QJEEAdv24Ma5559: JEE Advanced 2024 Mathematics - Class 11 JEEAdv24Ma5559 Solution: If the value of is , then is _________.

given $|a-b| \geq 2$ so if

i.e. Total elements in X is ${ }^{20} \mathrm{C}_6$

Now for $\mathrm{n}(\mathrm{Y})$ , range of R has exactly one element i.e. second elements must be constant in R and since R must have 6 element so it is not possible to satisfy both condition so $\mathrm{n}(\mathrm{Y})=0$ .

$\begin{aligned}\text { for } \quad \mathrm{n}(\mathrm{z}) \quad & 1 \rightarrow 3,4,5,6 \\& 2 \rightarrow 4,5,6 \\& 3 \rightarrow 1,5,6 \\& 4 \rightarrow 1,2,6 \\& 5 \rightarrow 1,2,3 \\& 6 \rightarrow 1,2,3,4\end{aligned}$

no. of relation that are function will be

$\begin{aligned}& ={ }^4 \mathrm{C}_1 \times{ }^3 \mathrm{C}_1 \times{ }^3 \mathrm{C}_1 \times{ }^3 \mathrm{C}_1 \times{ }^3 \mathrm{C}_1 \times{ }^4 \mathrm{C}_1 \\& =(4 \times 3 \times 3)^2=\mathrm{k}^2 \\& \text { i.e. } \mathrm{k}=36\end{aligned}$