Solution:

The full set of $10$ chairs qualifies; count the others by locating each maximal run of at least three adjacent chosen chairs at its clockwise start. Any such subset contains a block of four consecutive chairs that is empty-chosen-chosen-chosen. There are $10$ positions for this block, and the remaining $6$ chairs are free, giving $10 \cdot 2^6 = 640.$

This counts once for each maximal run of length at least $3.$ Two such runs require at least $3 + 3$ chosen chairs plus two gaps, so three runs are impossible, and subsets with exactly two runs are counted twice. To have two runs, place two disjoint empty-chosen-chosen-chosen blocks: $\frac{10 \cdot 3}{2} = 15$ ways (the second block fits in $3$ positions among the remaining $6$ chairs), with the last $2$ chairs free, for $15 \cdot 2^2 = 60$ subsets.

The total is $1 + 640 - 60 = 581.$