Number of Factional Combinations (SMACX)

[Petek]

If you like to solve combinatorial problems, compute the total number of possible combinations of factions in a SMACX game. Assume that either both alien factions must be present or neither (since some early versions allowed a single alien presence). Also assume that the same faction doesn't appear twice in the same game. I've posted my answer below, after some spoiler space.

S
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E
R

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A
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First count the number of combinations that include no aliens. That would be choosing 7 factions from 12, which is 12!/(7!5!), or 792. Next, if both alien factions are present, the remaining 5 slots can be filled from any of the twelve non-aliens. That's 12!/(5!7!), which also is 792. The sum 792+792=1584 is the total number of possible combinations.

[stuntman19]

Is this really the answer? For example are not these 2 combinations simply a repeat, but in different slots. So it would not truely be a different combination would it?

#1 Spartans #2 Believers
Hive Peacekeepers
Gaians Morgan
Believers Spartans
Morgan University
University Hive
Peacekeepers Gaians

Therefore, I would not concur with your results.

[Petek]

Well, if I'm wrong it wouldn't be the first time! However, in your example you've given two permutations of the same combination of factions. In a permutation, the ordering of the elements matters, whereas the order in a combination does not. The formula for the number of ways to choose r objects from a collection of n [n!/r!(n-r)!] gives the number of combinations of n objects taken r at a time. If you want the number of permutations of k objects taken r at a time, the formula is n!/(n-r)!

In short, I think that I only counted your #1 and #2 as one combination.

(Sorry if this reply sounds overly defensive. I still admit I could be all wet!)

[Deimos]

Petek,

You've got it right. This is basic prob/stats stuff. Apply this to a deck of cards, and you'll really see how hard it is to pull that inside straight on the tables in Vegas.