Letter arrangements of "AAABBC" with length two "AAABBC" has 3 repetitions of "A" and two repetitions of "B".
So total number of arrangements of it with length 2 will be:
AA,AB,AC,BA,BB,BC,CA,CB
Using permutation we need to do something like:
${P(6,2)\over 3!2!}$
because we have 3 repetitions of "A" and 2 "B".
but then we get ${6*5*4!\over4!*2!*3!}$ which is not divisible.
Just wondering what's wrong with my logic?
 A: How many $A$s and $B$s you have doesn't matter as long as you have at least two of them when you are making strings of length two.  If you had two $C$s there would be $3^2$ strings because you have three choices for each letter.  Since you have only one $C$ you can't make $CC$ so there are $3^2-1=8$ two letter strings.
A: If we were arranging all six letters in the string AAABBC, we would have to select three of the six positions in the string to fill with an A, two of the remaining three positions to fill with a B, and fill the final position with the C.  We can do this in
$$\binom{6}{3}\binom{5}{2}\binom{1}{1} = \frac{6!}{3!3!} \cdot \frac{5!}{2!3!} \cdot \frac{1!}{1!0!} = \frac{6!}{3!2!1!0!} = \frac{6!}{3!2!1!}$$
ways.  The factor of $3!$ in the denominator represents the number of ways we could permute the three A's among themselves without producing an arrangement distinguishable from the given arrangement; the factor of $2!$ represents the number of ways we could permute the two B's among themselves without producing an arrangement distinguishable from the given arrangement; the factor of $1!$ represents the number of ways we could permute the only C among the C's without producing an arrangement distinguishable from the given arrangement.  
The factor $P(6, 2)$ in your numerator represents the number of ordered arrangements of two elements of a set with six different elements.  However, we only have three distinguishable elements in the string AAABBC.
In a string of length $2$, we can have at most two identical letters, so it does not make sense to divide by the factor $3!2!$.  Moreover, it only makes sense to divide by the factor $2!$ if the two letters in the string are identical.  However, some of the strings we could form, such as AB, contain different letters.
There are two strings of length two we could form with identical letters, namely AA and BB.
There are $P(3, 2) = 3 \cdot 2 = 6$ strings we could form with two different letters chosen from A, B, and C.  
In total, there are $2 + P(3, 2) = 8$ strings of length two that can be formed from the letters of the string AAABBC.
