333for the first part,an element can go to a or to b or to none...so no of ways are 35-1 ways...
- Hardik Sheth i dint read unordered pairs..!!
\hspace{-16}$Let $\bf{S = \{1,2,3,4,5\}}$. Then the no. of unordered pairs $\bf{\{A,B\}.}$\\\\\ of Subsets of $\bf{S}$ such that\\\\ $\bf{(i)\;\;\;\; A\cap B=\phi}$. Where $\bf{A\neq B}$\\\\ $\bf{(ii)\;\;\;\; A\cap B=S}$. Where $\bf{A\neq B}$\\\\
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2 Answers
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!
333