# Probability Question

There 32 teams in the UEFA Champions league draw. Each team is placed in 1 of 4 pots (i.e. pots A, B, C or D). Each pot has 8 teams. Each team will now be drawn into 1 of 8 groups (A, B, C, D, E, F, G & H) and each group will have four teams. The teams in pot A are the seeded teams and cannot be drawn against each other. How many different draw combinations are possible?

24! 8! * ---------- (3!)^8 8! for the number of ways 8 teams in pot A can be assigned 8 different groups. 24! ----------- for the number of ways the remaining 24 teams can be assigned 8 groups. (3!)^8

fsa-sucker - I am not sure what the answer is but your answer looks a little large. Let us see if someone else confirm.

I don’t understand what the first four pots are for. It seems that there should only be two pots - a seeded and an unseeded pot.

I agree Joey. Blastt, can you confirm that apart from pot A (the seeded pot), the teams in pot B, C and D could end up in the same groups (i.e. two pot B teams end up in group H)?

aren’t you just drawing 24 teams into 8 groups of 3?

Guys, Thanks for your responce. This is how the draw works: Pot A - seeded teams CHELSEA LIVERPOOL REAL MADRID AC MILAN BARCELONA MANCHESTER UNITED ARSENAL INTER MILAN Port B (unseeded) Porto Marseille Besiktas Valencia Schalke 04 Rosenborg Werder Bremen Lazio Port C (Unseeded) Olympiakos Benfica Celtic Shakhtar Donetsk Lyon Stuttgart Rangers Roma Pot D (unseeded) Sporting Lisbon Dynamo Kiev PSV Eindhoven CSKA Moscow Fenerbahce Sevilla/AEK Athens Steau Bucharest Slavia Prague The draw will begin by initially assigning each to to seeded teams to one of each groups (A to H). Subsequently, every other team in the other 3 pots will be drawn into a group (which will each have a seeded team). The process will be repeated until the draw is complete (i.e. until each group have 4 teams). I hope this helps.

I think you aren’t seeing the point of our questions. Can Porto and Marseille end up in the same group?

tobias Wrote: ------------------------------------------------------- > I think you aren’t seeing the point of our > questions. > > Can Porto and Marseille end up in the same group? Yes they can.

So in Group A, we can assign 1 of 8 seeded teams and then 24 choose 3 unseeded teams, in B we can assign 1 of 7 remaining seeded teams and then 21 choose 3 unseeded teams, etc. So it’s 8!*Product(3k choose 3) where k goes from 1 to 8. Too lazy to plug in numbers.

JoeyDVivre Wrote: ------------------------------------------------------- > So it’s 8!*Product(3k choose 3) where k goes from > 1 to 8. Too lazy to plug in numbers. Can someone explain this so I can plug in the numbers?

8! * 4 = 161280

vstolin Wrote: ------------------------------------------------------- > 8! * 4 = 161280 this does not look right… 8! * 4 = 40,320 is this the answer?

joey’s right, and this is the same answer as fsa-sucker’s. the answer is 14894166022402867200000.

joey, the pi sum largely cancels. 24!/21!3! * 21!/18!3! * 18!/15!3! etc.

i’m totally lost here!

justin7 Wrote: ------------------------------------------------------- > joey’s right, and this is the same answer as > fsa-sucker’s. the answer is > 14894166022402867200000. So it is - I was having trouble reading fsa-sucker’s. We need a richer format here for this stuff (it would be cool if we could post html for example). >joey, the pi sum largely cancels. 24!/21!3! * 21!/18!3! * 18!/15!3! etc. That’s pleasing isn’t it? It’s also just a, um, what do you call the generalization of the hypergeometric, so if we were dividing up 24 into groups of 6 it would be 24!/(6!)^4 and the numbers downstairs can be any positive numbers that add up to 24.