一道令我费解的模考题

Xiaohuan_Mo

今日做PRINCETON模考3，一道数学题令我郁闷。请大牛们解救我：
Five people meet and exchange handshakes. If exactly 3 of the people each shake hands with 3 people, and if 1 person shakes hands with only 1 other person, then what is the least number of handshakes that could have been exchanged?
Answer: 6

How can we get the result from the conditions?
Thanks a lot in advance.

hz

ABCDE5人，假设ABCD4人之间两两握手，C(4,2)，这样有4人均与3个人握过手，为满足题目条件，取消其中C与D的握手，则只有AB两人与3人握过手；再让D与E握手，这样就有ABD与3人握手，E仅与1人握手。
所以，C(4,2)-1+1=6。

sxmeimei

I am not sure if my thoughts right or no, but post it anyway:

For the first condition: If exactly 3 of the people each shake hands with 3 people, the least number
should be 4 to satisfy this condition.

For the second condition: the least number should be 2 to satisfy this condition.

It's obvious that there is no overlap for the 4 and 2. So the answer should be 4+2=6.

I'd like to discuss more with you if you find something wrong with me.


[glow=255,red,2]650[/glow]

[此贴子已经被作者于2003-10-24 3:38:26编辑过]

hz

没理解。
由第一个条件怎么得出4？
由第二个条件得到的最小数不是2，应该是1啊？