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.

EDIT (in response to the comment):
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 $2$.

Have a look at the following for better intuition of why division works: Division using repeated subtraction

  • $\begingroup$ But how does the dividing operation actually remove the over-count? What I mean is why dividing works instead of subtraction? $\endgroup$
    – user390456
    Nov 17 '16 at 19:57
  • $\begingroup$ Because for every single arrangement, there is another arrangement which is similar after removing the subscripts. So, you need to subtract repeatedly for every double-counted arrangement. And may I ask you, what is repeated subtraction? - Division $\endgroup$ Nov 17 '16 at 20:00

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