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?

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 $\frac{10!}{2^4}$.
