There are $5$ $E$s in total, which have $4$ intervals leaving for the other $4$ letters. Thus, the answer is $\_4P\_4=24$. 
In how many ways can the letters in $BEEKEEPER$ be rearranged so that two or more $E$s do not appear together? 
$$1$$ 
$$4$$ 
$$12$$ 
$$24$$ 
$$120$$ 