What is the difference in these two logics used to solve this question? In a school committee comprising 12 members, there are exactly two students from each of the classes from class 7 to class 12. In how many ways can a sub-committee of four students be formed, such that no two students in the sub-committee belong to the same class?
First logic
There are 6 classes in total. We have to select 4 out of them which can be done in 6C4 ways. And from each of these classes we can select in 2 ways so 2^4.
So totally there would be (6C4)*(2^4) ways which comes out to be 240 ways.
Second logic
There are 12 members to choose from which can be done in 12C1 ways. Then for the second spot, we have 10 members(we can't choose the guy already chosen plus the guy in his class). third spot can be filled in 8C1 ways. Fourth spot can be filled in 6C1 ways. SO this can be done in 12C1*10C1*8C1*6C1 = 5760. 
The answers are starkingly different but I can't seem to find any flaw in both the logics.

Consider the following simpler argument: "There are two ways to pick a pair of two elements from a two-element set $\{a, b\}$: there are 2C1 choices for the first element, and 1C1 for the second."
Obviously this is wrong. So what's the mistake? (HINT: you only want to count distinct possible choices of pairs . . .)
