Split a parent–child–parent block by its table

$9!=362880$.

Arrange the six guests in $6!$ ways. Their seven linear gaps include both ends. Choose three gaps and order the three Tans: $6!\binom73\,3!=151200$.

Choose four of nine diners for the smaller table: $\binom94=126$. Arrange them circularly in $(4-1)!=3!$ ways. The remaining five form the other circle in $\binom55(5-1)!=4!$ ways. Total $126\times3!\times4!=18144$. No division by 2 is made because table capacities distinguish the tables.

Treat parent–David–parent as one consecutive block with $2!$ parent orders. Choose its one guest companion in $\binom61$ ways and arrange the resulting two objects circularly in $(2-1)!$ ways. The smaller-table count is 12; the remaining five have $4!=24$ orders. Contribution $12\times24=288$.

Choose two guest companions in $\binom62$ ways. The family block and two companions are three circular objects, with $(3-1)!$ orders and $2!$ internal parent orders. This gives 60. The remaining four have $3!=6$ orders. Contribution $60\times6=360$. Total $288+360=648$.