It means you have 12 equally likely possibilities (girls) and you want to know how many 3 people combinations you can make out of them where the order of the 3 doesnt matter.


If we were talking about a horse race where the order of the top 3 would matter the number would be 1320.

yo im down for some fake Quading..lol

ill want to go as well though...i am a terror on them, just ask plat dave

Not more than 34 combinations, maybe a bit less if grouped in a smart way, I didn't want to think about that :-)

Math is not your friend :(

Proof me wrong.

I already did. Here it is a 2nd time:

http://www.kickbackcash.com/LSstats.jpg

kard this isnt really a mathematical proof
but i dont feel like typing it out

well done showing your work teach these peon schlebs a lesson

I did it earlier because steve said show your work for bonus :1orglaugh

Its the retarded ass combination theory from a first year statistics class and I think its proof. When you use a theory or formula do you also prove it?

You are right i was just going to say that

I just realized how horrible my math skills are:(

50 Panamanian hookers.

Who needs a degree?

http://www.google.com/search?sourcei...&q=12+choose+3

Just imagine 12 girls, even if you'd only need to group them in groups of 2 you'd only need 121 combinations (11x11), if you can group them in groups of 3 then you'd need less combinations to ensure that "each girl performs at least once with each of the other girls", not more.

Or just imagine 4 Girls, to make it simplier let's call them A;B;C and D. To group them in groups of 3 so that each girl is ounce with each other you'd have to make the following combinations ABC; DAB and DC?. So you'd only need to make 3 combinations and you still have one place left for an additional girl (so you can save a few combinations if you group them smart) With your calculation you'd need 4 combinations. If you take 5 Girls A;B;C;D;E then you'd need ABC; DAB; EAB and ECD, 4 combinations, not 10 as with your calculation. So I don't know what you've calculated there, but 220 isn't right.

But I have to say that I can't mathematical proof my own number of combinations (34) at this time and it might be possible that I'm wrong too, but I can give you some kind of "circumstantial evidence"

If you'd just have that you'd have to group in groups of random 3 then you'd have with:

3 Girls, 1 combination neccessary
4 Girls, 3 combinations neccessary
5 Girls, 5 combinations neccessary
6 Girls, 8 combinations neccessary
7 Girls, 11 combinations neccessary
8 Girls, 15 combinations neccessary
9 Girls, 19 combinations neccessary
10 Girls, 24 combinations neccessary
11 Girls, 29 combinations neccessary
12 Girls, 35 combinations neccessary

If you can group them in a smart way you can save a few combinations:

3 Girls, 1 combination neccessary
4 Girls, 3 combinations neccessary
5 Girls, 4 combinations neccessary
6 Girls, 7 combinations neccessary
7 Girls, 9 combinations neccessary
8 Girls, 13 combinations neccessary
9 Girls, 16 combinations neccessary
10 Girls, 21 combinations neccessary
11 Girls, 27 combinations neccessary
12 Girls, 33 combinations neccessary





:glugglug

Ummmmmmmmmmmmm

hj,

Gonna pretend you aren't joking and post this. The answer is 220. Here are the first 55 combinations if you number the girls 1 through 12.


1 2 3
1 2 4
1 2 5
1 2 6
1 2 7
1 2 8
1 2 9
1 2 10
1 2 11
1 2 12
1 3 4
1 3 5
1 3 6
1 3 7
1 3 8
1 3 9
1 3 10
1 3 11
1 2 12
1 4 5
1 4 6
1 4 7
1 4 8
1 4 9
1 4 10
1 4 11
1 4 12
1 5 6
1 5 7
1 5 8
1 5 9
1 5 10
1 5 11
1 5 12
1 6 7
1 6 8
1 6 9
1 6 10
1 6 11
1 6 12
1 7 8
1 7 9
1 7 10
1 7 11
1 7 12
1 8 9
1 8 10
1 8 11
1 8 12
1 9 10
1 9 11
1 9 12
1 10 11
1 10 12
1 11 12

Steve want's "each girl to perform one triple blowjob with each of the other girls" which means to me that each girl only have to be ONCE with another, so as I see it you don't have to make 55 attempts to group the first girl with each other of the remaining 11 you' only have to make 6 combinations.

1 2 3
1 4 5
1 6 7
1 8 9
1 10 11
1 12 ?

In your example girl 1 is blowing 10 times together with girl 2, 9 times with girl 3 etc, I don't think Steve wants to waste his time :-)

Nice. I took high level calculus classes and I still can't do math to save my fucking life.

I cant believe people actually figured this out :1orglaugh

This thread shows how many high school kids we have here ;)

Anyway, question is not really formulated well... is it 2 or 3 girls and if its 3 do you want every possible combination or just each girl with 2 different ones but never repeating any of the 2 you already had?

ok here is the correct answer:

12 girls giving triple blowjobs in every possible combination with each of the other girls = one very sore little red nub where my dick used to be.

"NO MORE YANKY MY WANKY!!!"

I'm headed home now. End of the Panamanian Adventure Weekend for me.


Steve Lightspeed

Ok you math skils arent terrible, you are wrong because you mis interpreted the question.

You are wrong, the guy who listed the first 55 combos is right.

I shot the cat on this one.


Thank god it waS NOT my cat. I think it belongs to the people next door.,


Oh well.

I can't believe this thread is still going on, lol.
WG