Answer:
True
Step-by-step explanation:
[tex]A\cup B=A\cup(A-B)\\\\A\ and\ A-B\ are\ disjoint\ that\ means\ there\ is\ no\ common\ element\ in\ these\ two.\\\\n(A\cup B)=n(B\cup(A-B))\\\\n(A\cup B)=n(B)+n(A-B)\ \ \ \ \ \ \ as\ for\ disjoint\ set\ P,Q\ n(P\cup Q)=n(P)+n(Q)\\\\(n(P-Q)=n(P)-n(P\cap Q))\\\\n(AUB)=n(B)+n(A-B)\\\\n(AUB)=n(B)+n(A)-n(A\cap B)\\\\n(AUB)=n(A)+n(B)-n(A\cap B)\\\\Hence\ given\ statement\ is\ true[/tex]
