Probability of words The question is as follows:

A word of $6$ letters is formed from a set of $16$ different letters of English alphabet (with replacement). Find out the probability that exactly two letters are repeated.

What I think is there is a word that has six letters among which there are there are two that have repetitions i.e. among the 6 only 4 are different, rest are repetitions.
Based on this, I first found out the total number of 6 letter words possible from the 16 letters. It is, of course $16^6$. Now to find out the required no. of words, I have the following expression (E is event space),
$$ n(E) = {16\choose 4} \times \frac{6!}{2! \times 2!}$$
That's because, I need to choose only 4 letters from 16 and then the arrangements of 6 letters in the word among which 2 and 2 are same, e.g., fghkgk. So the probability comes out to be,
$$ P(E)= \frac{{16\choose 4} \times \frac{6!}{2! \times 2!}}{16^6}$$
However the answer given is,
$$ P(E)=\frac {18080}{16^6}$$
which is much less than the answer I am getting. I would like others to explain to me what is wrong in the logic that I have used.
 A: There are many possible interpretations of the problem. I will assume that $2$ letters are supposed to occur twice each, and the other $2$ letters are to occur once each. 
The lonely letters can be chosen in $\binom{16}{2}$ ways. For each such way, the doubled letters can be chosen in $\binom{14}{2}$ ways. Now that we have our letters, they can be arranged as you had it, in $\frac{6!}{2!2!}$ ways. Multiply. We get a big number. 
If each letter that we have $2$ of is to be next to its mate, then instead of multiplying by $\frac{6!}{2!2!}$ we multiply by $4!$. 
Your $\binom{16}{4}$ counts the number of ways to choose $4$ letters. However, there is a difference between having the letters a, a, b, b, c, d and having a, a, b, c, c, d. 
A: There are 16 different letters and we have to form words with 2 letters repeated.
Six different letters can be selected by 16C6 ways. Further, from these 6 different letters, 6! ways words can be formed.
Hence, the Exhaustive number of words = 16C6 x 6!
= 8008 x 720 words can be formed.
In the selection of 6 letters, we want exactly 2 letters (2 x 2) to be repeated. This amounts to selection of only 4 letters form the words with given condition.
Favorable cases = 16C4 x 6!/2!x 2! = 1820 x 720/4 = 1820 x 180
So, the required probability = 1820 x 180/8008 x 720 = 0.227
