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$$P(E)=\frac{{4 \choose 2}{38 \choose 1}}{{42\choose 3}}+\frac{{4 \choose 3}}{{42\choose 3}}= \frac{29}{1435}$$
 A: The $P(E)$ you've calculated is correct.
However, for solving the original problem, you'll have to calculate the probability of the event that there are $3$ Daniels in a group, and then add probabilities of both events.
Edit:
I re-read the problem, and your calculation seems a bit off.
For event $E$,
No of Daniels $= 4$;    ways of choosing $2$ from $4$ = $4 \choose 2$
No of "Non-Daniels" $= 42 - 4 = 38$; ways of choosing $1$ from $38$ = $38 \choose 1$.
Total groups of size $3 =$ ways of choosing $3$ from $42 =$ $42 \choose 3$
A: 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 $\frac{113}{410}$. 
A: We divide the whole class into 3 parts (14each) and now we find the probablity for one such group, that is:
P( 2 daniels in one group)= 14C2 * 28C2 / 42C4 ( here we are not concerned what happens in the other parts)
P (3daniels in one group) = 14C3 * 28C1 / 42C4
P (4 daniels in one group) = 14C4 * 28C0 / 42C4
now add up all the three for the atleast answer
