= Solution
{c}
Use $s=x_1$ and $r=x_2$. A putative <left Ore set> equality would be
$$
s'x_2=r'x_1,\qquad s'\in S,\quad r'\in R.
$$
The left side is a single <word> ending in $x_2$, with coefficient $1$. The right side is a finite <linear combination> of <words> ending in $x_1$. Distinct <words> are <linearly independent> in the <free associative algebra>, so equality is impossible. Therefore
$$
\boxed{S\text{ is not a left Ore set in }\mathbb Q\langle x_1,x_2\rangle.}
$$
The last letters matter because the denominators in the <left Ore set> equation occur on the right of the common left multiple.
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