Q:

Reason of dividing to n! ( repetition ) on Permutations with Repetitions

I'm trying to figure out the reason of diving the number of permutations by the number of repetitions (in factorial). Shouldn't it be without the factorial? I don't get why are there is a factorial in the denominator. For instance in the problem: 

How many anagram are there of the word "AMANHECE" ( portuguese )?

we have the solution:
$$\frac{8!}{2!2!}$$
but shouldn't it be:
$$\frac{8!}{2\cdot2}$$
-why do we get $2!$ instead of $2$?

A:

Here, the factorials don't really matter because $2!=2$, but in general, we do need to take factorials. Consider a simpler problem: How many anagrams does the word "AAAB" have? The answer is clearly $4$, since once we know where the "B" is, we know the rest of the letters are "A". To get there, we have to consider that there are $4!=24$ ways to arrange $4$ distinguishable letters. However, since there are $3$ of the letter "A", we can rearrange the "A"s in $3!$ ways without changing the word (e.g. swapping the first and third letters in "AAAB" has no effect). Thus, there are $\frac{4!}{3!}=4$ anagrams of "AAAB".
This is to say that the division represents accounting for a set of letters that can be freely rearranged - and we need to divide out the number "useless" rearrangements that we have. Factorials take in the size of a set a spit out how many ways we could rearrange it - so we get them both in numerator (total number of rearrangements) and denominator (number of useless rearrangements)

