Confused about order in probability 
A committee consists of five Mexicans, two Asians, three African Americans, and two Caucasians. A subcommittee of 4* is chosen at random. What is the probability that all the ethnic groups are represented on the subcommittee?

Note: original posting said 5 but I meant 4. Also, I know, the race thing is weird, but it was in my book.
It seems to me like the answer should be 
$$\frac{\frac{5*2*3*2}{4!}}{\binom{12}{4}}$$ 
because we want both the numerator and the denominator to be unordered selection. 
Alternatively, we would say that it is 
$$\frac{5*2*3*2}{12*11*10*9}$$
However, something in my reasoning is incorrect as the solution is just 
$$\frac{5*2*3*2}{\binom{12}{4}}$$
I am dreadfully confused.
source: http://www.math.illinois.edu/~psdey/stat20SU07/Solutions2_2007.pdf
 A: There are $\binom{12}{5}$ possible committees.
Committees of $5$ that include at least one member from each group necessarily have just one group having two members on the committee, all other groups have just one. So, the number of such committees is
$$\binom{5}{2}\cdot 2\cdot 3\cdot 2 + 5\cdot \binom{2}{2}\cdot 3\cdot 2
+ 5\cdot 2 \cdot \binom{3}{2}\cdot 2 + 5\cdot 2 \cdot 3
\cdot \binom{2}{2}$$
A: If we think of seating the people in 4 seats and insist that the first seat be occupied by a Mexican, the second by an Asian, the third by an African American and the fourth by a Caucasian, then the chance of this is: 5/12 x 2/11 x 3/10 x 2/9. In the problem you have set we don't care which person sits in which seat so long as one of each is present. There are 4! ways of re-arranging the people in the seats, so the answer to your question is 4! times the product I just gave, which once you re-arrange it is the same as the answer given in the book.
A: The numerator is the number of ways to choose $4$ people from $12$ people such that exactly one person is chosen from each ethnic group.  The numerator is thus ${5 \choose 1} {2 \choose 1}{3 \choose 2}{2 \choose 2}$.  The denominator is clearly ${12 \choose 4}$.  
Alternatively, you could assume that the subcommittee  has a name for each of the positions - say president, vice president, secretary and treasurer. Then, the numerator would be ${5 \choose 1} {2 \choose 1}{3 \choose 2}{2 \choose 2} 4!$ because once the 4 people are chosen so that each ethnic group is represented, they can be placed in various positions in $4!$ ways. For the same reason, the denominator is ${12 \choose 4} 4!$.  The $4!$'s cancel out and we get the same probability.
