There are $4$ $A$s in total, which have $3$ intervals leaving for the other $3$ letters. Thus, the answer is $\_3P\_3=6$. 
In how many ways can the letters in $AAABCDA$ be rearranged so that two or more $A$s do not appear together? 
  $$4$$ 
  $$6$$ 
  $$8$$ 
  $$12$$ 
  $$18$$ 