Re: Math Question

From: Martijn (Kw3Ls_at_hotmail.com)
Date: 03/29/05 

Date: Tue, 29 Mar 2005 06:21:49 +0200

a_weisman@yahoo.com wrote:

> There are 64 teams.
>
> In round 1 there are 32 games; 16 in round 2; 8 in round 3; 4 in round
> 4; 2 in the final round; and one national champion.
>
> There could be 64 different possibilities for round 6 (who is national
> champion).

Shouldn't it be: 2 (games) in round 5; 1 (game) in round 6?

>
> But in round 1 what I'm asking is how many different combinations could
> there be? Etc for the whole tournament?

Total possible combinations of teams joining a round:
join: 64! / ( (64-T)! x T! )
Total possible combinations of teams paring:
pair: T! / ( (2^G) x G! )

T=teams, G=games

Round 1: (64 teams, 32 games)
join: 1
pair: 64!/((2^32)x32!) = 1.12275575 × 10^44
total= 1.12275575 × 10^44

Round 2: (32 teams, 16 games)
join: 64!/((64-32)!x32!) = 1.83262414 × 10^18
pair: 32!/((2^16)x16!) = 1.91898784 × 10^17
total= 1.83262414 × 10^18 x 1.91898784 × 10^17 = 3.51678344 × 10^35

Round 3: (16 teams, 8 games)
join: 64!/((64-16)!x16!) = 4.88526937 × 10^14
pair: 16!/((2^8)x8!) = 2027025
total= 4.88526937 × 10^14 x 2027025 = 9.90256314 × 10^20

Round 4: (8 teams, 4 games)
join: 64!/((64-8)!x8!) = 4426165368
pair: 8!/((2^4)x4!) = 105
total= 4426165368 x 105 = 464747363640

Round 5: (4 teams, 2 games)
join: 4!/((64-4)!x4!) = 635376
pair: 4!/((2^2)x2!) = 3
total= 635376 x 3 = 1906128

Round 6: (2 teams, 1 game)
join: 64!/((64-2)!x2!) = 2016
pair: 2!/((2^1)x1!) = 1
total= 2016 x 1 = 2016

Winner: (1 team)
join: 64!/((64-1)!x1!) = 64

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