GWD MATH 5-3

qianyangnet · 发表于 2011-1-6 22:23:48

Q3:
If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?
A. 20
B. 40
C. 50
D. 80
E. 120
答案是D

如何解，向大家请教！谢谢

heywendy · 发表于 2011-1-6 22:33:25

我为啥算的是D呢。。。给出思路看看对不对··
因为不能是出自一个家庭，所以这3个人必须来自3个家庭。因此C53
3个家庭中，每个家庭有2个人，要从3个家庭各选1人，就是C21*C21*C21
所以答案是 C53*C21*C21*C21=5*4*2*2*2=120 选D

sdcar2010 · 发表于 2011-1-6 22:39:45

1st choice: 10 possibilities
2nd choice: 8 possibilities (removing one family)
3rd choice: 6 possibilities (removing the second possibilites)

Then among the chosen 3, no need to differentiate the 1st, 2nd and 3rd: final answer = (10*8*6)/3! = 80

heywendy · 发表于 2011-1-6 22:47:15

1st choice: 10 possibilities
2nd choice: 8 possibilities (removing one family)
3rd choice: 6 possibilities (removing the second possibilites)

Then among the chosen 3, no need to differentiate the 1st, 2nd and 3rd: final answer = (10*8*6)/3! = 80

-- by 会员 sdcar2010 (2011/1/6 22:39:45)

恩恩我看懂你的答案了但是我不知道为什么我的算法是错的呢？

sdcar2010 · 发表于 2011-1-6 23:06:38

C53=5*4*3.

Then you need to devide the product by 3! to remove the overcounting.

anono2010 · 发表于 2011-1-7 02:51:27

(10*8*6)看懂了，为什么要除以3！呢？
不好意思，我的排列组合很混乱。

heywendy · 发表于 2011-1-7 14:44:07

C53=5*4*3.

Then you need to devide the product by 3! to remove the overcounting.

-- by 会员 sdcar2010 (2011/1/6 23:06:38)

啊竟然犯了这么低级的错误。。。谢谢！！！

GWD MATH 5-3

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