Words with repeated blocks of letters If I have the word BARBARIANISM, how many arrangements of this word's letters contain two identical blocks of 3 letters (e.g. 'BAR' repeated twice in the original word, or 'AIR' repeated twice in 'BAIRBANSAIRM'). 
I think there are two ways in which we can create such a block of 3 letters. One include the letter 'A', and one without the letter 'A'. That is ${3\choose 1}\times3!$ and $3!$ ways respectively.
I think the number of ways we can arrange the other letters (not in the block of 3) with and without the letter 'A' being used in said block of 3 is $\frac{6!}{2!}$ and $\frac{6!}{3!}$ respectively.
There are ${7\choose2}$ ways we can arrange the 2 blocks of 3 within the new word, so by putting it all together I think that
\begin{equation*}
{7\choose2}\Big(\frac{6!}{3!}\times3! + \frac{6!}{2!}\times{3\choose 1}\times3!\Big) = 151200 \mbox{ ways}
\end{equation*}
is the answer. Is my method correct? I feel like this number is too large and I may have double counted. Thanks.
 A: Let's look at what we are dealing with a little more closely:
$B,B$
$A,A,A$
$R,R$
$I,I$
$N$
$S$
$M$
This means that if we want to have two identical blocks of $3$ letters, our "group of letters" has to be one of the four following forms: $BAR, BAI, BRI, ARI$ 
If we choose one of the $3$ combinations of letters where the $A$ is included, we have the number of ways to arrange is $$3!\frac{8!}{2!2!}$$
We have $3!$ for the ways to arrange the $3$ letters inside the "grouping"and the $2!2!$ in the denominator to account for different permutations being the same.
Example: $\color{red} {\boxed{BAR}}IANISM\color{red}{\boxed{BAR}}$
Counting the box as one letter, we have an $8$ letter word and we have two letters ($I$ and $BAR$) that appear twice so we account for that in the denominator
If our choice of the combination is the set of letters without the $A$, then we have the number of ways to be $$3!\frac{8!}{2!3!}$$
Example: $\color{red} {\boxed{BIR}}AANASM\color{red}{\boxed{BIR}}$ is an $8$ letter word, where we have the letter $A$ appearing three times and the "letter" $BIR$ appearing twice.
So the total number of ways at this point is $$3\left(3!\frac{8!}{2!2!}\right) + 3!\frac{8!}{2!3!} = 201600$$

Thanks for the note from @awkward,
However, we have overcounted. 
See the example: $\color{red}{BAR}IANSM\color{red}{BAR}I = B\color{green}{ARI}ANSMB\color{green}{ARI}$ 
To account for this, we need to subtract off all the words that have the grouping of $BARI$ shown twice. From here, we have $$4!\frac{6!}{2!} = 8640$$
So after subtracting, we get $$201600 - 8640 = 192960$$
