Why do we divide by the count of each letter in questions of the form "how many words can I form from the letters in"?

Question

For example, how many words can we form from the letters in the word google?

First I thought you counted how many different letters there are in this case 4, therefore in each spot (6 spots) there are 4 different choices. So the amount of words is $4^6$?

I found out this is wrong and instead you use the idea of a multinomial and calculate $\frac{n!}{a_1!a_2!,...,a_k!}$ where $n$ is the amount of letters in the word, here n = 6, and then $a_1="g"$ which appears twice and so forth.

Why do we divide by the $a_k!$ factorial terms? Do we not lose possible words?

Answer 1

The reason you divide by $a_k!$ is to avoid multiple counting for each of the arrangements of the identical letters.

Imagine the letters $AAAMNP$. Let's affix a subscript to each of the $A$s. So you can have, for example,

$PA_1NA_2MA_3$, or

$PA_2NA_1MA_3$, or

$PA_1NA_3MA_2$, etc.

The word in each case is $PANAMA$, but simply taking $6!$ words multiple-counts $PANAMA$ six times (and every other permutation, like $AAMPNA$).

That's why you divide by the factorials of the multiplicities of the identical letters.

(Also, small difference in all of the six-letter words that can be made from the letters in GOOGLE ($6!/2!2!$) and the number of six-letter words that can be made using only G, O, L, and E ($4^6$).)

Answer 2

From the word $\color{blue}G\color{red}o\color{orange}o\color{blue}g\color{green}l\color{red}e$ I rearrange that letters to get the word $\color{blue}g\color{orange}o\color{red}o\color{blue}G\color{green}l\color{red}e$.

Is that the same word or a different word?

Those are the same word. If we did straight $6!$ we'd have those having to count as different words. But for every word where we put the orange "O" in position $k$ and we put the red "O" is position $j$, it'd be the exact same result as if we had put the red "O" in position $k$ and the orange "O" in position $j$.

So to count those as one option and not $2$ options as there are $2!$ ways to arrange the $2$ "O"s but thar are considered to be the same, we must divide by the $2!$ ways or arrange the "O"s