求助一道manhattan数学ps难题！！

elaine0

In a room filled with 7 people, 4 people have exactly 1 sibling in the room and 3 people have exactly 2 siblings in the room. If two individuals are selected from the room at random, what is the probability that those two individuals are NOT siblings?
      A  5/21

      B  3/7

      C  4/7

       D 5/7

       E 16/21

答案是E，曼哈顿官方是这么解释的：
We are told that 4 people have exactly 1 sibling. This would account for 2 sibling relationships (e.g. AB and CD). We are also told that 3 people have exactly 2 siblings. This would account for another 3 sibling relationships (e.g. EF, EG, and FG). Thus, there are 5 total sibling relationships in the group.
Additionally, there are (7 x 6)/2 = 21 different ways to chose two people from the room.
Therefore, the probability that any 2 individuals in the group are siblings is 5/21. The probability that any 2 individuals in the group are NOT siblings = 1 – 5/21 = 16/21.
The correct answer is E.
但是我认为如果是A B C D E F G七个人的话，如果这么排列：AB CE DF EG FG也符合题目条件，这么一来官方解释就不对了。是不是我想错了，求nn解答！

elaine0

求解答啊 马上就要考试了很急啊

dldrzz000

非N试解答：

中间目标：求个体之间亲戚的关系数目。
如图，有二对关系的人物为红色，只有一对关系的人物为绿色。（LZ所说情况为图2）
画图法：这题感觉还是画图更容易理解些，如果题干再多个条件：无三角以上关系和无光棍，就更好得出无论怎么画都是五条关系的图（实际上被结果排除了）
总共选择方法有C 7|2=21对，亲戚关系为5对，结果很容易得出了。
数式法：从上面画图法不难看出这题本质上就是道魂淡的数字题，令关系总数为x，则有2x=4*1+3*2，x=5，意思就是亲戚关系总数*2=有2个亲戚关系的人数*2+有1个亲戚关系的人数


这种题目要是真出现考场中俺就要砸键盘了……
抛砖引玉，仅供参考，欢迎鸡蛋。

elaine0

dldrzz000 发表于 2014-4-19 23:42
非N试解答：

中间目标：求个体之间亲戚的关系数目。

说的对啊 就算还有别的分法也还是5对 是我没想到 谢谢~