i will do four calcs then
1. students w/ students
2. Rog w/ RoG
3. students or RoG w/ RoG or students
4. every character already used is bi
k. so first we make an assumption: each pairing of two partners of a valid combination of orientations is considered valid (e.g, straight chick and bi guy, but not straight guy and bi guy)
we make another that everyone who has not stated an orientation is straight
now, calc 1:
we have 15 students, 13 male and 2 female, 3 males are bi/pan, 1 is gay
so that makes 12 partners for each female for a total of 24 ships
the 3 bi guys have 3 additional partners, one another and the gay guy, so accounting for duplicates gives us 3+2+1=6 more ships
so 30 ships possible among students only
calc 2:
RoG only has 3 ships unless at least two of them are bi, so yeah 3 ships possible among RoG only
calc 3:
we'll take our 30+3 ships from the previous calcs, then account for new partners
we have 1 more female for the 12 interested males in the student group, so 12 more ships possible there
also, 2 more females for the interested parties in RoG, resulting in another 6 more ships
there are no additional male partners for interested males, 0 more ships
in total, 51 possible ships total
Calc 4:
we have 19 characters, nPr with n being 19 and r being 2 gives us an incredible 172 ships if everyone is bi