1. Interpret the condition on relations

A relation from $S$ to $S$ is any subset of $S\times S$.

Here, $S=\{1,2,3,4,5,6\}$ and $R\subseteq S\times S$ must satisfy:

• $R$ has exactly $6$ ordered pairs,
• for every $(a,b)\in R$, we have $|a-b|\ge 2$.

So first we count how many ordered pairs $(a,b)\in S\times S$ satisfy $|a-b|\ge 2$.

Then any valid relation in $X$ is simply a choice of $6$ such ordered pairs.

Hence, $n(X)=\binom{m}{6}$ where $m$ is the number of allowed ordered pairs.

2. Count total ordered pairs in $S\times S$

Since $|S|=6$, total ordered pairs are $6\times 6=36.$

3. Count forbidden ordered pairs

Forbidden pairs are those with $|a-b|<2,$ i.e. either $|a-b|=0$ or $|a-b|=1$.

(i) Pairs with $|a-b|=0$

These are $(1,1),(2,2),(3,3),(4,4),(5,5),(6,6),$ so there are $6$ such pairs.

(ii) Pairs with $|a-b|=1$

For consecutive numbers, both orders are allowed.

Adjacent pairs are: $(1,2),(2,1),(2,3),(3,2),(3,4),(4,3),(4,5),(5,4),(5,6),(6,5).$ So there are $10$ such pairs.

Therefore total forbidden pairs: $6+10=16.$

4. Count allowed ordered pairs

Thus the number of ordered pairs satisfying $|a-b|\ge 2$ is $36-16=20.$

So there are $20$ possible ordered pairs from which the relation $R$ can choose exactly $6$ elements.

Hence, $n(X)=\binom{20}{6}.$

Comparing with $n(X)=\binom{m}{6},$ we get $m=20.$

5. Final answer

$\boxed{20}$

6. Comparison with stored answer

Stored correct answer = $20$.

Our derived answer also is $20$, so it agrees.