If $\mathrm{A}=\{2,3\},\mathrm{B}=\{4,5\}$ and $\mathrm{C}=\{5,6\}$ , then $\mathrm{n}\left\{\right(\mathrm{A}\times \mathrm{B})\cup (\mathrm{B}\times \mathrm{C}\left)\right\}$ is

$\begin{array}{rl}\because \mathrm{A}\times \mathrm{B}& =\{2,3\}\times \{4,5\}\\ & =\left\{\right(2,4),(2,5),(3,4),(3,5\left)\right\}\\ \text{and}\mathrm{B}\times \mathrm{C}& =\{4,5\}\times \{5,6\}\\ & =\left\{\right(4,5),(4,6),(5,5),(5,6\left)\right\}\\ \therefore (\mathrm{A}\times \mathrm{B})& \cup (\mathrm{B}\times \mathrm{C})=\left\{\right(2,4),(2,5),(3,4),(3,5)\\ (4,5),(4,6),(5,5),\left(56\right)\}& \\ \text{Now,}\mathrm{n}\left\{\right(\mathrm{A}\times \mathrm{B})\cup (\mathrm{B}\times \mathrm{C}\left)\right\}=8& \end{array}$