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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.

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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.

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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.

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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 $AGA,AAG,GAA$.

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    $\begingroup$ That is not quite right: since he uses the choose function, he is not imposing any order on the elements that he picks. So, if he chooses $A_1, G_1, A_2$, then that does not necessarily represent sequence AGA, it could also be AAG or GAA. Also, notice that with your reasoning, any sequence gets counted at least 6 times ... So with 64 possible sequences, he would have to get at least 384 possibilities, and yet he only gets to 220. So again, your explanation for what goes wrong in his thinking is not correct. $\endgroup$
    – Bram28
    Nov 12, 2016 at 12:42
  • $\begingroup$ You are absolutely right. Thank you for your correction. $\endgroup$
    – Tengu
    Nov 12, 2016 at 12:49
  • $\begingroup$ Good to see how readily you realized and acknowledged the mistake: shows you are a good mathematician already! $\endgroup$
    – Bram28
    Nov 12, 2016 at 12:57
  • $\begingroup$ @Bram28 Thanks for the correction. $\endgroup$
    – Ivankovich
    Nov 12, 2016 at 13:02

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