Number of words such that only two $A's$ should be together Find number of words that can formed using letters of the word "$PARABOLA$" such that only two $A's$ should be together. I did using two methods:
Method $1.$ Let $AA=X$ 
$*P*R*B*O*L*$ now we have 6 places to keep $X$ and $A$ which can be done in $\binom{6}{2} \times 2!=30$ ways. Also $P,R,B,O,L$ can arrange themselves in $120$ ways. So Total words is $3600$
Method $2.$ if $AA=X$ total permutations of $P,X,R,B,O,L,A$ is $7!$. Among these i have to remove all words where $X$ and $A$ are together. That is if $XA=Z$ then total permutations of $P,Z,R,B,O,L$ is $6!$. hence total words with exactly two $A's$ together is $7!-6!=4320$.

To remove the confusion, here is a third method:
Total permutations - 3 A's together - no A's together
$= 8!/3! - 6! - 5!\binom63 = 3600$
