Question

if n(A) =43,n(B)=51and n(A u B)=75, then n(A-B)u n(B-A)=

Answer 1

Given,

• n(A) = 43
• n(B) = 51
• n(AUB) = 75

We know ,

$\large \boxed{n(A \cup B) = n(A) + n(B) - n(A \cap B) }$

So,

$\mathsf{75 = 43 + 51 - n(A \cap B) }$

$\mathsf{75 = 94 - n(A \cap B) }$

$\mathsf{n(A \cap B) = 94 - 75 }$

$\mathsf{n(A \cap B) = 19 }$

For n(A - B),

$\large \boxed{n(A-B) = n(A) - n(A \cap B) }$

$\mathsf{n(A-B) = 43 - 19 }$

$\mathsf{n(A-B) = 27 }$

For n(B-A),

$\large \boxed{n(B-A) = n(B) - n(A \cap B) }$

$\mathsf{n(A-B) = 51 - 19 }$

$\mathsf{n(A-B) = 32 }$

Answer 2

note: cap means union in the below lines ...

GIVEN :

n(A) = 43

n(B) = 51

n(AUB) = 75

{n(A \cup B) = n(A) + n(B) - n(A  \cap  B) }

So,  

{75 = 43 + 51 - n(A  \cap B) }

75 = 94 - n(A \cap B) }

{n(A \cap B) = 94 - 75 }

{n(A \cap B) = 19 }

For n(A - B),

{n(A-B) = n(A) -  n(A  /cap B) }

{n(A-B) = 43 -  19 }

n(A-B) = 27 }

For n(B-A),

{n(B-A) = n(B) -  n(A  \cap B) }

{n(A-B) = 51 -  19 }

{n(A-B) = 32 }