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RE: Horse Race Puzzle
 To: mathgroup at smc.vnet.net
 Subject: [mg9183] RE: [mg9162] Horse Race Puzzle
 From: Ersek_Ted%PAX1A at mr.nawcad.navy.mil
 Date: Tue, 21 Oct 1997 02:02:56 0400
 Sender: ownerwrimathgroup at wolfram.com
Seth J. Chandler wrote:

N horses enter a race. Given the possibility of ties, how many
different finishes to the horse race exist. Write a Mathematica
program that shows all the possibilities.

By way of example: here is the solution (13) by brute force for N=3.
The horses are creatively named a, b and c. The expression {{b,c},a}
denotes a finish in which b and c tie for first and a comes in next. 
{a, b, c}, {a, c, b}, {b, a, c}, {b, c, a}, {c, b, a}, {c, a, b},
{a,{b,c}}, {{b,c},a}, {b,{a,c}},
{{a,c},b},{{c,{a,b}},{{a,b},c},{{a,b,c}} 

I propose a different convention. Make a list where we have a list of
those in first, a list of those in second, and a list of those in
third. This gives the following for three horses:
{{a}, {b}, {c}},
{{a}, {c}, {b}},
{{b}, {a}, {c}},
{{b}, {c}, {a}},
{{c}, {b}, {a}},
{{c}, {a}, {b}},
{{a}, {b,c}, {}},
{{b,c}, {a}, {}},
{{b}, {a,c}, {}},
{{a,c}, {b}, {}},
{{c}, {a,b}, {}},
{{a,b}, {c}, {}},
{{a,b,c}, {}, {}}
I don't know but this may be easy to do using some of the 230 commands
in the package, DisctreteMath`CombinatorialFunctions`. See
Mathematica 3.0 Standard Packages (pp 83102).
Apparently an even better guide to the package is the following book:
Implementing Discrete Mathematics: Combinatorics and Graph Theory with
Mathematica, by Steven Skiena
On the other hand you could write the code yourself if you want to take
on the challenge.
However, I don't have the time to pursue this any further right now.
Ted Ersek
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