Easy counting problem I have been trying to learn combinatorics from a book. While doing so, I encountered a problem. Here it is:

A DNA chain is composed of blocks from four chemicals - $\mathrm{A}$, $\mathrm{T}$, $\mathrm{G}$ and $\mathrm{C}$. How many such chains of length three are present if repetitions are allowed?

So what I did was I assumed that I had $3~\mathrm{A}$'s, $3~\mathrm{G}$'s, $3~\mathrm{T}$'s and $3\mathrm{C}$'s. Now the problem is reduced to finding the number of ways to choose $3$ elements from $12$ which is equal to $[12~\mathrm{choose}~3]$. The answer for that is $220$.
However, the book says that there are $4$ ways to fill each of the three blocks so the answer is $4^3 = 64$.
I want to why my method is wrong in this problem.
 A: You picked 3 A's, 3G's, etc. because you are thinking: OK, the first element can be an A, the second can be an A, and the third can be an A ... And the same goes for the other letters.  So that means that you are basically starting out with the following 12 objects: an A in the first place, an A in the second, an A in the third place ... And again the same for the other 3 letters. We can look at this as having 12 different objects $A_1, A_2$, etc.  
But now, when you randomly choose 3 objects out of these 12, some possible outcomes would be something like $A_3, G_1, G_3$. But note, that is really not an allowed sequence, because you would have both an A and a G in the third position (and nothing in the second position). So, your method gets too many possibilities compared to what are the real possibilities.
A: In your counting argument, for example, $AGA$ is counted $3\times 3\times 2$ times: there are $3$ ways to pick $G$, $3$ ways to pick the first $A$, $2$ ways to pick the last $A$.
You've picked $3$ elements from $12$ elements: $A_1,A_2,A_3,G_1,G_2,G_3,T_1,T_2,T_3$ and $C_1,C_2,C_3$. That means you've included $A_1G_1A_2$ and $A_2G_1A_3$ in your counting but these are essentially the same and both are $AGA$.
EDIT: Wrong argument as pointed out below. $A_1G_1A_2,A_2G_1A_3$ could be $AGA,AAG,GAA$.
A: Your first assumption is wrong, if I got the question correctly, the question says you can use 4 chemicals and creat a chain of length 3. So , if repeatetion allowed 3 places can be filled 4*4*4 = 64.
