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Apolyton Civilization Forums : Powered by vBulletin version 2.0.3 Apolyton Civilization Forums > Miscellaneous > Archive > AC-General/Help/Strategy-Archive > Number of Factional Combinations (SMACX)
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Petek is offline Petek

Prince
Berkeley, CA
Jul 2000
time: 21:13
Post  Old Post 03-10-2000 03:24
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#1 Report this post to a moderator
Number of Factional Combinations (SMACX) Support Apolyton

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
P
O
I
L
E
R

S
P
A
C
E


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 offline stuntman19
Prince
Wa, usa
Aug 2000
time: 05:13
Post  Old Post 03-10-2000 04:23
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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 is offline Petek

Prince
Berkeley, CA
Jul 2000
time: 21:13
Post  Old Post 03-10-2000 05:46
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Support Apolyton, buy Alpha Centauri

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 is offline Deimos
Prince
of Mars, Son of Ares
Sep 2000
time: 00:13
Thumbs up  Old Post 03-10-2000 06:11
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Support Apolyton

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.

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