HardMathematics
+4 / -1
If A₁ ⊂ A₂ ⊂ A₃ ⊂ ⋯ ⊂ A₅₀, and n(Aᵢ) = i – 1, then n(⋂ᵢ₌₁₁⁵⁰ Aᵢ) is equal to:
A
49
B
10
C
11
D
0
Explanation
We know that, ⋂ᵢ₌₁ⁿ Aᵢ = A₁ ∩ A₂ ∩ A₃ .....∩ Aₙ. As per the question, A₁ ⊂ A₂ ⊂ A₃ ⊂ ⋯ ⊂ A₅₀. Thus, ⋂ᵢ₌₁₁⁵⁰ Aᵢ = A₁₁ ∩ A₁₂ ∩ A₁₃ .....∩ A₅₀ = A₁₁. Since, n(A₁₁) = 11 – 1 = 10. Hence, n(⋂ᵢ₌₁₁⁵⁰ Aᵢ) = 10.
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