@adatron  
Lol, yeah, but you know how shippers are.

@Cirrus Light  
Of course, this is theoretical. Some pairings would be more likely to happen than others. I wouldn’t be surprised if NL/TV had never met the other parents, especially BM/PB.

@Cirrus Light  
…What is all this science crap?

@laofuzi  
I just realized you could do that with every set of the mane 6’s parents. Pre-canon shipping.
 
Hmm… I wonder how many possible combinations there are if you explore alternate timelines where different ponies got together…
 
So, normally, to see how many possible combinations there are for n elements it’s  
1 * 2 * 3 * ... * (n - 2) * (n - 1) * n  
Which is just n factorial (which is written as “n!” ).
 
So the possible combinations for a set of 12 elements is  
1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10 * 11 * 12
 
Because first you can choose any of the 12 elements. Then once you’ve chosen one, you can choose any of the 11 remaining. Then when you’ve chosen that, you can choose any of the 10 remaining, so on and so forth until there’s just 1 left to choose.
 

 
At first I was thinking doing this with shipping the pony parents is a bit odd since there’s 6 couples and 12 ponies, but if you stick to mammalian reproduction, then I guess it’s still just 6 choices, since for each male you can only choose one of the six females and vice versa.
 
For example, I can list all their dads (the order doesn’t matter), so  
A = Bright Mac  
B = Twi’s Dad  
C = Rarity’s Dad  
D = Pinkie’s Dad  
E = RD’s Dad  
F = FlutterDad
 
And now I assign each a female partner
 
A pair with (6 to pick from)  
B pair with (5 to pick from)  
C pair with (4 to pick from)  
D pair with (3 to pick from)  
E pair with (2 to pick from)  
F pair with (1 left to pick)
 
And the “vice versa” is that you can do this with the moms and then pick dads. This doesn’t generate a second list because Bright Mac x Buttercup = Buttercup x Bright Mac.
 
I guess shipping is commutative xp
 

 
Now, if you allow magical lesbian offspring (and gay offspring) then things get more complicated and I’m less sure…
 
It’s not 12 factorial because the order the pairs are in don’t matter.
 
So if I choose 1 of 12, then 1 of 11, then 1 of 10… Then I I’ll generate lists like this
 
AxB, CxD, ExF, GxH, IxJ, KxL  
and  
KxL, IxJ, AxB, CxD, ExF, GxH
 
As separate lists, even though they’re the same pairings, so it’s not just the 12! approach, and the method I used for straight couples only works for that - note that way of counting them doesn’t count any homo ships.
 
I’m sure I can look up this logic problem - I think it’s the same as that problem of trying to get everyone in a party to do a quick meet and greet - but where’s the fun in not deriving the answer yourself? Xp

Pre-canon shipping intensifies!!

Once they go in for the kiss, every x seconds they get half as far apart.
 
Unfortunately, that means they’ll never touch xp