Probability

讓我們先將同區的４隊看為A組，而其他地區的４隊為Ｂ組
四個比賽均不可有同組出現的組合
＝（第一場比賽沒有同組的組合）＊（第二場比賽沒有同組的組合）＊（第三場比賽沒有同組的組合）＊（第四場比賽沒有同組的組合）
＝（從Ａ組抽出一隊的組合＊從Ｂ組抽出一隊的組合）＊（從餘下Ａ組抽出一隊的組合＊從餘下Ｂ組抽出一隊的組合）＊（從再下的Ａ組抽出一隊的組合＊從再餘下的Ｂ組抽出一隊的組合）＊（最後Ａ組的一隊組合＊最後Ｂ組的一隊組合）
＝(4*4)(3*3)(2*2)(1*1)=576
而４個比賽的總組合＝8*7*6*5*4*3*2＝40320
所以，四隊不會碰頭的機會率
＝（四場比賽沒有同組相對的組合）／（四場比賽的總組合）
＝576/40320
=0.0143
希望幫到你！

2008-03-14 12:22:11 補充：
若果你看四場比賽沒有分別時，那麼就要除4 *3*2*1
即總組合為40320／24＝1680
但同樣四場比賽沒有同組相對的組合也要除4*3*2*1
所以，答案是沒有改變
＝0.0143

2008-03-14 12:32:26 補充：
對不住因為１對３與３對１是一樣的，所以要再除２
所以，總組合＝１６８０／２＝８４０

但沒有同組的組合，因早以先考慮從Ａ組抽出，所以，不會出現這情況，因此，機會率是

（５７６／２４）／（１６８０／２）
＝０．０２８６

Let's change the question to 8 teams arrange in a line, 1x2, 3x4, 5x6 and 7x8 is arrange to have a game.
no. of permutation = 8!
There are 4! for same location teams and there are 4! for others.
Therefore, the probability = 4! x 4! / 8! = 576 / 40320 = 1/70

2008-03-12 16:59:27 補充：
As there are two positions for each match, which are 1x2 and 2x1.
Hence, the probability = 2 x 4! x 4! / 8! = 2 x 576 / 40320 = 1/35.

2008-03-13 13:22:28 補充：
Let's change the question to 8 teams arrange in a line, 1x2, 3x4, 5x6 and 7x8 is arrange to have a game.
no. of permutation = 8!
There are 4! for same location teams and there are 4! for others.

As there are two positions for each match, which are 1x2 and 2x1.
Hence, the probability = 2 x 4! x 4! / 8! = 2 x 576 / 40320 = 1/35.

假設有ABCDEFGH 8 隊
ABCD來自同一地區
Let the teams be A,B,C,D,E,F,G & H
where A,B,C,D come from the same region

比賽形勢可能是:
The match pair might be:
AB AC AD AE AF AG AH
BC BD BE BF BG BH
CD CE CF CG CH
DE DF DG DH
EF EG EH
FG FH
GH

可能性總數= 28
Total no. of possible match pairs = 28
碰頭次數= 3+2+1 = 6
No. of times the 2 teams in the pair are from the same regoin = 6

唔碰頭既概率
= 1 - 碰頭概率
= 1 - 6/28
= 11/14
P(teams not from the same region)
= 1 - 6/28
= 11/14

希望幫到你

If 4 teams will not meet. Each of the 4 teams will be allocated to 4 different match.
No. of choice for first match = 4
No. of choice for second match = 3
No. of choice for third match = 2
No. of choice for fourth match = 1
Therefore, no. of combination = 4x3x2x1 = 24

If there is no limitation,
no of combination = 8C2 x 6C2 x 4C2 x 2C2
= 28x15x6x1 = 2520

Therefore the required probability = 24/2520 = 1/105 or 0.009524

我覺得題目不清楚. 比賽是如何安排的? 是四場比賽同時舉行, 只舉行一次, 還是四場初賽 - 兩場半決賽 - 決賽? 因為會影響答案... 請說明