$T1\rightarrow T2$
Because of $W_{1}(Y)$ and $R_{2}(Y)$ conflict there is one serializable schedule which is also a serial schedule.
$T2\rightarrow T1$
$R_{2}(Y)$ $W_{2}(Y)$ $R_{2}(Z)$ $W_{2}(Z)$ $\rightarrow$ $R_{1}(X)$ $W_{1}(X)$ $R_{1}(Y)$ $W_{1}(Y)$
Let's assume $R_{2}(Z)$ $W_{2}(Z)=Q$ and $R_{2}(Z)=M$ and $W_{2}(Z)=N$
$\underbrace{R_{1}(X)W_{1}(X)}$ $R_{2}(Y)$ $W_{2}(Y)$ $Q$ $R_{1}(Y)$ $Q$ $W_{1}(Y)$ $Q=3$
$\underbrace{R_{1}(X)W_{1}(X)}$ $R_{2}(Y)$ $W_{2}(Y)$ $M$ $R_{1}(Y)$ $N$ $W_{1}(Y)=1$
$\underbrace{R_{1}(X)W_{1}(X)}$ $R_{2}(Y)$ $W_{2}(Y)$ $M$ $R_{1}(Y)$ $W_{1}(Y)$ $N=1$
$\underbrace{R_{1}(X)W_{1}(X)}$ $R_{2}(Y)$ $W_{2}(Y)$ $R_{1}(Y)$ $M$ $W_{1}(Y)$ $N=1$
$Total =6$
$R_{2}(Y)$ $\underbrace{R_{1}(X)W_{1}(X)}$ $W_{2}(Y)$ $Q$ $R_{1}(Y)$ $Q$ $W_{1}(Y)$ $Q=3$
$R_{2}(Y)$ $\underbrace{R_{1}(X)W_{1}(X)}$ $W_{2}(Y)$ $M$ $R_{1}(Y)$ $N$ $W_{1}(Y)=1$
$R_{2}(Y)$ $\underbrace{R_{1}(X)W_{1}(X)}$ $W_{2}(Y)$ $M$ $R_{1}(Y)$ $W_{1}(Y)$ $N=1$
$R_{2}(Y)$ $\underbrace{R_{1}(X)W_{1}(X)}$ $W_{2}(Y)$ $R_{1}(Y)$ $M$ $W_{1}(Y)$ $N=1$
$Total=6$
$R_{2}(Y)$ $W_{2}(Y)$ $Q$ $\underbrace{R_{1}(X)QW_{1}(X)}$ $Q$ $R_{1}(Y)$ $Q$ $W_{1}(Y)$ $Q=5$
Similarly start putting M and N, you will get $10$ more ways.
$Total=15$
$\underbrace{R_{1}(X)}$ $R_{2}(Y)$ $\underbrace{W_{1}(X)}$ $W_{2}(Y)$ $Q$ $R_{1}(Y)$ $Q$ $W_{1}(Y)$ $Q=3$
Similarly start putting M and N, you will get $3$ more ways.
$Total=6$
$\underbrace{R_{1}(X)}$ $R_{2}(Y)$ $W_{2}(Y)$ $Q$ $\underbrace{W_{1}(X)}$ $Q$ $R_{1}(Y)$ $Q$ $W_{1}(Y)$ $Q=4$
Similarly start putting M and N, you will get $6$ more ways.
$Total=10$
$R_{2}(Y)$ $\underbrace{R_{1}(X)}$ $W_{2}(Y)$ $Q$ $\underbrace{W_{1}(X)}$ $Q$ $R_{1}(Y)$ $Q$ $W_{1}(Y)$ $Q=4$
Similarly start putting M and N, you will get $6$ more ways.
$Total=10$
$Finally\ for\ T2\rightarrow T1=6+6+15+6+10+10=53$
$Ans: 54$
