Using Draw Blanks approach to solve a probability question of drawing 2 red, 2 blue and 1 green marbles. Q: "There are 6 children at a family reunion, 3 boys and 3 girls. They will be lined up single-file for a photo, alternating genders. How many arrangements of the children are possible for this photo?" A: 
if we use the drawing 6 blanks approach _ _ _ _ _ _, we get (3*3*2*2*1*1) + (3*3*2*2*1*1)
= 36 + 36 = 72.
so I want to try to use this same approach to answer this below 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: _ _ _ _ _
(4*3*5*4*2) divide by total probability which is 5!
, which is wrong.
I know that we can easily answer this question using: $$\dfrac{(_4C_2) \cdot (_5C_2) \cdot (_2C_1)}{_{11}C_5}$$
I can also use permutations to answer this:
$$\binom{5}{2}\binom{3}{2}\binom{1}{1}\left(\frac{4}{11}\right)\left(\frac{3}{10}\right)\left(\frac{5}{9}\right)\left(\frac{4}{8}\right)\left(\frac{2}{7}\right) = \frac{20}{77}$$
But I am trying to learn how I can use different methods to solve the same problem.
please help me in solving this problem using the method from my first question.
Any help is greatly appreciated.
A: There are $\binom{5}{2}$ possible positions for the blue balls in the sequence.  That leaves three positions in the sequence, of which two must be filled with red balls, so there are $\binom{3}{2}$ positions for the red balls in the sequence.  The remaining position in the sequence must be filled with the green ball.  Consequently, there are
$$\binom{5}{2}\binom{3}{2}\binom{1}{1} = 10 \cdot 3 \cdot 1 = 30$$
possible sequences with two blue balls, two red balls, and one green ball.
The probability of obtaining blue, blue, red, red, green in that order is 
$$\frac{5}{11} \cdot \frac{4}{10} \cdot \frac{4}{9} \cdot \frac{3}{8} \cdot \frac{2}{7}$$
The probability of obtaining blue, green, red, red, blue in that order is 
$$\frac{5}{11} \cdot \frac{2}{10} \cdot \frac{4}{9} \cdot \frac{3}{8} \cdot \frac{2}{7}$$
Notice that the denominators are the same and that the terms in the numerator have just been permuted.  Therefore, the probabilities of these sequences are equal.  In fact, this is true of each of the $30$ sequences consisting of two blue balls, two red balls, and one green ball.  
Therefore, while the probability of any particular sequence is 
$$\frac{5}{11} \cdot \frac{4}{10} \cdot \frac{4}{9} \cdot \frac{3}{8} \cdot \frac{2}{7}$$
the probability that a sequence will contain two blue balls, two red balls, and one green ball can be obtained by multiplying the probability that any particular sequence will appear by the 
$$\binom{5}{2}\binom{3}{2}\binom{1}{1} = 30$$
possible orders in which these balls could appear.
Thus, the probability of obtaining two blue balls, two red balls, and one green ball when five balls are selected from a bag containing five blue balls, four red balls, and two green balls is
$$\binom{5}{2}\binom{3}{2}\binom{1}{1}\frac{5}{11} \cdot \frac{4}{10} \cdot \frac{4}{9} \cdot \frac{3}{8} \cdot \frac{2}{7}$$
Since we actually care about which balls are selected rather than the order in which they are selected, we do not need to consider the sequence in which the balls are selected.  We just need to find the probability of selecting two of the five blue balls, two of the four red balls, and one of the two green balls when selecting five of the eleven balls in the bag, which is
$$\frac{\dbinom{5}{2}\dbinom{4}{2}\dbinom{2}{1}}{\dbinom{11}{5}}$$
