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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- 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.

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
**Posts:**665**Joined:**Thu Dec 09, 2010 5:34 pm**Location:**Khulna,Bangladesh.