Let's say I need to find total number of words I can form by using the letters of the word "ROOT".
Since $2$ O's are present. They would arrange amongst themselves in $2!$ ways, so we have $2!-1$ ways being over-counted.
So my answer then would be $4! -(2!-1) = 23$
My question : Exactly what cases are still being over-counted?
Important Note : I'm aware that you need to divide $4!$ by $2!$ as each case is being counted $2!$ times. But this is not my question. I want to know what exactly is going on with my first method. I just can't seem to get a "feel" for it.
Thanks :)
You forget about those partners of O's, that's the 'R' and 'T':
$$(\textrm{Choose position for R})(\textrm{ " for T})(2!-1),$$
so the part over-counted is
$$(4)(3)(2!-1)=12.$$
That is: you have to also think about how O's repetition influences the others.
For the case of type $ABAB$, as @Jürg Merlin Spaak mension in the comment, then two separate steps are needed in this special case
(1) First see two A's as different
$$(\textrm{pos. for} A_1)(\textrm{pos. for}A_2)(\textrm{this part same idea as above}),$$
so
$$(4)(3)(2!-1)=12,$$
this is the part over-counted but what we want is $4!-12=12.$
(2) Now let's say they are identical:
$$(\textrm{ans from(1)})=(\textrm{other factors})(\textrm{factors of}A_1A_2)$$
so
$$12=(\textrm{other factors})(2!),$$
and $(\textrm{other factor})=12/2!=6$, now use the same idea
$$(\textrm{over-counted})=(\textrm{other factors=6})(2!-1)=6.$$
so total = $12-6=6.$