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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 $$|a - b| \geq 2$$.
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 $$A$$.
We first list all ordered pairs $$(a,b)$$ in $$S\times S$$ that satisfy the restriction $$|a-b|\ge 2$$.
Total ordered pairs in $$S\times S$$ are $$6\times 6 = 36$$.
Forbidden pairs:
• Diagonal pairs with $$a=b$$ → 6 pairs.
• Pairs with $$|a-b|=1$$.
When $$b=a+1$$: $$(1,2),(2,3),(3,4),(4,5),(5,6)$$ → 5 pairs.
When $$b=a-1$$: $$(2,1),(3,2),(4,3),(5,4),(6,5)$$ → 5 pairs.
So forbidden pairs = $$6+5+5 = 16$$.
Allowed pairs (with $$|a-b|\ge 2$$) = $$36-16 = 20$$.
Every relation $$R\in X$$ is a subset of these 20 allowed pairs containing exactly 6 elements. Choosing any 6 of the 20 allowed pairs gives such a relation, and every such choice is valid.
Hence $$n(X)=\binom{20}{6}= {}^{m}C_6$$ which implies $$m=20$$.
Final Answer: 20
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