Schedules conflict equivalent to serial schedule $T1 \rightarrow T2 :$
$R2(A) $ must occur after $W1(A),$ and $R2(B)$ must occur after $W1(B)$ and within a transaction, order of operations must remain same, hence, The only possibility is $R1(A), W1(A),\_,\_,\_,\_,R2(B), W2(B)$, in those four blank spaces, we can put remaining operations, hence, $4!/(2! \times 2!)$
Similarly for Schedules conflict equivalent to serial schedule $T2 \rightarrow T1 .$
Hence, answer 12.