The problem is equivalent to counting the rearrangements of PPPPPPPPEEEEEEEE (standing for people and empty seats) where no three Ps appear together. Suppose such an arrangement contains $q$ pairs $PP$ and $8 - 2q$ Ps not adjacent to another P, where $0 \le q \le 4$. There are 9 spaces before, after, or between the 8 Es, and $PP$s will be placed in $q$ of those 9 spaces, while Ps will be placed in $8 - 2q$ of the remaining $9 - q$ spaces. Hence the number of arrangements for a given value of $q$ is $\binom{9}{q}\binom{9-q}{8-2q}$, and the total number of ways to select chairs is
$$\begin{aligned} \sum_{q=0}^{4} \binom{9}{q} \binom{9-q}{8-2q} &= \binom{9}{0} \binom{9}{8} + \binom{9}{1} \binom{8}{6} + \binom{9}{2} \binom{7}{4} + \binom{9}{3} \binom{6}{2} + \binom{9}{4} \binom{5}{0} \\ &= 1 \cdot 9 + 9 \cdot 28 + 36 \cdot 35 + 84 \cdot 15 + 126 \cdot 1 = 9 + 252 + 1260 + 1260 + 126 = 2907. \end{aligned}$$
The requested remainder is 907.