Let $S=\{1,2,3,4,5,6\}$ and $X$ be the set of all relations $R$ from $S$ to $$ that satisfy both the following properties:

i. $R$ has exactly 6 elements.

ii. For each $(a,b) \in R$, we have $\mathrm{\mid }a-b\mathrm{\mid }\ge 2$

Let

$$Y=\{R\in X:\text{ The range of }R\text{ has exactly one element}\}$$

and

$$Z=\{R\in X:R\text{ is a function from }S\text{ to }S\}.$$

Let $n(A)$ denote the number of elements in a set $A$.

If the value of n(Y) + n(Z) is k², then ∣k∣ is ______.

The correct answer is 36.

Step 1: Understand the Given Set and Relations
We are given the set $S=\{1,2,3,4,5,6\}$ containing 6 elements.
$X$ is the set of all relations $R$ from $S$ to $S$ satisfying two conditions:
1. $R$ contains exactly 6 ordered pairs $(a,b)$.
2. For every ordered pair $(a,b)\in R$, we have $|a-b|\ge <!--\; \ge \; -->2$.

Step 2: Determine the Allowed Pairs for Each Element in S
Let us find all possible values of $b\in S$ for each given $a\in S$ such that $|a-b|\ge <!--\; \ge \; -->2$:
• For $a=1$: $b\in \{3,4,5,6\}$   ⇒   4 choices
• For $a=2$: $b\in \{4,5,6\}$   ⇒   3 choices
• For $a=3$: $b\in \{1,5,6\}$   ⇒   3 choices
• For $a=4$: $b\in \{1,2,6\}$   ⇒   3 choices
• For $a=5$: $b\in \{1,2,3\}$   ⇒   3 choices
• For $a=6$: $b\in \{1,2,3,4\}$   ⇒   4 choices

Step 3: Calculate n(Y)
The set $Y$ consists of all relations $R\in X$ whose range contains exactly one element.
For the range of $R$ to have exactly one element, say $b_0$, all 6 pairs in $R$ must share this same second element $b_0$, meaning $R=\left\{\right(1,b_0),(2,b_0),(3,b_0),(4,b_0),(5,b_0),(6,b_0\left)\right\}$.
However, for $a=b_0$, we have $|b_0-b_0|=0<2$, which violates the condition $|a-b|\ge <!--\; \ge \; -->2$.
Thus, no such relation exists, so:

$n\left(Y\right)=0$

Step 4: Calculate n(Z)
The set $Z$ consists of all relations $R\in X$ such that $R$ is a function from $S$ to $S$.
Since $R$ contains exactly 6 elements and must be a function defined on all 6 elements of $S$, each element $a\in S$ must be mapped to exactly one valid $b\in S$ such that $|a-b|\ge <!--\; \ge \; -->2$.
Using the number of choices determined in Step 2, the total number of functions is:

$n\left(Z\right)=4\times <!--\; \times \; -->3\times <!--\; \times \; -->3\times <!--\; \times \; -->3\times <!--\; \times \; -->3\times <!--\; \times \; -->4=1296$

Step 5: Find |k|
We are given that $n\left(Y\right)+n\left(Z\right)=k^2$:

$k^2=0+1296=1296$

$\left|k\right|=\sqrt{1296}=36$