Re: Arrangements

*To*: mathgroup at smc.vnet.net*Subject*: [mg108911] Re: Arrangements*From*: Bob Hanlon <hanlonr at cox.net>*Date*: Tue, 6 Apr 2010 08:43:46 -0400 (EDT)*Reply-to*: hanlonr at cox.net

lst = Table[{"A", "B"}, {8}] // Flatten {A,B,A,B,A,B,A,B,A,B,A,B,A,B,A,B} p = Permutations[lst]; Length[p] == Binomial[16, 8] True There are no duplicates Length[Union[p]] == Length[p] True p[[RandomInteger[{1, Length[p]}]]] {B,A,B,A,B,A,B,A,B,A,A,B,A,B,B,A} Bob Hanlon ---- John <jwa0 at lehigh.edu> wrote: ============= 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