If the value of \(n(Y)+n(Z)\) is \(k^2\), then \(|k|\) is
Let \(S=\{1,2,3,4,5,6\}\) and \(X\) be the set of all relations \(R\) from \(S\) to \(S\) that satisfy both the following properties:
i. \(R\) has exactly 6 elements.
ii. For each \((a, b) \in R\), we have \(|a-b| \geq 2\).
Let \(Y=\{R \in X\) : The range of \(R\) has exactly one element \(\}\) and \(Z=\{R \in X: R\) is a function from \(S\) to \(S\}\).
Let \(n(A)\) denote the number of elements in a set \(A\).

given \(|a-b| \geq 2\) so if

i.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 \(\mathrm{n}(\mathrm{Y})=0\).
\(\begin{aligned} & \text { for } \quad n(z) \\ & 1 \rightarrow 3,4,5,6 \\ & 2 \rightarrow 4,5,6 \\ & 3 \rightarrow 1,5,6 \\ & 4 \rightarrow 1,2,6 \\ & 5 \rightarrow 1,2,3 \\ & 6 \rightarrow 1,2,3,4 \\ & \end{aligned}\)
no. of relation that are function will be \(\quad={ }^4 \mathrm{C}_1 \times{ }^3 \mathrm{C}_1 \times{ }^3 \mathrm{C}_1 \times{ }^3 \mathrm{C}_1 \times{ }^3 \mathrm{C}_1 \times{ }^4 \mathrm{C}_1\)
\(\begin{aligned} & =(4 \times 3 \times 3)^2=k^2 \\ & \text { i.e. } \mathrm{k}=36\end{aligned}\)