**Solution:** Fix the two specific objects opposite each other. Now, arrange the remaining $\textbf{n}-2$ objects around the circle, which can be done in $(\textbf{n}-2-1)! = (\textbf{n}-3)!$ ways. Since the two are fixed opposite, total arrangements = $$(\textbf{n}-2) \times (\textbf{n}-3)! = (\textbf{n}-2)!$$ due to the symmetry, but fixing their positions simplifies the calculation, leading to $(\textbf{n}-2)!$ arrangements.