kievkeepsitreal: obsessingmuch: ...

kievkeepsitreal:

obsessingmuch:

kievkeepsitreal:

obsessingmuch:

Word from the man himself!!!

So, 25.

I think you’re in for a shock tbh Kiev when you realise how many ships you’ve missed! :O

Fill me in, then! I know the law, not maths!

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.

The maths.

~~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.~~

~~5+10+10+5+1=31~~

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)

c=n!/(x!*(n-x)!

Which means, when we want to know how many possible threesomes there are in the Grimes family, we go about it like so:

threesomes=5!/(3!2!)=(1*2*3*4*5)/[(1*2*3)*(1*2)]=120/12=10

Eudaemonia

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