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?
Best Answer
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: $$\frac{10!}{4!2!}\cdot3$$