Given 5 children and 8 adults, how many ways can they be seated so that there are no two children sitting next to each other. [duplicate]

We have $8$ nice comfortable chairs for the adults, separated by some space. This determines $9$ "gaps" where a kid can drag a stool. (It is $9$ because a kid can drag as stool between two adult chairs, or to the left end or to the right end.)

The seating arranger chooses $5$ of these places to put a stool into. This can be done in $\binom{9}{5}$ ways. For each of these ways, the adults can be seated in $8!$ orders, and for every way to do this, the children can occupy the stools in $5!$ orders. The number of ways is therefore $$5!8! \binom{9}{5}.$$

The solution below assumes the seats are in a row:

This is a stars and bars problem. First, order the children (5! ways). Now, suppose the adults are identical. They can go in any of the places on either side or between of the children. Set aside 4 adults to space out the children, and place the other 4 in any arrangement with the 5 children; there are $\binom{9}{4}$ ways to do this. Finally, re-order the adults. So we get $$8!5!\binom{9}{4}$$