Weekly Challenges
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I went with an iterative macro to dynamically handle different numbers of horses.
Spoiler
Extra credit:
The number of possible outcomes is n! (n factorial). So, if there were five horses, the number of outcomes would be 5! = 5 x 4 x 3 x 2 x 1 = 120.
Outer workflow:
Iterative macro:
This seems way too long to do something so simple, however, removing the duplicates (e.g. horse 1 = horse 2) after each iteration was MUCH faster than removing them after all of the iterations completed (because those unnecessary records multiply quickly!)
The number of possible outcomes is n! (n factorial). So, if there were five horses, the number of outcomes would be 5! = 5 x 4 x 3 x 2 x 1 = 120.
Outer workflow:
Iterative macro:
This seems way too long to do something so simple, however, removing the duplicates (e.g. horse 1 = horse 2) after each iteration was MUCH faster than removing them after all of the iterations completed (because those unnecessary records multiply quickly!)
The extra credit was easy, but I though and hoped there would be a more elegant solution to the main task... :S
Spoiler
Extra credit: 5! = 120, n!
