Given n(U) = 20 , n(A) = 12 , n(B) = 9 , n(A cap B) = 4 , where U is universal set, A and B are subs

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1.Set Theory

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Given $n(U) = 20$, $n(A) = 12$, $n(B) = 9$, $n(A \cap B) = 4$, where $U$ is the universal set, $A$ and $B$ are subsets of $U$, then $n({(A \cup B)^C}) = $

A

$17$

B

$9$

C

$11$

D

$3$

Solution

(d) $n(A \cup B) = n(A) + n(B) – n(A \cap B) = 12 + 9 – 4 = 17$

Now, $n({(A \cup B)^C}) = n(U) – n(A \cup B) = 20 – 17 = 3$.

Standard 11

Mathematics

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