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

**References**:**Arrangements***From:*John <jwa0@lehigh.edu>