Number of words with two "$A$" using two letters from "$RATA$" and three letters from "$TIERRA$" 
Find the number of words with two "$A$" using two letters from "$RATA$"
  and three letters from "$TIERRA$".

What I did:
There are two cases, one where I choose both $A$ from $RATA$ and the rest of the letters from $TIERRA$, and the other one where I choose an $A$ from $RATA$ and the other $A$ from $TIERRA$:
1. Choosing both $A$ from $RATA$:
There are ${5}\choose{2}$ ways to put the $A$s on the word, and then I choose the other three letters from $TIERRA$: $\frac{5\cdot4\cdot3}{2!}$
So there are $5\choose2$$\frac{5\cdot4\cdot3}{2!}=300$ words using both $A$ from $RATA$.
2. Choosing an $A$ from $RATA$ and the other one from $TIERRA$:
Again, there are ${5}\choose{2}$ ways to put the $A$s on the word, there are only two options to choose a letter from $RATA$ (either $R$ or $A$), and then there are $\frac{5\cdot4}{2!}$ ways to choose the remaining two letters from $TIERRA$.
So there are ${5}\choose{2}$$\frac{5\cdot4}{2!}=100$ words in this case.
By the rule of sum, there are $300+100=400$ words for this problem.
But the solution to this problem is $430$. What am I doing wrong?
 A: I think it is easier to approach this problem by listing all five-letter combinations (not permutations just yet).  Aside from the exactly two A's, which are mandated, there are only three letters to be chosen from a repertoire of I, E, R, and T, some of which can be repeated.  There just aren't that many possible combinations; the list begins


*

*AAEIR

*AAEIT

*AAERR

*AAERT

*AAETT

*$\ldots$
There are fewer than a dozen of these; be sure to eliminate the ones that cannot be made by selecting two from TARA and three from TIERRA.  Then list the number of permutations of each combination.  For instance, AAEIR and AAEIT each have $\frac{5!}{2!} = 60$ permutations, while AAERR has $\frac{5!}{2!2!} = 30$ permutations.  And so on.
Like Daniel Mathias in the comments, I also arrive at an answer of $430$.  Maybe there was a typo.
A: notice that if you take two $T$'s or two $R$'s so you need to divide by 2, because there is no difference which of those $T$'s/$R$'s comes first. i would advise to separate into sub-cases, and rethink and approach from another angle, like: which are the 3 other letters in used in every permutation, beside the $A$'s. are them the set {$T$,$T$,$R$}? are them {$T$,$R$,$R$}? or is it a {$R$,$T$,$E$}, or any other combination?
cause it doesn't really matter which letter you took from each word (at the ends whats important is the unique word that created), and in the way you approached before you can over counting by mistake
A: The apparent complication of having two sources for letters is nullified by the restrictions, assuming that we require exactly two $A$s: the letters then are "AA" + three from "RRRTTEI".
If we allow three $A$s (and arguably a word with three $A$s has two $A$s), we only need to add two from "RRTIE" (all from TIERRA) once the three As are assumed chosen. Note the different remaining choices.
For a given unordered multiplicity pattern in the additional chosen letters, say $AA+xxy$, we can fill this from the choice of letters based on which letters occur at a suitable multiplicity. For the example $AA+xxy$, $x$ can be chosen form $R$ or $T$, then $y$ is a free choice from the remaining $3$ letters. Then the arrangements of this pattern are the multinomial $\binom {5}{2,2,1} = 30$.
So once we identify all patterns we can assess each in turn to get a total answer:
$$\begin{array}{|c|c|} \hline
\text{For this pattern:} & \text{options to fill} & \text{arrangements} &  \text{total options} \\[1ex] \hline
 AA+xxx & \binom 11  = 1  & \binom{5}{3,2} = 10  &  10 \\[1ex]
 AA+xxy & \binom 21 \binom 31  = 6  & \binom{5}{2,2,1} = 30  &  180 \\[1ex]
 AA+xyz & \binom 43 = 4  & \binom{5}{2,1,1,1} = 60  &  240 \\[1ex]\hline
 &   & \text{total for }AA &  430 \\[1ex]\hline
 AAA+xx & \binom 11 = 1  & \binom{5}{3,2} = 10  &  10 \\[1ex]
 AAA+xy & \binom 42 = 6  & \binom{5}{3,1,1} = 20  &  120 \\[1ex] \hline
 &   & \text{total, }AA\text{ or }AAA &  560 \\[1ex]\hline
\hline
\end{array}$$
give a base answer of $\fbox{430}$ options for exactly two $A$s and an extended possibility for the case when three $A$s are also allowed of $\fbox{560}$ options
