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If n[(A×B ) (A×C )]=8 and n(B∩C )=2 then n(A) is |
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Answer» Consider, n[(A x B)∩(A x C)]=8 n[A x (B∩C)] = 8 But n(A x B) = n(A) x n(B) => n(A) x n(B∩C) = 8 => n(A) x 2=8 => n(A) = 8/2 => n(A) = 4 |
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