Question 14

To find $n(X)$, we first count the number of ordered pairs $(a, b)$ from $S \times S$ that satisfy $|a - b| \geq 2$. For $a = 1$, $b \in \{3, 4, 5, 6\}$ (4 pairs) For $a = 2$, $b \in \{4, 5, 6\}$ (3 pairs) For $a = 3$, $b \in \{1, 5, 6\}$ (3 pairs) For $a = 4$, $b \in \{1, 2, 6\}$ (3 pairs) For $a = 5$, $b \in \{1, 2, 3\}$ (3 pairs) For $a = 6$, $b \in \{1, 2, 3, 4\}$ (4 pairs) Total number of available pairs $(a, b)$ is $4 + 3 + 3 + 3 + 3 + 4 = 20$. A relation $R$ in $X$ is a subset of these available pairs containing exactly 6 elements. Therefore, $n(X) = {}^{20}C_6$. Comparing with $n(X) = {}^mC_6$, we get $m = 20$.