Very silly permutation question Okay let me briefly explain my doubt.
I'll explain some easy problems,so that you can study easily my mind and you can guess what confusion i might be going through right now.
This may be silly.But please help me out.
How do you permute the letters ABC with no character can be repeated?
We can do $3!$ arrangements.
Like, 3 ways to choose the first character, 2 ways for the second one,one way for the last character.
Okay,so far so good.
Now,how do you permute $AAB$?
$$AAB,ABA,BAA,BAA,ABA,AAB.$$
The formula will be like dividing the number of times each letter repeats.
Like, $3!/2!$ (Since $A$ is repeated).
I don't understand this part.Why we need to divide by $2!$? Reason is being told as $A$ is repeated twice. So what? What happens internally? What happens when we divide ? Subtraction should make sense here since we are eliminating the redundancy. But dividing gives the actual answer! Can somebody explain me clearly why dividing with repeated terms makes sense?
Please and thanks.

This can be understood by renaming the second A to A'. This gives you obviously 3!=6 possible permutations:
AA'B
ABA'
A'AB
A'BA
BAA'
BA'A
Now you can group these into pairs whose elements are equal except of A and A' being swapped:
AA'B, A'AB
ABA', A'BA
BAA', BA'A  
Since A and A' shall not be distinguished we only keep one permutation from each pair, effectively dividing by two.
The same argument works for any number of A, A', A'', ... (k times) grouping them into tuples of k! elements.
