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Party concept, curious about how many possible combinations of the below there are.
But I don't know off the top of my head how to math it and don't have the time to dig around on Google right now to try to figure it. So maybe one of y'all can help me:
Concept: Party of six, with all 6 Ancestries and all 12 classes! (6 base, each with one of the other 6 as a multiclass archetype.)
Wondering how many possible parties that would make. (Don't want to mess with heritage, subclass, background, or other choices bc that gets too big and the point here is just to capture all Ancestries and all classes bc of the convenient 6/12 number.)
And, of course, submit your party ideas!!
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Start with the more complicated part, which is figuring out which classes are base. That's 12 choose 6, or 924. There are 6 factorial (or 720) ways to assign the ancestries to the base classes, and also 720 ways to assign the multiclasses. Multiply that out and it leads to 479,001,600 parties that fit these criteria.
Looks like a situation where factorials would apply, at least for the class combinations. (Character 1 would have 12 choices for classes, 11 choices for their multiclass option, Character 2 would have 10 choices for classes, 9 choices for multiclass options, etc...)
I'm not sure if it would be 6! or 12! though. 12! seems like too huge of a number to be possible, I think because factorials don't account for duplicate data sets in different arrangements.
Edit: Ninja'd by Mark!
Multiply that out and it leads to 479,001,600 parties
I guess it is 12 factorial.
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And because math is cool: Here's another way to calculate it that's more efficient:
Start alphabetically with the dwarf. Choose a base class, then a multiclass, and continue on. There are twelve slots to fill, so there are 12 factorial possibilities, which is also.....479,001,600. (I might use an exclamation point to indicate excitement, but then it would look like factorial)
EDIT: We each ninjaed the other. There's beauty in the symmetry!
Start with the more complicated part, which is figuring out which classes are base. That's 12 choose 6, or 924. There are 6 factorial (or 720) ways to assign the ancestries to the base classes, and also 720 ways to assign the multiclasses. Multiply that out and it leads to 479,001,600 parties that fit these criteria.
How many of those succeed in defeating Thanos?
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Start with the more complicated part, which is figuring out which classes are base. That's 12 choose 6, or 924. There are 6 factorial (or 720) ways to assign the ancestries to the base classes, and also 720 ways to assign the multiclasses. Multiply that out and it leads to 479,001,600 parties that fit these criteria.How many of those succeed in defeating Thanos?
Only 1, or about half of them if you just subtract Starlord.
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Only 1, or about half of them if you just subtract Starlord.Start with the more complicated part, which is figuring out which classes are base. That's 12 choose 6, or 924. There are 6 factorial (or 720) ways to assign the ancestries to the base classes, and also 720 ways to assign the multiclasses. Multiply that out and it leads to 479,001,600 parties that fit these criteria.How many of those succeed in defeating Thanos?
If it were that easy, Dr. Strange would've just knocked Starlord's punk ass out.
Wow you guys. Someone call Mathmuse, they're going to be disappointed to miss this happening in real time.
I spotted this thread earlier and saw that Mark Seifter had already given the correct answer.
But Mathmuse, I love your mathematical musing.
Unfortunately, this time it was a very simple question of math and someone beat you to it :(
But Mathmuse, I love your mathematical musing.
Unfortunately, this time it was a very simple question of math and someone beat you to it :(
I still play with heavy mathematical musing in the playtest forum. That forum has become a ghost town as people look ahead to the previews and release of the definitive version of Pathfinder 2nd Edition. I can't muse on the real Pathfinder 2nd Edition before I see the numbers.
Are we sure there's no derangement in there?
Shouldn't it be: 110,880
12!/(6!(12-6))
I'm far too out of practice to figure out the correct math myself, but I've been on this forum long enough to know that if Mathmuse says the answer is correct, it's correct. XD
Are we sure there's no derangement in there?
Shouldn't it be: 110,880
12!/(6!(12-6))
I think is only correct if you're not including iterations for players with different PCs.
If we're assuming that each race, class, player combo is unique (and not just looking at overall party composition).
But I could also be remembering my math incorrectly.
Well, that's the formula for what Mark said it was, "12 choose 6."
12! is not "12 choose 6."
Marching order doesn't matter.
No class can multiclass with itself.
Once one base class is chosen, no other class can multiclass with it.
Once one multiclass is chosen, no other race can choose that as class or multiclass.
...per Joe's rules of the thought experiment.
You are forgetting, I think, that each multiclass pair has six possible races, so 12-choose-6 is not the end of it. For that matter, 12-choose-6 is only the base classes.
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Time to explain the math from the beginning.
n choose k, abbreviated n C k or by a notation with n above k and surrounded by large parentheses, is the number of different ways to chose k distinct objects from a set of n distinct objects. The formula is n choose k = n!/(k!×(n-k)!).
The formula is fairly easy to derive. Choosing the first of the k objects has n choices. Chosing the second of the of the k objects has n-1 choices, because we cannot re-use the first choice. Chosing the third of the k objects has n-2 choices, because we cannot re-use the first two choices. And so on until chosing the last of the k objects has n-k+1 choices, because we cannot re-use the previous k-1 choices. The number of combinations of choices is the product of the number of choices at each step: (n)(n-1)(n-2)...(n-k+1). That product has a name, (n)_k, but most non-mathematicians don't know that notation, so we instead use n!/(n-k)!.
However, I overcounted because the n choose k does not care about the order of the k distinct objects. I counted each set of k distinct objects once for each possible order they could be selected in. Fortunately, we know the number of orders in which we can arrangek distinct objects. The number of orders is k!. Thus, the real value of n choose k is n!/(k!×(n-k)!).
For this problem, Mark Seifter started with 12 choose 6, which is 924. He counted the ways to choose 6 classes out of 12. But order matters in this case, so he multiplied by 6!, which is 720, to count the ways to put those 6 classes in order. He did not need to chose any classes for the multiclasses, because those are the remaining 6 classes. He did have to put them in order, so he multiplied by 6! again. Thus, his result was (12!/(6!×6!))×6!×6! = 12! = 479,001,600. The second method given by Thebazilly and Mark Seifter realizes that the problem can be set up to never divide by 6! and gets 12! directly.