There is only one way to have (conflict) serializable schedule as $T1 \rightarrow T2$, because last operation of $T1$ and first operation of $T2$ conflicts with each other.
Now See How many schedules are conflict serializable to $T2 \rightarrow T1$.
I am writing $T1-$
$R(A)$ $W(A)$ $R(B)$ $W(B)$
If you notice, I wrote $T1$ with space in between operation.
Now See $T2$ from right, if we see $T2$ from right, then tell me first operation of $T2$ that conflicts with any operation of $T1$.
$W(C)$ and $R(C)$ do not have any conflict with any operation, but $W(B)$ has.
Pick $W(B)$ and see, at how many places it can be there.
Case1: $\bf{W(B)}$ $R(A)$ $W(A)$ $R(B)$ $W(B)$
Case2: $R(A)$ $\bf{W(B)}$ $W(A)$ $R(B)$ $W(B)$
Case3: $R(A)$ $W(A)$ $\bf{W(B)}$ $R(B)$ $W(B)$
Pick each case and see,how many positions other operation of $T2$ can take.
Case1: $\bf{W(B)}$ $R(A)$ $W(A)$ $R(B)$ $W(B)$
How many positions $W(C)$ and $R(C)$ can take ?
(note that these $W(C)$ and $R(C)$ cant come before $\bf{W(B)}$)
that is $5C1 + 5C2 = 15$ (either both can take same space or two different spaces)
Now see, for each of these $15$ positions, how many can $R(B)$ take ?
Obviously, $W(B)$ cant come before $R(B),$ therefore one position.
$15 \times 1 = 15$ total possible schedules from case $1.$
Case2: $R(A)$ $\bf{W(B)}$ $W(A)$ $R(B)$ $W(B)$
How many positions $W(C)$ and $R(C)$ can take ?
that is $4C1 + 4C2 = 10$ (either both can take same space or two different spaces)
Now see, for each of these $10$ positions, how many can $R(B)$ take ?
Only $2$ positions, because it has to come before $\bf{W(B)}$.
$10 \times 2 = 20$ total possible schedules from case $2.$
Case3: $R(A)$ $W(A)$ $\bf{W(B)}$ $R(B)$ $W(B)$
How many positions $W(C)$ and $R(C)$ can take ?
that is $3C1 + 3C2 = 6$
Now see, for each of these $6$ positions, how many can $R(B)$ take ?
Only $3$ positions, because it has to come before $\bf{W(B)}$.
$6 \times 3 = 18$ total possible schedules from case $3.$
total schedules that are conflict serializable as $T2 \rightarrow T1 = 15+20+18 = 53.$
total schedules that are conflict serializable as $T1 \rightarrow T2 = 1.$
total schedules that are conflict serializable as either $T2 \rightarrow T1$ or $T1 \rightarrow T2 = 53+1 = \bf{54}$.