Question:
If $A$ and $B$ are two sets such that $n(A)=24, n(B)=22$ and $n(A \cap B)=8$, find:
(i) $n(A \cup B)$
(ii) $n(A-B)$
(iii) $n(B-A)$
Solution:
Given:
$n(A)=24, n(B)=22$ and $n(A \cap B)=8$
To Find:
(i) $n(A \cup B)$
$n(A \cup B)=n(A)+n(B)-n(A \cap B)$
$=24+22-8$
$=38$
Therefore,
$n(A \cup B)=38$
(ii) $n(A-B)$
We know that,
$n(A-B)=n(A)-n(A \cap B)$
$=24-8$
$=16$
Therefore,
$n(A-B)=16$
(iii) $n(B-A)$
We know that,
$n(B-A)=n(B)-n(A \cap B)$
$=22-8$
$=14$
Therefore,
$n(B-A)=14$