Probability of two Daniels in one group my professor gave us a "fun" problem to work on at home, but I am relatively new to probability. 
The question is as follows: There are four Daniels in a class of 42 students. If the class breaks into groups of three what is the probability that there is at least two Daniels in one group?
I am trying to solve the easier problem "What is the probability that there is two Daniels?" However, I am stuck even here. I will show that I have this far and why.
$$P(E)=\frac{{4 \choose 2}{40 \choose 1}}{{42\choose 3}}$$
I am using ${4 \choose 2}$ for choosing two daniels from 4, ${40 \choose 1}$, 1 student from the remaining forty, and  ${42\choose 3}$ for groups of 3 possible given 42 students. I may have a deep misunderstanding of probability, however I feel that the thing I feel most unsure about is using ${40 \choose 1}$. Part of me feels like it should be ${42 \choose 1}$, but I am not entirely sure why. 
Any help, or clarifications of things I may be confused about will be greatly appreciated.  
Edit after feedback: Attempted solution to original question$$P(E)=\frac{{4 \choose 2}{38 \choose 1}}{{42\choose 3}}+\frac{{4 \choose 3}}{{42\choose 3}}= \frac{29}{1435}$$

We will assume that the question asks for the probability that there is at least one group that contains two or more Daniels.  (Note that there can be two such groups.) We will first find the probability $p$ that the four Daniels are in different groups. Then the answer to the question is $1-p$. 
We can assume that the placement into (labelled, but it doesn't matter) groups takes place as follows. The $42$ people are lined up in a row. The first three go into Group 1, the next three into Group 2, and so on. The number of labelled groupings is then $\frac{42!}{(3!)^{14}}$, but it is convenient not to divide by the $(3!)^{14}$. 
The oldest Daniel can be placed in $42$ positions, the next oldest in $39$, the next in $36$, and the next in $33$. Now the rest of the people can be lined up in the remaining slots in $38!$ ways. Thus
$$p=\frac{(42)(39)(36)(33)(38!)}{42!}.$$
The above expression for $p$ can be simplified considerably! Finally, compute $1-p$. It is $\frac{113}{410}$. 
