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?

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!}$
