For three non empty finite sets $A, B, C$ the following relations hold:
$B \cap (C – A) = \{ \} $
$B \cap C \not = \{ \}$
$A,B,C \subset N$ ($N$ is the set of natural numbers)
Given that $A = \{1,2,3,4\}, C = \{3,4,5,6\}$ and no element of $B$ is greater than the largest element of $C$ how many possible options are there for $B$?
Dhaka Secondary 2011/10
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Please don't post problems (by starting a topic) in the "Secondary: Solved" forum. This forum is only for showcasing the problems for the convenience of the users. You can post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
- leonardo shawon
- Posts:169
- Joined:Sat Jan 01, 2011 4:59 pm
- Location:Dhaka
Re: Dhaka Secondary 2011/10
$B \cap (C-A) = \{\}$
$B\cap {5,6} = \{\}$
[value of (C-A) ]
so,, $B\not = {5,6}$ [not equal]
cause $B\cap C \not =\{\}$
so the possible value of B is {3,4}
$B={3,4}$
$B\cap {5,6} = \{\}$
[value of (C-A) ]
so,, $B\not = {5,6}$ [not equal]
cause $B\cap C \not =\{\}$
so the possible value of B is {3,4}
$B={3,4}$
Ibtehaz Shawon
BRAC University.
long way to go .....
BRAC University.
long way to go .....
Re: Dhaka Secondary 2011/10
@Shawon, you have some problem with set theoretic notations, work on that.
You solution is close, think a bit more.
You solution is close, think a bit more.
"Je le vois, mais je ne le crois pas!" - Georg Ferdinand Ludwig Philipp Cantor
- Tahmid Hasan
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