Okay so for anyone that actually gives a damn (and how could I blame you if you don’t?) here’s the total number of combinations and how to figure them out. (Of course you can skip down to the end if you just want a number)
First let me say every calculation I’ve done thus far has been off. Even the whole 565554 thing was off. EC was right, without taking assists into account you have 27,720 character combinations.
First of all this calculation is based on the DC version where you can double and triple select the same player.
I’m going to agree with EC and say that Strider, Doom, Sent. Is the same as Sent, Doom, Strider.
The dominant term will be [56!/( 3!(56-3!))]. This accounts for each unique team (assists will be factored in later)
For those of you familiar with basic stats or know the term “Binomial Coefficient” you’ll recognize that that term only accounts for unique teams as when you say, “56 choose 3” you’re not replacing each one of your choices. Therefore you have to add on the two remaining cases
Case I.
Double picking. If I pick character #1, but choose to double up on #2 and #3 there are 56 characters I can select for #1, but only 55 remaining for my second choices. So there are simply 56*55 cases of double picking.
Case II.
Triple picking. This is easy. There are 56.
Assists.
Now let’s factor in assists. There are 3^3 = 3*3 =27 ways to pick them. i.e. AAA AAB, AAC etc.
So to start we’ll take:
([56!/( 3!(56-3!))] + 56*55 + 56) *27 = 831,656
But wait
Multiplying everything by 27 isn’t quite right because the second and third factor clearly don’t have 27 unique assist combinations, and there are also Guile, Iceman, and Charlie to account for. To do this we now subtract off these terms
Subtraction Case I.
Double picking
Let’s say I pick Strider, Doom, Doom. Well Strider’s got 3 assits he’s free to pick from. Doom Doom however are going to eventually have to pick the same one. For this we only need to realize that A<->A, B<->B, and C<->C are not unique as would be the case for a team consisting of 3 unique players so in this case there are only 24 combinations and not 27 (. Which means we’ll have to take 831,656 - (56553) = 822416. This term is the term figured above for double picking minus the 3 double picked cases.
Subtraction Case II.
This one accounts for picking 3 of the same character. In this case there are 9 different combinations. So we do 822416-(56*9) = 821912
Subtraction Case III.
Iceman Guile and Charlie really fuck things over.
It’s best to do them one at a time. And I’M SHAKING MY FIST REALLY HARD at whomever brought up this exception, because it’s really tedious to incorperate, but for the sake of accuracy…
So you want to pick a team that only has Iceman OR Guile OR Charlie. After you pick one that leaves 53 characters and a team with only 18 unique assist combinations so you have 53!/(2!(53-2)! for the combinations of remaining team members *3(Charlie, Guile, Iceman) *9 (Initially these teams were counted as having 27 combinations of assists, but they actually have 18, so we’re subtracting off the 9 remaining). We’re now down to 747500 teams
You know… I’m going to stop myself here. I’m perfectly capable of finishing this post, but… right now I want to play some fucking SFIV, and I doubt anyone but me and possibly EC even gives a damn. If you want me to finish it you can post here or PM me, but until further notice I’m going to assume no one cares and leave it here.
