Number of possible $4$ letter words from a set of $3$ letters, where only one letter repeats twice. I have a set $A$ of $3$ letters, $A=\{a,b,c\}$. How many $4$ letter words are there that be formed where exactly $1$ letter repeats twice. I think that I have to work out the number of such combinations where $a$ repeats twice, then again for $b$ and then $c$ and add these up. But I'm not sure how. The possible solutions are:


*

*$2 \cdot 4!$

*$1/2 \cdot 3! \cdot 3 \cdot 4$

*$2 \cdot 3 \cdot {4 \choose 2}$

*$2^4 \cdot {3 \choose 2}$
Edit 1 - The 3rd option is $4 \choose 2$ and not $4 \choose 3$.
Edit 2 - I did it the same way @fleabood did. As @Arthur suggested, it is easier to consider the case where only $a$ repeats itself twice since the number of way for such an arrangement is the same for when only the $b$ or $c$ repeats itself. If only the $a$ repeats itself twice, since we have $4$ available slots of where to place the $2$ $a's$, we have that the number of ways to place the $2$ $a's$ is $4 \choose 2$. The remaining two slots can be used for $b$ or $c$, to which only $2$ such arrangements exist. Therefore, for the case where only the $a$ repeats itself twice, we have that the number of such arrangements is ${4 \choose 2} = 6$. The same logic works for if only the $b$ or $c$ repeats itself twice, therefore the total number of possibilities is $2 \cdot 3 \cdot {4 \choose 2} = 36$.
There is however another answer: Option 2 $1/2 \cdot 3! \cdot 3 \cdot 4$, since this also equals 36. However, I'm not sure I understand the logic behind that. What I think is happening, is that in any $3$ of the $4$ slots, we do not place a repeated letter, i.e. only place one $a,b,c$. For such an arrangement there are $3!$ possibilities. Now for the $4^{th}$ slot we can choose any one of the letters to place. If we say that we want to repeat only $a$, then this $a$ can go in any one of the $4$ slots, therefore, we have $3! \cdot 4$ such possibilities for repeating only the letter $a$ twice. Applying the same logic for the previous answer we have $3 \cdot 4 \cdot 3!$. But I'm not sure why the $1/2$ is present there.
PS - Sorry for the long explanation. I'm just trying to make sure that I understand it fully.
 A: There are four letters in the string.  Exactly two of those letters are the same letter repeating twice. The other two letters must be distinct and different.  So we use three different types of letters.  And we only have three types of letters to choose from so we use all types.
We have a $1$ out of $3$ choice,( ${3\choose 1} = 3$), for which letter is repeated.  Then we have a $2$ out of $4$ choice, (${4\choose 2}=\frac {4!}{2!2!}=\frac {3*4}2=6$), for where to position those letters.  And of the remaining two slots  we can arrange the remaining two distinct letters $2!=2$ ways.
So the solution is ${3\choose 1}{4\choose 2}2!$ or $3*2!{4\choose 2}$ or $3*6*2=36$ or some other similar representation.
So the answer is $c$.

To follow your idea:
You hae to place the $a$s.
You can do $aa--$ or $a-a-$ or $a--a$ or $-aa-$ or $-a-a$ or $--aa$. So there are $6$ ways.
This is small enough be do by hand but it may be better to get used to think:  I have $4$ options to place the $a$ and I must choose $2$ of the.  That is ${4\choose 2}$.  I can have for options to choose the first $a'$  and then I have three remaining slots for the $a^*$. So that is $4*3$ ways.  But the $a'$ and $a^*$ are indistinguishable so I overdid it.  There for whatever order I put $a'$ and $a^*$ in I could switch them around so we have $\frac{4*3}2 =6$.  In general... $n$ choose $r$ will be ${n\choose r} = \frac {n*(n-1)*.....*(n-r+1)}{r*(r-1)*....*2*1}=\frac {n!}{(n-r)!r!}$.
Then in the remaining two remaining spaces you must place the $b$ and $c$.  You have two options:  Place the $b$ in the first available slot and $c$ in the second, or ..... place the $b$ in the second and place the $c$ in the first.  So that is $2$ ways
So $6*2=12$ ways to place the $2$ $a$s.
Then be symmetry, two $b$s and two $c$s are the same.
So there are $3*12 =36$ ways.
A: Consider the 3 letter $\{a,b,c\}$. Note  these letters can be permuted in $3!=6$ ways. 
Now  consider the spaces  found within a given three letter word, for example 
$$-a-b-c-$$ 
In those four spaces you can place  any one of the 3 letters, therefore you have 12 combinations  between the four spaces and the three letters  
$$\{a1,b1,c1,a2,b2,c2,a3,b3,c3,a4,b4,c4\}$$
where $a1$ represents letter a in space 1. 
This means we have a total of 12*6=72  letters we can form. 
Since two letters are repeated we have over counted by the permutations of those two letters so we divide our total number of words $72$ by $2!=2$. So we have a total of $\frac{72}{2}=36$   four letter words that can be formed from the letters $\{a,b,c\}$ where  only two letters can be repeated. 
