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).