$\begin{array}{l}|a-b| \geq 2 \text { or }|b-a|=2 \\ \text { Total } \\ a =1 \quad b =3,4,5,6 \quad 8 \\ a =2 \quad b =4,5,6 \quad 6 \\ a =3 \quad b =5,6 \quad 4 \\ a =4 \quad b =6 \quad 2 \\ \overline{\operatorname{sum}=20} \\ n ( X )={ }^{20} C _6={ }^{ m } C _6 \\ m =20 \\\end{array}$

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

$a=1  b=3,4,5,6  \rightarrow  4 \times 2=8$

$a=1  b=4,5,6 \rightarrow  3 \times 2=6$

$a=1  b=5,6  \rightarrow 2 \times 2=4$

$a=1  b=6  \rightarrow  2 \times 1=2$

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

Now for $n ( Y )$,

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 $n ( Y )=0$.

$n ( z ) \quad  1 \rightarrow 3,4,5,6$

$2 \rightarrow 4,5,6$

$3 \rightarrow 1,5,6$

$4 \rightarrow 1,2,6$

$5 \rightarrow 1,2,3$

$6 \rightarrow 1,2,3,4$

no. of relation that are function will be

$={ }^4 C _1 \times{ }^3 C _1 \times{ }^3 C _1 \times{ }^3 C _1 \times{ }^3 C _1 \times{ }^4 C _1$

$=(4 \times 3 \times 3)^2= k ^2$

i.e. $k =36$