Question
If $\mathrm{n}(\mathrm{A} \cap \mathrm{B})=5, \mathrm{n}(\mathrm{A} \cap \mathrm{C})=7$ and $\mathrm{n}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})=3$, then the minimum possible value ofn $(B \cap C)$ is(1) 0(2) 1(3) 3(4) 2
Step 1
First, we know that n(A ∩ B ∩ C) = 3, which means that there are 3 elements that are common to sets A, B, and C. Show more…
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