A family consists of a grandfather, $6$ sons and daughters and $4$ grandchildren. They are to be seated in a row for dinner. The grandchildren wish to occupy, the two seats at each end and the grandfather refuses to have a grandchild on either side of him. In how many ways can the seating arrangements be made for the dinner?

Total no. of seats:
$=1grandfather+6sonsanddaughters+4grandchildren=11.$

The grandchildren can occupy the $2$ seats on either side of the table in $2!=2$ ways.

The grandfather can occupy a seat in $(6-1)=5$ ways ($5$ gaps between $6$ sons and daughter).

And, the remaining seats can be occupied in $4!=24$ ways ($4$ seat for sons and daughter).

Hence, total number of required ways $=6!\times (4!\times 5)\Rightarrow 6!\times 120.$