Re: Combinations

*To*: mathgroup at smc.vnet.net*Subject*: [mg30828] Re: Combinations*From*: "Ian McInnes" <ian at whisper-wood.demon.co.uk>*Date*: Thu, 20 Sep 2001 03:51:20 -0400 (EDT)*References*: <9o96q8$cev$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

The number of permutations (possible orderings) of a list of n objects is n! only if all items are different. There are fewer permutations if the list contains repeated items. For example: Length[Permutations[{a, b, c, d}]] is 24 whereas: Length[Permutations[{a, b, b, c}]] is 12 (the examples assume that the letters a, b, c... are undefined). In fact the number of permutations is Factorial[n] divided by the product of the factorials of the numbers involved in each repeated group. For example, the number of permutations of: {a, a, b, b, b, c}is: Factorial[6] / (Factorial[2] * Factorial[3] * Factorial[1]) = 60 The following one-liner computes the number of permutations of a list as above: In[1]:= permCount[x_]:=If[ListQ[x], Length[x]! / Apply[Times, Map[Count[x,#]!&, Union[x]]],1] and can be efficiently applied to lists of arbitrary length: In[2]:= permCount[{a,g,e,d,d,e,a,f,a,c,e,f,e,d,b,a,d,c,a,b,b,a,g,e}] Out[2]:= 6233607431251200 Checking this result by the use of Length[Permutations[]] is not recommended ;-) Ian McInnes. "David Park" <djmp at earthlink.net> wrote in message news:9o96q8$cev$1 at smc.vnet.net... > Dana, > > The number of permutations of n objects is just n! (n factorial) which you > can enter in just that way in Mathematica. For a set with 4 elements: > > 4! > 24 > > Which checks with > > Permutations[{1, 2, 3, 4}] > Length[%] > {{1, 2, 3, 4}, {1, 2, 4, 3}, {1, 3, 2, 4}, {1, 3, 4, 2}, {1, 4, 2, 3}, {1, > 4, > 3, 2}, {2, 1, 3, 4}, {2, 1, 4, 3}, {2, 3, 1, 4}, {2, 3, 4, 1}, {2, 4, 1, > 3}, {2, 4, 3, 1}, {3, 1, 2, 4}, {3, 1, 4, 2}, {3, 2, 1, 4}, {3, 2, 4, > 1}, {3, 4, 1, 2}, {3, 4, 2, 1}, {4, 1, 2, 3}, {4, 1, 3, 2}, {4, 2, 1, > 3}, {4, 2, 3, 1}, {4, 3, 1, 2}, {4, 3, 2, 1}} > 24 > > The number of k element subsets picked from n items is just Binomial[n, k]. > For example: > > Binomial[4, 2] > 6 > > Needs["DiscreteMath`Combinatorica`"] > > KSubsets[{1, 2, 3, 4}, 2] > Length[%] > {{1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4}} > 6 > > How precisely did I find this out? By looking up Permutations in my > "Mathematics Handbook for Science and Engineering" and then looking up > Binomial and Factorial in Help. Finding KSubsets is not that easy because > the Combinatorica Help is not very good or convenient. > > There was a book "Implementing Discrete Mathematics: Combinatorics and Graph > Theory with Mathematica" by Steven Skiena the author of Combinatorica. > Unfortunately, it has gone out of print. It would be a great thing if WRI > could somehow arrange to bring it back into print, and also improve the Help > documentation for Combinatorica because it is a great package. > > New users of Mathematica are often disappointed that the precise command or > function they want is not there, or not there under the name they might > expect. But you will quite often have to hit the math books to find how to > calculate something and you will also quite often have to write your own > small routines. If Mathematica tried to have direct routines for everything > it would end up with millions of commands and still fall short. > > David Park > djmp at earthlink.net > http://home.earthlink.net/~djmp/ > > > From: Dana [mailto:ng_only at hotmail.com] To: mathgroup at smc.vnet.net > > > > Hello. I have Mathematica 4.1 > > > > In a Program like Excel, or a hand-held calculator, one can return the > > number of combinations and permutations. > > > > However, I can not find an equivalent function in Mathematica. > > For example, Permutations[ ] returns a long list of all the "Actual" > > permutations. > > I am looking for just the final number. > > If there is one, could you include 'how' you found it. I have looked > > everywhere. > > I know DiscreteMath`Combinatorica` has some stuff in it, but the > > Help system > > appears not to explain many of them. > > > > I can write a custom function, but I am curious to find out if > > this function > > is built in to Mathematica. > > > > I hope the answer is not to take the Length[ ] of a rather long list. > > > > TIA. Dana > > > > (I posted this question a month ago, but it never showed up in the > > newsgroup, or in the archives. > > I hope I am not doing something wrong.) > > > >

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