How many arrangements of letters in REPETITION are there with the first E occurring before the first T? The question is:
How many arrangements of letters in REPETITION are there with the first E occurring before the first T?
According the book, the answer is $3 \cdot \frac{10!}{2!4!}$, but I'm having trouble understanding why this is correct.
I figured there are $4$ positions for the $2$ E's and $2$ T's, and one of the E's must be placed before any of the T's. $P(3;2,1)$ -> remaining $3$ positions which can be filled with $1$ E and $2$ T's. -> $3!/2!$ -> $3$. So, I have this part correct I believe.
Where I'm having trouble is figuring out why $\frac{10!}{2!4!}$ is correct. Is it this $P(10; 4, 2, 1, 1, 1, 1)$? What are the different types of letters represented here?
 A: First we'll arrange the letters like this:
$$\text{E E T T}\quad\text{R P I I O N}$$
Now replace $\text{EETT}$ with $****$:
$$\text{* * * *}\quad\text{R P I I O N}$$
The number of arrangements is $$\frac{10!}{4!2!}$$ because four symbols are the same, two other symbols are the same, and all the rest are different.
Now consider any one of those arrangements. We can replace the asterisks $****$, from left to right, with $\text{EETT}$ or $\text{ETET}$ or $\text{ETTE}$. There are three possibilities. This results in the answer given in the book: $$\frac{10!}{4!2!}\cdot3$$
A: Alternative appraoch
Write the ten letters as
$$
R,E_1,P,E_2,T_1,I_1,T_2,I_2,O,N
$$
then permute them in $10!$ ways. Divide by $2!$ for each of the pairs $(E_1,E_2),(T_1,T_2)$ and $(I_1,I_2)$. By symmetry, in half of the cases the first $E$ occurs before the first $T$, so we have
$$
\frac12\cdot\frac{10!}{(2!)^3}=3\cdot\frac{10!}{2!4!}
$$
but I would rather have simplified it as $\frac{10!}{2^4}$.
A: Another way to get this answer is to first choose the 4 places for the E's and T's, 
which can be done in $\binom{10}{4}$ ways.  
Since an E must be in the first of these places, there are $\binom{3}{1}$ ways to place the second E.
Then there are $\frac{6!}{2!}$ ways to place the remaining 6 letters,
so this gives an answer of
$ \hspace{.4 in}\binom{10}{4}\cdot3\cdot\frac{6!}{2!}=3\cdot\frac{10!}{4!2!}$
