Going Back to University!

[Solly]

Has anyone spent a few years away from their uni (college) town and then gone back? We're planning a trip down to Wales this summer, but I've been hearing that a couple of our favourite places have closed etc etc. There'll always be the beach of course unless Global Warming takes it away, but what are your experiences?

[KrazyHorse]

If I were you I would ask for my money back while I was there...

[VetLegion]

ouch!

[Dinner]

Sometimes the truth hurts.

[Ninot]

I'm going back to university tommorow! It's been days!

[Solly]

Originally posted by Oerdin
Sometimes the truth hurts.

KH is still smarting from when I pwned him in a stats question the dumb ****.

[KrazyHorse]

Originally posted by Doddler


KH is still smarting from when I pwned him in a stats question the dumb ****.

a) It was a probability/combinatorics question, not statistics

b) You're an idiot. Dauphin and I were the only ones talking truth in that thread. Dumbass

[Solly]

a) combinatorics is not a word
b) no-one showed that my answer was wrong

[Kuciwalker]

Originally posted by Doddler
combinatorics is not a word

[Kuciwalker]

My math and CS professors use it an awful lot for a non-word

[Solly]

Originally posted by Kuciwalker
My math and CS professors use it an awful lot for a non-word

Then it must be a word!

[Solly]

[Kuciwalker]

Guys, should I even bother?

[KrazyHorse]

Originally posted by Doddler
b) no-one showed that my answer was wrong

What, the answer to the first question?

No, it's not...

List the teams in some permuted order. Pair up the teams in positions 1&2, 3&4 etc
The number of permutations of 8 teams is 8! = 40320

However, the pairings have some symmetries associated with them. Flipping team 1 with team 2 does not have an effect on the actual pairings. Therefore you first divide by 2^4 as a symmetry factor

The next symmetry is on any permutation of the 4 pairs. i.e. 12345678 is the same as 56347812. This next symmetry gives a factor of 4! = 24

So the number of distinct sets of pairings is 8!/(2^4 * 4!) = 105

Your answer is thus ruled out (there does not exist n an integer s.t. n/105 = 19/42)

It's pretty simple to do this, actually. Take any two British teams. Their probability of playing each other is 1/7. There are three British teams, so there are three different Brit pairings. Those are mutually exclusive, because ManU can't play both Chelsea and Liverpool at the same time.

So it's just a 3/7 chance, which off the top of my head is about 43%.

3 in 7

Combination 8 / 2 = 28

Divide by the 4 games in the round = 7, 3 of which are the English clubs playing each other.

[KrazyHorse]

As for (a):

http://www.answers.com/combinatorics&r=67

You're a total ****ing idiot, Doddler.