jmc

We know that a subset with less than $3$ chairs cannot contain $3$ adjacent chairs. There are only $10$ sets of $3$ chairs so that they are all $3$ adjacent. There are $10$ subsets of $4$ chairs where all $4$ are adjacent, and $10\cdot 5$ or $50$ where there are only $3.$ If there are $5$ chairs, $10$ have all $5$ adjacent, $10\cdot 4$ or $40$ have $4$ adjacent, and $10\cdot \left(\genfrac{}{}{0ex}{}{5}{2}\right)$ or $100$ have $3$ adjacent. With $6$ chairs in the subset, $10$ have all $6$ adjacent, $10(3)$ or $30$ have $5$ adjacent, $10\cdot \left(\genfrac{}{}{0ex}{}{4}{2}\right)$ or $60$ have $4$ adjacent, $\frac{10\cdot 3}{2}$ or $15$ have $2$ groups of $3$ adjacent chairs, and $10\cdot \left(\left(\genfrac{}{}{0ex}{}{5}{2}\right)-3\right)$ or $70$ have $1$ group of $3$ adjacent chairs. All possible subsets with more than $6$ chairs have at least $1$ group of $3$ adjacent chairs, so we add $\left(\genfrac{}{}{0ex}{}{10}{7}\right)$ or $120$, $\left(\genfrac{}{}{0ex}{}{10}{8}\right)$ or $45$, $\left(\genfrac{}{}{0ex}{}{10}{9}\right)$ or $10$, and $\left(\genfrac{}{}{0ex}{}{10}{10}\right)$ or $1.$ Adding, we get $10+10+50+10+40+100+10+30+60+15+70+120+45+10+1=\boxed{581}.$