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a.since A_{1} \cap A_{2} \neq \phi so A_{1} is connected to A_{2} and A_{1} \cap A_{3}=\phi so A_{1} is not connected to A_{3} similarly we get for all the pairs.so we get thatA_{1} is connected to A_{2}, A_{4}, A_{5}A_{2} is connected to A_{1}, A_{3}, A_{5}A_{3} is connected to A_{2}, A_{4}, A_{5}A_{4} is connected to A_{1}, A_{3}, A_{5}A_{5} is connected to A_{1}, A_{2}, A_{3}, A_{4}so graph looks like following______________________________________________________________________b.similar to part a, hereA_{1} is connected to A_{2}, A_{3}, A_{4}, A_{5}A_{2} is connected to A_{1}, A_{3}, A_{4}, A_{5}A_{3} is connected to A_{1}, A_{2}, A_{5}A_{4} is connected to A_{1}, A_{2}, A_{5}A_{5} is connected to A_{1}, A_{2}, A_{3}, A_{4}so graph looks like followingusing same procedure here graph we obtain is ...