# Let A and B be tw

Here, n(A)=3 and n(B)=6

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

=3+6-n(AB)

=9-n(AB)

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

Now,AB then max(n(AB))=n(A)=3.

min(n(AB) )=9-3=6

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