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Consider the following Entity-Relationship $(E-R)$ diagram and three possible relationship sets I, II and III) for this $E-R$ diagram:

 

If different symbols stand for different values (e.g., $t _{1}$ is definitely not equal to $t_{2}$ ), then which of the above could not be the relationship set for the $E-R$ diagram ?

  1. I only
  2. I and II only
  3. II only
  4. I, II and III
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Answer would be $I$ and $II$ relationship. i.e. $I$ and $II$ are Invalid Tables for the given ER Diagram.

Because There is Many to One relationship from $P$ to $T$ in the ER Diagram. So, An element $p_1$ of $P$ can be related to at most One element of $T$ and An Element of $T$ can be related to any number of elements of $P$. So, Seeing $I$ and $II$ relationships, $p_1$ is related to $t_1$ and $t_2$. Thus, These are Invalid RDBMS Tables for the Given ER Diagram.

Coming to $III$, It is fine. There is Many to Many relationship from $P$ to $Q$ in the ER Diagram. So, An element $p_1$ of $P$ can be related to  any number of  elements of $T$ and An Element of $T$ can be related to any number of elements of $P$.

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