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.

What you calculated is the probability of selecting two red marbles, two blue marbles, and one green marble in that order.  However, if we select blue, green, red, red, blue, we still get two red marbles, two blue marbles, and one green marble.  
To correct your attempt, we must multiply by the number of orders in which we could get two red marbles, two blue marbles, and one green marble.  Choose two of the five positions for the blue marbles and two of the remaining three positions for the green marbles.  The only green marble must go in the remaining position.  Then we obtain
$$\binom{5}{2}\binom{3}{2}\binom{1}{1}\left(\frac{4}{11}\right)\left(\frac{3}{10}\right)\left(\frac{5}{9}\right)\left(\frac{4}{8}\right)\left(\frac{2}{7}\right) = \frac{20}{77}$$
Since we do not care about the order in which the marbles are selected, it is simpler to calculate the probability of selecting two of the four red marbles, two of the five blue marbles, and one of the two green marbles when we select five of the eleven marbles, which yields
$$\frac{\dbinom{4}{2}\dbinom{5}{2}\dbinom{2}{1}}{\dbinom{11}{5}} = \frac{20}{77}$$
