Different results for choosing committee with at least two architects A committee of four needs to be chosen from a group of 3 engineers, 5 economists and 4 architects. In how many ways (people are unique) can it be formed, such that there are at least 2 architects in it?
I have reasoned in two different ways and got different answers. Where is the flaw in my reasoning?
The first way is to split the possibilities into those with 2, 3 and 4 architects. This yields:  

$ C(4,2) \cdot C(8,2) + C(4,3) \cdot C(8,1) + C(4,4) \cdot C(8,0) = 201$

The other is to choose first two architects and then 2 people from the remaining 10:

$ C(4,2) \cdot C(10,2) = 270$

One of this solutions was arrived at by other people, so I have reason to think that one is correct but I am not certain so I let it to you to resolve this matter. Thanks!
Also, if my interpretation is correct the second method is used in this question. What is the difference?

The first method is correct. The second method is not correct. 
Letting $A,B,C,D$ be four architects, the second method counts a set $\{ABCD\}$ multiple times.
1) First you choose $AB$, then you choose $CD.$
2) First you choose $AC$, then you choose $BD$.
and so on...
