Let ${\text{S}}_1=\{1,2,3,\dots ,20\} , {\mathrm{S}}_2=\{\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d},\mathrm{e}\} , {\mathrm{S}}_3=\{\mathrm{a},\mathrm{c},e,f\}$ then the number of elements of $\left({\mathrm{S}}_1\times {\mathrm{S}}_2\right)\cap \left({\mathrm{S}}_1\times {\mathrm{S}}_3\right)$ is 

$$\begin{gathered} S_{1}X{S}_2=100 and S_{1}X{S}_3=80 \\ \left(S_1\times S_2\right)\cup \left(S_1\times S_3\right)=120 \\ \left(S_1\times S_2\right)\cap \left(S_1\times S_3\right)=S_1\times S_2+S_1\times S_3-\left(S_1\times S_2\right)\cup \left(S_1\times S_3\right) \\ \therefore \left(S_1\times S_2\right)\cap \left(S_1\times S_3\right)=100+80-120 \end{gathered}$$

No of elements of $\left({\mathrm{S}}_1\times {\mathrm{S}}_2\right)\cap \left({\mathrm{S}}_1\times {\mathrm{S}}_3\right)\text{ is }60$