Unique combinations of strings If I have the string Delaware and I want to figure out how many unique strings can be made from the letters in this word, I know that the answer is 8!/(2!)(2!) and that the reason we divide by 2! and 2! is because the letters a are the same and the letters e are the same, and those strings will be counted twice.
For example,
e(1)e(2)a(1)a(2)Dlwr is equivalent to e(2)e(1)a(1)a(2)Dlwr (I'm using the numbers in the string to represent to same letter in the next string)
I don't understand why we divide by the number of repeats though? If someone could provide an answer that's easily understood by someone who hasn't done too much combinatorics that would be amazing.
 A: Imagine a simpler example.  
TOtE
If all arrangements were unique, we would go with $4!$.
But we know, TOtE is the same as tOTE. In fact, there will always be $2!$ ways the t's can be arranged that are the same.  We only want to count one of them, so we divide by 
$2!$.  Similarly, if a letter is repeated 3 times, it can be arrange $3!$ ways, but we only want to count 1 of them, so we divide by $3!$
For a letter repeated 3 times, lets go with $CH\color{red}E\color{blue}ES\color{green}E.$
$CH\color{red}E\color{blue}ES\color{green}E,\space CH\color{red}E\color{green}ES\color{blue}E,\space CH\color{blue}E\color{red}ES\color{green}E,\space CH\color{blue}E\color{green}ES\color{red}E,\space CH\color{green}E\color{red}ES\color{blue}E,\space CH\color{green}E\color{blue}ES\color{red}E\space $
$HC\color{red}E\color{blue}ES\color{green}E,\space HC\color{red}E\color{green}ES\color{blue}E,\space HC\color{blue}E\color{red}ES\color{green}E,\space HC\color{blue}E\color{green}ES\color{red}E,\space HC\color{green}E\color{red}ES\color{blue}E,\space HC\color{green}E\color{blue}ES\color{red}E\space $
Each of the sets of 6 is really identical so we will need to divide by $3!$
A: Consider first the following different but related problem: how many words can be made by arranging the eight different letters of $DE_1LA_1WA_2RE_2$?  The answer is $8!$ because we are ordering $8$ different "letters" (or symbols if you don't want to call $A_1$ a letter).  The words we obtain include, for example
$$A_1DRE_2A_2E_1WL\quad\hbox{and}\quad A_2DRE_2A_1E_1WL\ .$$
For your problem, however, the subscripts on the $A$s are not there, so these words are both the same, and we have counted them twice when we should have counted them once.  So $8!$ is incorrect for this problem and we have to divide by $2!$.  In fact then we still have
$$ADRE_2AE_1WL\quad\hbox{and}\quad ADRE_1AE_2WL\ ,$$
so we have still counted every word twice on account of $E_1$ and $E_2$, and we have to divide by $2!$ again.
