Calculate number of (four-letter) strings that contain exactly two matching characters (s) The following problem refers to strings in A, B, ..., Z.
Question:

How many four-letter strings are there that contain exactly two S's?

I used the formula in this answer to come up with the following: 
$\left(26^4-25^4\right)\cdot 2\space =132,702$
However, this seems to be incorrect... any tips?

 A: There are $\binom42=6$ choices for the locations of the $2$ S's.  The remaining two letters can be assigned arbitrarily, and there are $25$ choices for each.  So the total number of $4$-long strings containing exactly $2$ S's is
$$ 6\times 25^2 = 3750.$$
A: If a string contains exactly two S's, then there are $25^2$ possibilities for the other two letters. For $25*24$ of these possibilities, the two letters that are not S's will be distinct, and for the other 25, the two non-S letters will be identical. Fixing some position for the S's, there are $25*24*2$ strings with distinct non-S letters, and there are $25$ strings with identical non-S letters.
Now, there are ${4 \choose 2} = 6$ possible positions of the S's, so the total number of strings is
$$
6(25*24*2 + 25) = 7350.
$$
A: I would like to suggest as follows, however it is better to double check !!!
Lets consider possible number of three-letter strings which include $S$ from $A,B,\cdots,S,\cdots,Z$:
$${_{25}}C_2\cdot {_1}C_1$$
Now we can add additional $S$ to each set, e.g., $A,B,S$ with $A,B,S,S$, which is a four-letter string. 
Then, we consider possible number of combinations of each four-letter string, which gives
$$\frac{4!}{2!}$$ (Note that we have to divide by $2!$ since we have two identical letters)
Then, we may have all possible number of strings
$${_{25}}C_2\cdot {_1}C_1\cdot \frac{4!}{2!}=3600.$$
