Question

if n (A) =8 , n (B) =14 , n (AuB) =18 , then find n (AnB) = ?

Answer 1

$\red{\large\underline{\sf{Given- }}}$

$\rm :\longmapsto\:n(A) = 8$

$\rm :\longmapsto\:n(B) = 14$

$\rm :\longmapsto\:n(A \: \cup \: B) = 18$

$\pink{\large\underline{\sf{To\:Find - }}}$

$\rm :\longmapsto\:n(A \: \cap \: B)$

$\green{\large\underline{\sf{Solution-}}}$

Given that

$\rm :\longmapsto\:n(A) = 8$

$\rm :\longmapsto\:n(B) = 14$

$\rm :\longmapsto\:n(A \: \cup \: B) = 18$

We know,

$\rm :\longmapsto\:n(A \: \cup \: B) = n(A) + n(B) - n(A \: \cap \: B)$

$\rm :\longmapsto\:18 = 14 + 8 - n(A \: \cap \: B)$

$\rm :\longmapsto\:18 = 22 - n(A \: \cap \: B)$

$\rm :\longmapsto\:18 - 22 = - n(A \: \cap \: B)$

$\rm :\longmapsto\: - 4 = - n(A \: \cap \: B)$

$\rm :\longmapsto\: n(A \: \cap \: B) = 4$

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More to know

$\red{\rm :\longmapsto\:n(A \cap \: B') = n(A) - n(A \: \cap \: B) \: }$

$\red{\rm :\longmapsto\:n(B \cap \: A') = n(B) - n(A \: \cap \: B) \: }$

$\red{\rm :\longmapsto\:n(B' \cap \: A') = n( \cup) - n(A \: \cup \: B) \: }$

$\red{\rm :\longmapsto\:n(B' \cup \: A') = n( \cup) - n(A \: \cap \: B) \: }$