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The Answer is 3. But I think only 2 are required as R2 is many(E3) to one(E1) with total participation.

Thus tables formed E2R1(DFA) with pk as DA. R2(GHABC) with pk as G.

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Ans can't be two table, because all are strong entity representing.

But there is a case possible where 1:1 with total participation in E3,E2 then we could make 1 table for both E2 and E3 combined one.

As per your concept for 1:M is not possible, because it violates the relational model property which is multi valued attributes can't holds.

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