1057They are asking the no of ways of dividing 5 numbers into two groups having no element in common keeping in mind that each of the five numbers is present in either of the groups.
I may be wrong but i think the answer is 25=32.
Shaswata Roy I had thought of that before.But you see the problem is that there might be a case where A = {1,2,3} and B = {2,3,4}. A and B are not equal but they can have common elements.Upvote·0· Reply ·2013-02-24 23:13:04- Shaswata Roy Oh yeah...then the intersection would not be 0.(Sorry)
Hardik Sheth but it is not mentioned that a union b should be s..- Hardik Sheth it shud be something like 2power5+2power4+2power3...
- Hardik Sheth nw wen we r taking 4 or 3 elements...combinations have to b considered..@man111 singh,is my approach ok or have i misinterpreted the question
- Shaswata Roy I think (i) and (ii) are 2 cases of the same question and it should be union of A and B = S (and not intersection).
- Hardik Sheth we can have a discussion on all three types...two given above and the recommended change one!
