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