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        riddles >> easy >> NBA teams
        
(Message started by: Altamira_64 on Nov 8^th, 2012, 9:09am)

Title: NBA teams
Post by Altamira_64 on Nov 8^th, 2012, 9:09am

We have 20 NBA players and want to create 4 teams of 5 players each. How many different ways are there?

Title: Re: NBA teams
Post by Altamira_64 on Nov 8^th, 2012, 9:14am

I thought it was 20!/15! but apparently it isn't...

Title: Re: NBA teams
Post by Grimbal on Nov 8^th, 2012, 9:31am

Maybe 20!/(5!^4)/4! = 488864376

Title: Re: NBA teams
Post by rmsgrey on Nov 9^th, 2012, 4:25am

on 11/08/12 at 09:31:07, Grimbal wrote:

Maybe 20!/(5!^4)/4! = 488864376

I agree with the formula (I haven't checked the arithmetic)

Take your 20 players and line them up (20! possibilities) Take the first five for team A, the next five for team B, 11-15 for team C and the remaining five are team D.

There are 5! ways of ordering the players on each time, so 5!⁴ ways of getting the same four teams assigned in the same order, but if you swap the labels of the four teams around, you end up with the same actual teams, just in a different order, and there are 4! ways of ordering the four teams.

So there are 5!⁴*4! ways of getting each (unordered) set of teams out of the original 20! orders of the players.

If it makes a difference which team is which, then it's 20!/(5!⁴) possible teams; if the teams are equivalent apart from their players, then it's Grimbal's figure.

Title: Re: NBA teams
Post by pex on Nov 9^th, 2012, 7:11am

Yet another way to get the same result:
- pick five people for Team A (20!/(5!15!) possibilities)
- from the remaining fifteen, pick five for Team B (15!/(5!10!))
- again for Team C (10!/(5!5!))
- Team D is fixed now, because only five people are left.
Overall, 20!/(5!15!) * 15!/(5!10!) * 10!/(5!5!) = 20!/(5!⁴) possibilities if it matters which team has which name. If not, divide by another factor of 4!.