In how many ways can you select a committee of 3 persons, so that no two are from the same department? The problem asks the following:

A certain company has 4 departments, with 100, 200, 300, and 400 employees respectively.
  In how many ways can you select:
  (a) a committee of 4 persons, so that no two are from the same department?
  (b) a committee of 3 persons, so that no two are from the same department?

I think I've solved (a). There are four positions, and for the first position you can choose from the whole pool of the four departments {100,200,300,400} which gives you a 1000 total options. For the second position, lets say someone was chosen from the biggest department. We'll exclude that from the pool, leaving us with {100,200,300} people, i.e. 600 options. For the third and fourth positions we exclude the next two largest departments, leaving us with 300 and 100 options respectively. Multiply these values and you get the total number of options for part a. Does that make sense?
For the second question (b) things get more complex. I think it's safe to say that for the first position, there are still a 1000 options for committee members. But for options b and c, what options are available depend on what you've chosen for the first option. In total, there are 4 * 3 * 2 total configurations for the departments in problem (b) (i.e. 24). I have no idea how to work out this problem. Should I manually calculate all 24 configurations? I think there needs to be a more elegant solution but I can't quite think of one.
EDIT: I also just realized that you technically don't have to figure out every configuration, as many would amount to the same number of options - but I'm still stuck. Any insight would be most welcome! 
 A: There is a problem with your solution for (a). When you say "let's say someone was chosen from the biggest department" you are throwing away all the options where it was not the case. One possible solution would be to add other cases to account for the possibilities where the first person chosen is from one of the smaller department. An easier solution is to consider that you need to chose one person per department, so you have $400$ possibilities for the first, $300$ for the second, $200$ for the third and $100$ for the last. This gives you
$$400 \cdot 300\cdot 200 \cdot 100 = 24\cdot 10^{8}$$
combinations.
For the (b) you have $4$ different cases, depending on which department is not represented in the commitee. For example, if the biggest department is not represented then you have $300\cdot 200\cdot 100$ possibilities (by a similar reasoning as above). Adding all the different possible cases gives you
\begin{align}300\cdot 200 \cdot 100 & + 400\cdot 200 \cdot 100 +400\cdot 300 \cdot 100 +400\cdot 300 \cdot 200 \\ & =(6+8+12+24)\cdot10^6= 50\cdot 10^6\end{align}
possible combinations.
