Q:

Probability box of cards

I have to solve next task: There is box with 12 red cards, 8 blue cards and 7 green cards. Four cards are taken from it, and must find:
a) probability to has 2 red, 1 blue and 1 green card
b) all 4 cards to be the same color:
My logic: a)
$$\frac{C\frac{2}{12}.C\frac{1}{8}.C\frac{1}{7}}{C\frac{4}{27}} = \frac{66.8.7}{17550} = \frac{3696}{17550} = 0.21$$
b) $$\frac{ C\frac{4}{12} }{ C\frac{4}{27} } + \frac{C\frac{4}{8}}{\frac{4}{27}} + \frac{C\frac{4}{7}}{C\frac{4}{27}} = \frac{11880}{421200} + \frac{1680}{421200} + \frac{840}{421200} = \frac{14400}{421200} = 0.034$$
Are my calculations correct?

A:

There are some mistakes in combinatorics notation, but the final answers are OK. In question b), you shouldn't use the combinatorics notation in the second step and on, just in the first expression. 
I mean:
a) $\frac{\binom{12}2\binom{8}1\binom{7}1}{\binom{27}4}=\frac{3969}{17550}=0.2105$ 
b) $\frac{\binom{12}4+\binom{8}4+\binom{7}4}{\binom{27}4}=\frac{495+70+35}{17550}=\frac{600}{17550}=0.034$

