If the value of $$n(Y) + n(Z)$$ is $$k^2$$, then $$|k|$$ is ______.

The set $$S = \{1,2,3,4,5,6\}$$ has six elements. Every relation under consideration must satisfy the two fixed conditions:

(i) it contains exactly $$6$$ ordered pairs, and
(ii) each ordered pair $$(a,b)$$ in it obeys $$|a-b|\ge 2$$.

We study the two subsets of relations, $$Y$$ and $$Z$$, separately.

By definition, each relation $$R\in Y$$ has a range that is a single element, say $$\{b\}$$. Hence the only possible pairs in such an $$R$$ are of the form $$(a,b)$$ with $$a\in S$$. Because $$R$$ must contain exactly six distinct ordered pairs, every first component $$a=1,2,3,4,5,6$$ must occur once:

$$R=\{(1,b),(2,b),(3,b),(4,b),(5,b),(6,b)\}.$$

For this relation to be admissible, each pair must satisfy $$|a-b|\ge 2$$. In particular, $$|b-b|=0\lt 2$$ fails for the pair $$(b,b)$$. Therefore no value of $$b$$ can make all six pairs valid, and hence

$$n(Y)=0.$$

A relation $$R\in Z$$ is a function from $$S$$ to $$S$$. Thus for every $$a\in S$$ there is exactly one ordered pair $$(a,f(a))$$ in $$R$$. The requirement $$|a-f(a)|\ge 2$$ must hold for each $$a$$.

We list the permissible images for every element of $$S$$:

For $$a=1$$: $$|1-b|\ge 2\Longrightarrow b=3,4,5,6$$  ⇒  4 choices.
For $$a=2$$: $$|2-b|\ge 2\Longrightarrow b=4,5,6$$  ⇒  3 choices.
For $$a=3$$: $$|3-b|\ge 2\Longrightarrow b=1,5,6$$  ⇒  3 choices.
For $$a=4$$: $$|4-b|\ge 2\Longrightarrow b=1,2,6$$  ⇒  3 choices.
For $$a=5$$: $$|5-b|\ge 2\Longrightarrow b=1,2,3$$  ⇒  3 choices.
For $$a=6$$: $$|6-b|\ge 2\Longrightarrow b=1,2,3,4$$  ⇒  4 choices.

Because the six choices are independent (the function need not be one-to-one), the total number of such functions equals the product of the individual counts:

$$n(Z)=4\cdot 3\cdot 3\cdot 3\cdot 3\cdot 4 =4^{2}\,3^{4}=16\cdot 81=1296.$$

$$n(Y)+n(Z)=0+1296=1296=k^{2}\; \Longrightarrow\; |k|=\sqrt{1296}=36.$$

Hence the required value is 36.