Basic combinations logic doubt in probability 
"If $3$ students are chosen at random from a class with $6$ girls and $4$ boys, what is the probability that all $3$ students chosen will be girls?"
$\left(\dfrac{6}{10}\right)\left(\dfrac{5}{9}\right)\left(\dfrac{4}{8}\right)$

So why can't we use that logic to answer this question?
"A bag holds $4$ red marbles, $5$ blue marbles, and $2$ green marbles. If $5$ marbles are selected one after another without replacement, what is the probability of drawing $2$ red marbles, $2$ blue marbles, and $1$ green marble?"
My answer: $\left(\dfrac{4}{11}\right)\left(\dfrac{3}{10}\right)\left(\dfrac{5}{9}\right)\left(\dfrac{4}{8}\right)\left(\dfrac{2}{7}\right)$
But the correct answer is $\dfrac{(_4C_2) \cdot (_5C_2) \cdot (_2C_1)}{_{11}C_5}$ (where $C$ is a combination).
Why doesn't the logic from the first problem work here?
The draws are without replacement in all cases.

The first question solved in the second method looks as follows:
$$\frac{{6\choose 3}}{{10\choose 3}}=\frac{6\cdot 5\cdot 4}{10\cdot 9\cdot 8}$$
Interpretation: There are ${6\choose 3}$ ways to choose $3$ girls out of $6$ and there are ${10\choose 3}$ ways to choose $3$ students out of $10$, hence the probability is the ratio of number of favorable outcomes to the total number of possible outcomes. 
Now compare it with the answer of the second problem and try to interpret the selections.
