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 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
