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>
- Arrangements