PARAGRAPH I
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 $|a-b| \ge 2$.
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 $k^2$, then $|k|$ is ___.
Step-by-Step Solution
Key Concept: Applying range and function definition constraints to a relation, and using multiplication principle of counting for independent choices.
1) To find $n(Y)$:
The range of $R$ must have exactly one element, say $b_0$.
This means all 6 elements of $R$ must be of the form $(a, b_0)$ for $a \in S$.
Since $R$ has exactly 6 elements, all 6 elements of $S$ must map to $b_0$.
Thus, we must have $|a - b_0| \ge 2$ for all $a \in \{1, 2, 3, 4, 5, 6\}$.
However, for any choice of $b_0 \in S$, the element $a = b_0$ gives $|b_0 - b_0| = 0 < 2$. Thus, no such relation exists.
So, $n(Y) = 0$.
2) To find $n(Z)$:
$R \in Z$ is a function, meaning each of the 6 elements in the domain $S$ has exactly one image in $S$ satisfying $|a - f(a)| \ge 2$.
Let's count the choices of images for each element:
- For $a = 1$: $f(1) \in \{3, 4, 5, 6\}$ (4 choices)
- For $a = 2$: $f(2) \in \{4, 5, 6\}$ (3 choices)
- For $a = 3$: $f(3) \in \{1, 5, 6\}$ (3 choices)
- For $a = 4$: $f(4) \in \{1, 2, 6\}$ (3 choices)
- For $a = 5$: $f(5) \in \{1, 2, 3\}$ (3 choices)
- For $a = 6$: $f(6) \in \{1, 2, 3, 4\}$ (4 choices)
Since the choices for each element of the domain are independent, the total number of functions is:
$$n(Z) = 4 \times 3 \times 3 \times 3 \times 3 \times 4 = 16 \times 81 = 1296$$
3) Calculate $n(Y) + n(Z)$:
$$n(Y) + n(Z) = 0 + 1296 = 1296$$
We are given $n(Y) + n(Z) = k^2 \implies k^2 = 1296 \implies |k| = 36$.
Thus, the answer is 36.
Correct Answer: 36