Let A, B and C be finite sets such that A∩B∩C=ϕ and each one of the sets AΔB,BΔC and CΔA has 100 elements. The number of elements in A∪B∪C is

A
250
B
200
C
150
D
300

Solution:

Let n(X ) denote the number of elements in X.
Then

$\begin{aligned} n (A \cup B \cup C) & = n (A) + n (B) + n (C) - n (A \cap B) - n (B \cap C) - n (C \cap A) + n (A \cap B \cap C) \\ = Σn (A) - Σn (A \cap B) (since A \cap B \cap C = ϕ) \\ AΔB & = (A - B) \cup (B - A) = (A \cup B) - (A \cap B) \end{aligned}$

Therefore,

$\begin{aligned} n (AΔB) & = n (A \cup B) - n (A \cap B) \\ = n (A) + n (B) - 2 n (A \cap B) \\ and 300 & = Σn (AΔB) = Σ [n (A) + n (B) - 2 n (A \cap B)] \\ = 2 [Σn (A) - Σn (A \cap B)] \end{aligned}$

Therefore, $n (A \cup B \cup C) = Σn (A) - Σn (A \cap B) = \frac{300}{2} = 150$

Let A, B and C be finite sets such that $A \cap B \cap C = ϕ$ and each one of the sets $AΔB, BΔC and CΔA$ has 100 elements. The number of elements in $A \cup B \cup C$ is

Solution:

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