Q:

Combinations Problem

 
I first calculated the total amount of ways to have a committee of 6 members without the given restrictions and found that to be $(12*11*10*9*8*7/6!)$ = 924 
Then I subtracted from 924 the total amount of ways that I could arrange of committee of 6 members that fails to meet the restrictions. 
I calculated the number of ways that a committee could be arranged in which the restrictions are not met to be equal to $4(9*8*7*6*5*4/6!)$ + 2 = 338 
After subtracting 338 from 924 I obtained 586 as my final answer, however, the answer is D. 592 
Would anyone like to take a crack at this? I have not taken Discrete Math, so if possible try to keep the solutions somewhat intuitive. 

A:

Hints:

There are only $2$ ways to choose a committee of $6$ members such that one of the majors is not represented. In both these cases the $4$ classes are represented. So firstly disregard the condition that the majors must be represented and if  that is solved then subtract $2$. 
Disregarding majors $6=1+1+1+3$ and $6=1+1+2+2$ cover all possibilities. Handle these possibilities separately.

$592=\binom41\times\binom31\binom31\binom31\binom33+\binom42\times\binom31\binom31\binom32\binom32-2$

