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I can't pick between men and women. I'm gonna have to be in a thruple to get one of each. But what if my partners were also bi? The man of the relationship would only have 2 girls, so we need to add an extra man, and be in a 4-couple.
It would be mathematically perfect. I can see no flaws. 4 is not too many relationship dynamics to keep track of either.
The power set of 4 is 16. That means that there are 16 possible subsets you can make out of 4 people. So you can get a group of 3 people, or 2 people, and each such group can have its own unique dynamic.
But 4 of those subsets are just the individuals themselves. And one of those sets is the empty set. So really there are 11 possible distinct dynamics.
If you want to know how many of those possible dynamics involve you personally (in a 4-couple), it is the powerset of 3, minus 1 (the empty set). The powerset of 3 gives every group that doesn't contain you in it. So you can have a unique dynamic with each of these groups. This number turns out to be 7.
I believe that is just at the cusp of what a person can handle. 7 is not too many dynamics.
With 5 people on the other hand, the same formula gives 26 whole dynamics for everyone! And 15 of them involve you! It's too many dynamics. Most of them will just end up being copies of one another, or unexplored.
BTW, to get the powerset of a number, just take 2^number. Also, this is all based on what I think is reasonable about what a person can handle. Not to mention, that not every poss
Came for the polyam talk, but fuckyes i am so here staying for the set theory! I miss discrete maths tbh. Hardest maths course i ever took but absolutely the most rewarding and useful.
Could get someone who is genderfluid if you wanna stick to monogamy
Brilliant idea. But this requires more specialised circumstances.
To get the number of relationships youre interested in, its 2^n - n - 1 (which is also the sum of the number of binomial coefficients for a given n, and not for no reason). You could also set it up as a graph, then you have a complete graph (K n) and count subsets of at least size 2
Isn't the sum of binomial coefficients of n equal to 2^n? And then you remove (n choose 1) and (n choose 0) to get 2^n - n -1
Yeah, what the fuck was I thinking of??