Mathematics - Sets and Relations Question with Solution | TestHub

given $|a-b|\ge 2$ so ifi.e. Total elements in $\mathrm{X}$ is ${}^{20}{\mathrm{C}}_6$ Now for $\mathrm{n}(\mathrm{Y})$, range of $\mathrm{R}$ has exactly one element i.e. second elements must be constant in $\mathrm{R}$ and since $\mathrm{R}$ must have 6 element so it is not possible to satisfy both condition so $\mathrm{n}(\mathrm{Y})=0$. $\begin{array}{rl}& \text{ for }n(z)\\ & 1\to 3,4,5,6\\ & 2\to 4,5,6\\ & 3\to 1,5,6\\ & 4\to 1,2,6\\ & 5\to 1,2,3\\ & 6\to 1,2,3,4\\ & \end{array}$ no. of relation that are function will be $={}^4{\mathrm{C}}_1\times {}^3{\mathrm{C}}_1\times {}^3{\mathrm{C}}_1\times {}^3{\mathrm{C}}_1\times {}^3{\mathrm{C}}_1\times {}^4{\mathrm{C}}_1$ $\begin{array}{rl}& =(4\times 3\times 3{)}^2=k^2\\ & \text{ i.e. }\mathrm{k}=36\end{array}$