Question

$\large{ \underline{ \underline{ \red{ \sf{Question:-}}}}}$
If A, B, C are finite set, prove that:-

$\tt{n(A \cup B \cup C)= n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(C \cap A) + n(A \cap B \cap C) }$
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Answer 1

$\red{\large \mathfrak {qestion}}$

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$\tt{n(A \cup B \cup C)= n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(C \cap A) + n(A \cap B \cap C) }$

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$\green{\large \mathfrak {solution}}$

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We know that,

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$\tt \: n(A \cup B \cup C) = n(A \cup (B \cup C))$

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$\tt\: = n(A) + n(B \cup C) - n(A \cap (B \cap C))$

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$\tt∵ n(A \cup B) = n(A) + n(B) - n(A \cap B)$

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$\tt∴ A \cap (B \cup C) = (A \cup B) \cup (A \cap C) \: \: \: \: \: \: \sf \: ..(distributie \: property)$

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$\tt= n(A) + n(B \cup C) - n(A \cap (B \cap C))$

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$\tt= n(A) + [n(B) + n(C) - n(B \cap C)] - n(A \cap (B \cap C)$

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We know that,

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$\tt(B \cup C) = n(A) + n(B) + n(C) - n(B \cap C) - [n(A \cap B) + n(A \cap C) - n((A \cap B) \cap (A \cap C))]$

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$\tt= n(A) + n(B) + n(C) - n(B \cap C) - [n(A \cap B) + n(A \cap C) - n(A \cap B \cap C]$

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$\tt= n(A) + n(B) + n(C) - n(B \cap C) - n(A \cap B) - n(A \cap C) + n(A \cap B \cap C) = RHS$

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$\sf∴ \: LHS = RHS$

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Hence Proved

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$\pink{\large \mathfrak { \: !! \: hope \: it \: helps \: you \: !! }}$

Answer 2

Answer:

n(A∪B∪C)=n(A∪(B∪C))

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\tt\: = n(A) + n(B \cup C) - n(A \cap (B \cap C))=n(A)+n(B∪C)−n(A∩(B∩C))

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\tt∵ n(A \cup B) = n(A) + n(B) - n(A \cap B)∵n(A∪B)=n(A)+n(B)−n(A∩B)

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\tt∴ A \cap (B \cup C) = (A \cup B) \cup (A \cap C) \: \: \: \: \: \: \sf \: ..(distributie \: property)∴A∩(B∪C)=(A∪B)∪(A∩C)..(distributieproperty)

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\tt= n(A) + n(B \cup C) - n(A \cap (B \cap C))=n(A)+n(B∪C)−n(A∩(B∩C))

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\tt= n(A) + [n(B) + n(C) - n(B \cap C)] - n(A \cap (B \cap C)=n(A)+[n(B)+n(C)−n(B∩C)]−n(A∩(B∩C)

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We know that,

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\tt(B \cup C) = n(A) + n(B) + n(C) - n(B \cap C) - [n(A \cap B) + n(A \cap C) - n((A \cap B) \cap (A \cap C))](B∪C)=n(A)+n(B)+n(C)−n(B∩C)−[n(A∩B)+n(A∩C)−n((A∩B)∩(A∩C))]

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\tt= n(A) + n(B) + n(C) - n(B \cap C) - [n(A \cap B) + n(A \cap C) - n(A \cap B \cap C]=n(A)+n(B)+n(C)−n(B∩C)−[n(A∩B)+n(A∩C)−n(A∩B∩C]

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\tt= n(A) + n(B) + n(C) - n(B \cap C) - n(A \cap B) - n(A \cap C) + n(A \cap B \cap C) = RHS=n(A)+n(B)+n(C)−n(B∩C)−n(A∩B)−n(A∩C)+n(A∩B∩C)=RHS

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\sf∴ \: LHS = RHS∴LHS=RHS