单场淘汰赛 - 可能的组合数量
单场淘汰赛中 8 人参加的组合有多少种?比赛总数为 7 场,但我还需要这组比赛的组合数量
What are the number of combinations in which 8 persons taking part in a single elimination tornament play? Total no of matches played would be 7 but I also need the number of combinations that can for this set
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如果玩家在树中的哪个位置开始并不重要,而只关心他/她与哪些对手战斗以及他/她能坚持多久,我们可以说左边的玩家总是获胜,然后只需计算创建的方法数量最下面一行,是 8! 40320.
第一种可能:
第二种可能:
If it doesn't matter where in the tree a player starts, but only which opponents he/she fights, and how long he/she gets, we can say that the left player always wins and then just calculate the number of ways to create the bottom most row, which is 8! 40320.
The first possibility:
The second possibility:
有 (8 * 7) / 2 种组合 = 28
[ 换句话说,8!/(2! * (8-2)!) ]
使用 Perl 中的 Set::Partition 我可以这样写:
您
可以解释两个玩家在玩,而另外六个玩家则站在周围欢呼喝酒啤酒。
There are (8 * 7) / 2 combinations = 28
[ in other words, 8!/(2! * (8-2)!) ]
With Set::Partition in Perl I can write:
which gives
which you can interpret two players playing, and the six others standing around cheering and drinking beer.
如果您的意思是,8 名玩家中有多少可能的 2 人比赛,那么答案是 28 (8x7/2)。如果您有其他意思,请稍微澄清一下您的问题。
If you mean, how many possible 2 player matches are there in a pool of 8 players then the answer is 28 (8x7/2). If you mean something else, then clarify your question a bit.