Question:
If $A$ and $B$ are two sets such that $n(A-B)=24, n(B-A)=19$ and $n(A \cap B)=$ 11, find:
(i) $n(A)$
(ii) $n(B)$
(iii) $n(A \cup B)$
Solution:
Given:
$n(A-B)=24, n(B-A)=19$ and $n(A \cap B)=11$
To Find:
(i) n(A)
We know that,
$n(A)=n(A-B)+n(A \cap B)$
$=24+11$
$=35$
Therefore, $n(A)=35 \ldots(1)$
(ii) $n(B)$
We know that,
$n(B)=n(B-A)+n(A \cap B)$
$=19+11$
$=30$
Therefore,
$n(B)=30 \ldots(2)$
(iii) $n(A \cup B)$
We know that,
$n(A \cup B)=n(A)+n(B)-n(A \cap B)\{$ From $(1) \&(2) n(A)=35$
and $n(B)=30\}$
$=35+30-11$
$=54$
Therefore,
$n(A \cup B)=54$