Dhaka Secondary 2011/10

Problem for Secondary Group from Divisional Mathematical Olympiad will be solved here.
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Dhaka Secondary 2011/10

Unread post by BdMO » Fri Jan 28, 2011 10:13 pm

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 $B$?

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leonardo shawon
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Location: Dhaka

Re: Dhaka Secondary 2011/10

Unread post by leonardo shawon » Sat Jan 29, 2011 12:45 pm

$B \cap (C-A) = \{\}$
$B\cap {5,6} = \{\}$
[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}
Ibtehaz Shawon
BRAC University.

long way to go .....

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Avik Roy
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Re: Dhaka Secondary 2011/10

Unread post by Avik Roy » Sat Jan 29, 2011 12:59 pm

@Shawon, you have some problem with set theoretic notations, work on that.

You solution is close, think a bit more.
There are $12$ possible $B$
"Je le vois, mais je ne le crois pas!" - Georg Ferdinand Ludwig Philipp Cantor

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Tahmid Hasan
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Location: Khulna,Bangladesh.

Re: Dhaka Secondary 2011/10

Unread post by Tahmid Hasan » Sat Jan 29, 2011 6:48 pm

my solution is a set where one of 3 and 4 is mandatory and the elements are 1,2,3,4.
so the sets are{3},{4},{1,3}{1,4}{2,3}{2,4}{3,4}{1,3,4}{1,2,3}{1,2,4}{2,3,4}{1,2,3,4}
ans. 12.remember $B$ may not be equal to $a$
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