# How does dividing remove repetitions?

I've been trying to visually wrap my head around how in problems involving combinations and permutations, dividing removes repetitions?

Like for example a simple question like: In how many ways can you re-arrange the letters in the word "SEE". It would be 3!/2!

But I can't visually understand how dividing by 2! works? How does it eliminate the identical possibilities?

Let's distinguish the two $E$s, lets say as $E_1$ and $E_2$. Then we take all possible permutations of the 3 (now distinct letters). If we consider any such permutation, say $E_1 \ S \ E_2$, then there is always another distinct permutation obtained by exchanging the positions of the indexed $E's$ that will give me the same word if I ignore the subscripts $1$ and $2$. So counting every permutation of the three letters has counted every word twice, and so we have to divide by two to get the right answer.
Suppose you have $10$ numbers as follows: $1,1,3,3,6,6,9,9,15,15$ (think of every number as an arrangement without subscripts).
Now you decide to remove the repetition corresponding to $1,3,6,9,15$. So, as you mention, we resort to subtraction. So out of the 2 copies of every number, we remove one and keep the other. That is the frequency of every number is halved. And so the overall division by $2$.