An urn has $4$ red, $6$ white and $3$ blue marbles, what is the probability of drawing with replacement $2$ red, $2$ white and $2$ blues? An urn has $4$ red, $6$ white and $3$ blue marbles, what is the probability of drawing with replacement $2$ red, $2$ white and $2$ blues? The order for marbles been picked up does not matter.
For example, I know the possible combination of selecting $6$ marbles from the total set should be ${13+6-1 \choose 6}$ because the selection comes with replacement. But I am not sure the numerator term. I have thought it is possible combinations from each kind of marbles and then multiply them all together: ${4+2-1 \choose 2}{6+2-1 \choose 2}{3+2-1 \choose 2}$.
 A: The selections you are making are not equally likely to occur.  For instance, there are $3^6 = 729$ ways for all six marbles to be blue since there are three ways to select a blue marble on each draw.  However, there are
$$\binom{6}{2}4^2\binom{4}{3}6^3\binom{1}{1}3^1 = 1,036,800$$
ways to select two red, three white, and two blue marbles since we must choose two of the six positions for the two red marbles, one of the four red marbles for each of those positions, three of the remaining four positions for the three red marbles, one of the six red marbles for each of those positions, and choose one of the three blue marbles for the remaining position.
There are $4 + 6 + 3 = 13$ choices for each of the $6$ selections, so there are $13^6$ possible sequences of marbles.  Of these,
$$\binom{6}{2}4^2\binom{4}{2}6^2\binom{2}{2}3^2$$
contain exactly two marbles of each color.  To see this, observe that we must choose which two of the six positions will be filled with red marbles, one of the four red marbles for each of those two positions, two of the remaining four positions for the white marbles, one of the six white marbles for each of those two positions, both of the remaining two positions for the blue marbles, and one of the three blue marbles for each of those positions.  Hence, the desired probability is
$$\frac{\dbinom{6}{2}4^2\dbinom{4}{2}6^2\dbinom{2}{2}3^2}{13^6}$$
