Question 15

Step 1: Find $n(Y)$. $Y$ is the set of relations $R \in X$ where the range has exactly one element. Let the range be $\{b\}$. Then all 6 elements of $R$ must be of the form $(a, b)$ where $a \in S$. For $R$ to have 6 distinct elements, all 6 values of $a \in \{1, 2, 3, 4, 5, 6\}$ must satisfy $|a - b| \geq 2$. Checking available $a$'s for each $b$:

• If $b=1$, $a \in \{3, 4, 5, 6\}$ (4 elements)
• If $b=2$, $a \in \{4, 5, 6\}$ (3 elements)
• If $b=3$, $a \in \{1, 5, 6\}$ (3 elements)
• If $b=4$, $a \in \{1, 2, 6\}$ (3 elements)
• If $b=5$, $a \in \{1, 2, 3\}$ (3 elements)
• If $b=6$, $a \in \{1, 2, 3, 4\}$ (4 elements) In no case can we choose 6 distinct values for $a$. Thus, $n(Y) = 0$.

Step 2: Find $n(Z)$. $Z$ is the set of relations $R \in X$ that are functions from $S$ to $S$. For $R$ to be a function, every $a \in S$ must be mapped to exactly one $b \in S$ such that $|a - b| \geq 2$. Number of choices for each $a$:

• For $a=1$: 4 choices ($b \in \{3, 4, 5, 6\}$)
• For $a=2$: 3 choices ($b \in \{4, 5, 6\}$)
• For $a=3$: 3 choices ($b \in \{1, 5, 6\}$)
• For $a=4$: 3 choices ($b \in \{1, 2, 6\}$)
• For $a=5$: 3 choices ($b \in \{1, 2, 3\}$)
• For $a=6$: 4 choices ($b \in \{1, 2, 3, 4\}$) $n(Z) = 4 \times 3 \times 3 \times 3 \times 3 \times 4 = 16 \times 81 = 1296$.

Step 3: Calculate $n(Y) + n(Z)$. $n(Y) + n(Z) = 0 + 1296 = 1296$. Given $k^2 = 1296$, we have $|k| = \sqrt{1296} = 36$.