Find all possible combinations of $A, A, A, B, B$ 10 year old daughter has this problem.
She knows that all possible combinations of $A,B,C,D$ are $4! = 24  $
She figured it like this:
If I write down $A$ first, it has $4$ possible places.
If I write down $B$ next, it has $3$ possible places for each of the places of $A (4.3)$
If I write down $C$ next, it has $2$ possible places for each of the above placings $(4.3.2)$
If I write down $D$ last, it has $1$ possible place for each of the above placings $(4.3.2.1)$  
How can she extend this method to find out all possible combinations of $A,A,A,B,B$ ?
 A: We are making a $5$-letter word, so we have $5$ slots to fill using $3$ A's and $2$ B's.
Maybe the leftmost B is in the first (leftmost) slot. Then there are $4$ places to put the second B.
Maybe the leftmost B is in the second slot. Then there are $3$ places to put the second B.
Maybe the leftmost B is in the third slot. Then there are $2$ places to put the second B.
Maybe the leftmost B is in the fourth slot. Then there is only $1$ place to put the second B.
Add up. We get $4+3+2+1$.
A: Hint. 
First pretend that the $A$'s and $B$'s are actually different - for example, say they're numbered $A_1$, $A_2$, $A_3$, $B_1$, $B_2$. Then this is the same problem you mentioned, and there are $5!$ ways to arrange the letters.
The problem is that you'll obtain many arrangements that differ only because of the numbering. So your problem is to determine how many arrangements there are differing only by the numbering that correspond to a single arrangement where the numbering is ignored.
For example, how many arrangements including numbers correspond to ABAAB? Can you get this by a calculation?
A: There are $5!$ combinations of $A_1A_2A_3B_1B_2$ if we first assume the $A$'s and $B$'s to be distinct. In all these combinations $A_1, A_2$ and $A_3$ can appear in $3!$ different ways, but they would all give the same word, so we'll have to divide by $3!$. Given the same argument for $B_1$ and $B_2$ we have to divide once again with $2!$, so the answer would be $\frac{5!}{3!2!}$.
