戰友們 !!fight!一起加油求助!!

hijenny

A committee of three people is to be chosen from four married couples.What is the number of different committees that can be chosen if two people who are married to each other cannot both serve on the committee?



(A)16

(B)24

(C)26

(D)30

(E)32

這題我做錯了
不過在看解答時仍然不懂
【答案】E

【思路】<作法1>committe中不能有couple的組合即 "全部的組合-committe中有couple的組合"
C(8,3)-4*6   (剪去4*6不就要算進去的人也剪掉了嗎??)
=56-24
=32



懇請賜教

hijenny

不理解...

yuyi325

C（4,1）*2^3=32

首先选出一对夫妻，两个人都不能入选committee。
然后剩下三对，每对夫妻里面选出一人，就是2^3，不晓得这样解释明白么？

danvirgil

C81*C61*C41=8*6*4=192

然后这样选出来的情况会有重复 比如A B C, A C B, B A C, B C A, C A B, C B A这六种是一样的

于是将192/6=32