During March Madness,64 college basketball teams try to battle their way into the Final Four,or the semifinals of the national tournament.From the original 64 teams,how many different Final Four groupings can there be?答案是635,376

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During March Madness,64 college basketball teams try to battle their way into the Final Four,or the semifinals of the national tournament.From the original 64 teams,how many different Final Four groupings can there be?答案是635,376
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During March Madness,64 college basketball teams try to battle their way into the Final Four,or the semifinals of the national tournament.From the original 64 teams,how many different Final Four groupings can there be?答案是635,376
During March Madness,64 college basketball teams try to battle their way into the Final Four,or the semifinals of the national tournament.From the original 64 teams,how many different Final Four groupings can there be?
答案是635,376

During March Madness,64 college basketball teams try to battle their way into the Final Four,or the semifinals of the national tournament.From the original 64 teams,how many different Final Four groupings can there be?答案是635,376
来给你通俗的解法,我上次考试唯一sat数学800
首先,64个队要选出4个队出来的概率是?
不管了,我们假设ABCD队晋级!
也就是A进入四强的几率是：1/64
B进入四强的几率是：1/63
C进入四强的几率是：1/62
D进入四强的几率是：1/61
也就是按照这个顺序ABCD进入四强的的几率是他们的乘积：15249024
但是,这个顺序可以变化啊,我们少算了另外的几种可能,
即：ABCD
ABDC
ACBD
ACDB
ADCB
ADBC
BACD
BADC
BCDA
BCAD
BDAC
BDCA
.
我就不算了,用6X4=24来算,这个你懂吧?
用刚才15249024除以可能性24=635376
就是答案了.
如果你学过排列组合,可以用排列组合来解.
类似的题目如：有8个人,要配对问有多少种配对方法等.

64*63*62*61/4*3*2*1=635376

不懂，答案怎么会是635?

原理就是从 64 个大学篮球队中任意抽出 4 个队的组合，公式为：
64!/(60! * 4!) = 635,376

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