Fill me in, then! I know the law, not maths!
Word from the man himself!!!
I think you’re in for a shock tbh Kiev when you realise how many ships you’ve missed! :O
Good point, Kiev, good point.
The math is here and here for 26 ships (the most likely way in which ships should be considered.) Did you forget a fivesome?
Seems I must have missed one threesome, then. Good calculating, birdie.
It’s 31 if you count in self-pleasuring ships. And why not, let’s not discriminate on what could arguably be considered the greatest form of love. But otherwise 26.
Say you have n people.
Singles: n. In this case 5.
Pairs: Arichmetric progression (1 step) from 1 to (n-1). In this case: 1+2+3+4=10
Threesomes: Arichmetric progression (1 step) from 1 to (n-2) plus from 1 to (n-3) and so forth till you only have one to add (n-(n-1). In this case: (1+2+3)+(1+2)+1=10
Foursomes: Otherwise you’d continue with the same pattern. But since we have 5 people (and we’re at n-1), we simply add n. In this case 5.
Fivesomes: Otherwise, you’d continue with the same pattern. But since we have 5 people, we simply add 1.
I’m sorry, I’m still at it. While my method is better than counting the combinations, it’s still really cumbersome and you have to remember a lot of mental pyramids. I just came up with it experimentally/logically.
Conventionally geometric progression is used for calculating maximum number of combinations. It’s really easy to use with any number of objects.
n - number of objects (people in the Grimes family in this case)
c - number of combinations
x - number of objects chosen at a time (threesomes, pairings etc in this case)
Which means, when we want to know how many possible threesomes there are in the Grimes family, we go about it like so: