一道机井不明白,求助![转帖]

sunny888

一社区有45户, 订报纸, 有 R, S, T 三种报,给出了三种报的订户人数分别是, S1=28, S2=17, S3=12,then, subscribe, both R&S, b1=9, both S&T b2=8, both R&T b3=7; Now 给出了一个重要条件, 订了三种的和什么都没有订的人,一样. 然后说, at least subscribe one kind of newspaper,29.  问: Subscribe only one kind Newspaper 的人数 X=?   [讨论]：X=S1+S2+S3-2B1-2B2-2B3+3a, 其中a是三种报纸都订了的户数 a=45-29=16

怎么理解at least subscribe one kind of newspaper,29

sunny888

I think the number in this question is wrong..

gernerally, suppose that there are A people subscribe 3 kind of newspaper. S1= a1 S2=a2 and S3=a3, S1&S2=b1, S1&S3=b2, S2&S3=b3,  if the number of people who subscribe none of the 3 newspaer is the same as the numuber of people who subscribe all of the 3 newspaper, then how many people subscribe only one kind of newspaper?

my solution to this problem is:
suppose that the number of people who subscribe all of the 3 kinds of newspaper is x,
A-x= a1+a2+a3-(b1+b2+b3)+ x
x= [A+(b1+b2+b3)-(a1+a2+a3)]/2
then people subscibe only one kind of newspaper = A-x-(b1+b2+b3)+ 2x
= A+x-(b1+b2+b3)

sunny888

这是一道比较好的题目，我想有关集合关系方面的题目再难也就这样了。在考场上
做是比较难。
机经上的答案没错，X=S1+S2+S3—2（B1+B2+B3）+3A。这里我也不知道怎么解释，
自己画下图吧。
不过题目是有点小错误，主要是有关数字记忆有误，题目说，订了三种的和什么都
没有订的人一样.  然后说, at least subscribe one kind of newspaper,29.这样订了三种的
人数是A=45-29=16。但题目告诉了我们，both S&T b2=8, both R&T b3=7。矛盾就
出现在这里，因为A不可能大於S&T，也不可能大於R&T。想一想，怎么可能订了三种
反而比只订了两种的多呢？
除了具体数据有错误外，我认为这道题本身还是不错的，可以给我们一些启发。