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Let $A$ and $B$ be two sets such that$n\left( A \right) = 20,n\left( {A \cup B} \right) = 42$ and $n\left( {A \cap B} \right) = 4$:
Find $n\left( {B - A} \right)$.

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Hint: In this question we have been given with two sets A and B and the values of some set operations are given, using these values we are supposed to find the number of elements that are in B but not in A, we can simply use the formula of intersection to obtain the elements in set B and then subtracting that with the intersection will give us the answer.
It is already given that,
$n\left( A \right) = 20$,
$n\left( {A \cup B} \right) = 42$
And,
$n\left( {A \cap B} \right) = 4$
Using the formula,
$n\left( B \right) = n\left( {A \cup B} \right) + n\left( {A \cap B} \right) - n\left( A \right)$
Therefore, on putting the values in the formula, we get,
$n\left( B \right) = 42 + 4 - 20$
$n\left( B \right) = 26$
$n\left( {B - A} \right) = n\left( B \right) - n\left( {A \cap B} \right)$
$n\left( {B - A} \right) = 26 - 4$
$n\left( {B - A} \right) = 22$
Note: While solving these questions, it is very important to apply the correct formula to obtain the answer. In this question, we used the formula of intersection first to obtain the elements in B and then subtracted with the intersection to get the answer.
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