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. 
Possible Duplicate:
How many ways are there for 8 men and 5 women to stand in a line so that no two women stand next to each other? 

Given 5 children and 8 adults, how many different ways can they be seated so that no two children are sitting next to each other.
My solution:
Writing out all possible seating arrangements:
tried using $\displaystyle \frac{34*5!*8!}{13!}$ To get the solution, because $13!$ is the sample space. and $5!$ (arrangements of children) * $34$ (no two children next to each other) * $8!$ (# of arrangements for adults).

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 
$$5!8! \binom{9}{5}.$$
