
  • From: Ingolf Dahl <ingolf.dahl at physics.gu.se>
  • Date: Tue, 6 Apr 2010 08:21:24 -0400 (EDT)
  • References: <201004061123.HAA25461@smc.vnet.net>
  • Reply-to: <ingolf.dahl at physics.gu.se>

Try with

Permutations[{a, a, a, a, a, a, a, a, b, b, b, b, b, b, b, b}]     

Then we should get a list of all variants

If we check 

Length[Permutations[{a, a, a, a, a, a, a, a, b, b, b, b, b, b, b, b}]]

we obtain 

12870

so that seems OK.


Best regards

Ingolf Dahl
Sweden

-----Original Message-----
From: John [mailto:jwa0 at lehigh.edu] 
Sent: den 6 april 2010 13:23
To: mathgroup at smc.vnet.net
Subject: [mg108909] [mg108886] Arrangements

Google 6, Windows-Xp

The number of different arrangements of sixteen symbols -- eight
letters A and eight letters B -- is Binomial[16,8]=12870. Is there a
command that will generate all 12870 arrangements one at a time
without duplication? Any order is acceptable.

I used RandomSample[Range[16]] to select a random permutation of
sixteen different symbols. Assigning the letter A to the first eight
numbers in the random permutation and the letter B to the last eight
letters in the permutation created a randomly selected arrangement.
Awkward, but it works because Binomial[2n,n](n!)(n!)=(2n)!).

John



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