Q. 75.0( 1 Vote )

# Let A and B be tw

Here, n(A)=4 and n(B)=7

Now, n(AB) =n(A)+n(B)-n(AB)

=4+7-n(AB)

=11-n(AB)

So, n(AB) is maximum whenever n(AB) is minimum and it is possible only when AB=ϕ

Now, AB=ϕ then min(n(AB)) =0.

min(n(AB))=11-0=11

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