maryland

Anyway, thank u very much! I will take the exam tomorrow. Whatever the result would be, I enjoy the time spent here and really appreciate your help and others' as well. Good luck to you all!!!  [em23][em12][em06]

tongxun

maryland：谨祝你成功！

pgclt

Both formulars are correct.

1) T=AUBUC+（非A非B非C的元素个数）=A+B+C-(AnB)-(AnC)-(BnC)+(AnBnC) +（非A非B非C的元素个数）
------------------------------------
(AnB), (AnC), (BnC) 求有且只有2种

2) T=AUBUC+（非A非B非C的元素个数）=A+B+C-(AnB)-(AnC)-(BnC)-2(AnBnC) +（非A非B非C的元素个数）
-----------------------------------------
(AnB), (AnC), (BnC) 求有2种or3种(AnBnC).

maryland

谢谢pgclt!

但是这两个式子的算法明显不同啊,如果是"求有2种or3种(AnBnC)"的话,把第一个式子中的
两两相交和三三相交加起来不是就行了吗? 能否举个例子说明它们的不同呢?谢!!

maryland

啊，我明白了。其实公式不冲突，只是问题不同，所以求解的式子不同。现在才想明白，不好意思。[em09] 再次谢谢！

pgclt

You are welcome.  It's easier to see if you draw a graph with 3 circles. Good luck with your exam!

pgclt

maryland, I was wrong about the two formulars.  the conditions should reverse. hope it didn't confuse you more. thx jupiter for his correction. sorry!

1) T=AUBUC+（非A非B非C的元素个数）=A+B+C-(AnB)-(AnC)-(BnC)+(AnBnC) +（非A非B非C的元素个数）
------------------------------------
(AnB), (AnC), (BnC) 有2种or3种(AnBnC).

2) T=AUBUC+（非A非B非C的元素个数）=A+B+C-(AnB)-(AnC)-(BnC)-2(AnBnC) +（非A非B非C的元素个数）
-----------------------------------------
(AnB), (AnC), (BnC) 有且只有2种