1. Tardigrade
  2. Question
  3. Mathematics
  4. Let X, Y, Z be subsets of U, where n(U) = 35, n(X) = 15, n(Y) = 22, n(Z) = 14 and n(X ∩ Y) = 11, n(Y ∩ Z) = 8, n(X ∩ Z) = 5, n(X ∩ Y ∩ Z) = 3 then n(X ∪ Y ∪ Z)' equals

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Q. Let $X$, $Y$, $Z$ be subsets of $U$, where $n(U) = 35$, $n(X) = 15$, $n(Y) = 22$, $n(Z) = 14$ and $n(X \cap Y) = 11$, $n(Y \cap Z) = 8$, $n(X \cap Z) = 5$, $n(X \cap Y \cap Z) = 3$ then $n(X \cup Y \cup Z)'$ equals

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Solution:

We know that $n(X \cup Y \cup Z) = n(X) + n(Y) +$
$n(Z) - n (X \cap Y) - n(Y \cap Z) - n(X \cap Z) + n (X \cap Y \cap Z)$
$= (15 + 22 + 14)-(11 + 8 + 5 )+ 3 = 30$
$\therefore n (X \cup Y \cup Z)' = n (U ) - n (X \cup F \cup Z) = 35-30 = 5$