You haven't told us how many couples are same sex. Obviously for every F-F pair there has to be a M-M pairing, so question works for 0,4 or 8 people paired with a member of the same sex. For 0 the answer is 2 possible combinations? I have no real proof because I have no idea how to write a proper mathematical proof. Each man can only be paired with 1 or 2 women on the table according to the rules. If he is paired with the left one then every other man has to be paired with the left. I wont bother solving for more same sex couples because im fairly sure you dont care and its hard. my question is solve seahawks problem with 2 same sex couples in the mix. I dont know the answer, i am sure its not hard, but it is boring. If you want to just lie and hit us with another question I wont call you out on it.

1 *of 2 women

Is it 480?

Start by putting the men in odd seats, (1,3,5,7,9)

there are 5! possibilities here (five possible men to be chosen for seat 1 then four for seat 2 etc.)
Then for each combination there are only two women who can sit in seat 2 without breaking the rules (the wives of the men in seats 5 and 9). Also there are only two seats each woman may sit in (three places either side of their husband). This means there are only two seating arrangements of women for each of the 5! arrangements of men.

5! * 2 = 240
Double this for the case where women are in odd seats to get 480.

^ Assuming all 5 are heterosexual couples

This isnt exactly a proof, but sze did ask that we link LaTeX if we used it: http://gifsoup.com/webroot/animatedgifs/980465_o.gif

couples are all 1 male 1 female, sorry if there was confusion dead piggy, and nich was correct so its his turn

I like this thread :)

Problem number 4, secondary alcohols (Chemistry)

What will you get if you oxidize a secondary alcohol?

(Lead: secondary alcohols are alcohols which the OH-group bind to a carbon atom, who bonds with two alkyle-groups (sorry for spelling)

http://en.wikipedia.org/wiki/Alcohol_oxidation a ketone?

Yes, a ketone

question 5: prove 3.333recurring=1/3

*ahem* question 5: prove 3.3recurring=10/3 and I will need to see you LaTeX for an answer to be considered valid.

Well, since 3 fits three times in 10, with a 1 left.
And 0,3 fits three times in 1 as well, with 0,1 left.
And because the laws of maths say that if you divide something the comma (or whatever sign people use to separate decimals and whole numbers) moves one spot to the right, this continues forever. Thus 10/3=3,3333...
Good?

Nope. Prove the "laws of maths". And do it quick before someone just googles the answer. This is actually an interesting fundamentally important tenet of mathematics, and noone ever knows it. It also creates arguments =D

Sorry, I forgot this: http://www.latex-project.org/

http://www.latex-project.org/

Noone wants to see that, you sicko. If you arent going to play the game then I wont give you any more hints.

Haha, found it:

Actually, that link will only lead you to a site describing LaTeX ;) You're being the sick one here.

Haha, you think i didn't click a link called latex project? Great work, you are correct.

Okay, here's a new problem:
What's the number of possible ways you can put notes and rests in a single 4/4 measure of sheet music? Not counting different clefs, and no notes or rests shorter than hemidemisemiquavers. And you can't just put lots of rests in different orders, but right next to each other, because when combined they last the same time. (I mean you can't have two quarter rests next to each other for example, because they would sound the same as a halve rest)

Good luck with that!