Let first relation $R_1$ be $R_1(A_{11}, A_{12}, A_{21}, A_{22})$
Let’s write its Functional Dependencies
$A_{11} → A_{12} \\ A_{21} → A_{22}$
The Candidate Key here is $A_{11}A_{21}$ as $(A_{11}A_{21})^{+} = R_1$
$A_{11} → A_{12} \\ A_{21} → A_{22}$
dependencies violate the rule of 3CNF, So easiest way is to append $A_{21}$ to first and ${A_{11}}$ to second dependency.
Now we have
$R_3(A_{11}, A_{21}, A_{12}) \\ R_4(A_{11}, A_{21}, A_{22})$
Note that here for every table the CK is $(A_{11}A_{21})$ which is its super key. So these are in 3CNF.
Coming to the second relation in the ER Diagram
$R_2(A_{11}, A_{12}, A_{31}, A_{32})$
Here we have
$A_{31} → A_{32}$
$A_{31} → A_{11}A_{12}$ Using $A_{31}$ you can get the one row in $E_1$ that you need.
$A_{11} → A_{12}$,
Candidate Key and Super Key is $A_{31}$
Problematic Dependency is $A_{11} → A_{12}$ as it is a partial dependency
So decompose it into a separate table
$R_5(A_{11}, A_{12})$ with $A_{11}$ as Super Key
and
$R_6(A_{31}, A_{32}, A_{11})$ with $A_{31}$ as Super Key
Therefore Valid BCNF decompositions are $R_3, R_4, R_5$ and $R_6$.
Answer is 4
