How many ways you arrange $12$ marbles where two marbles of the same color are considered the same? Problem:
Suppose that we have $6$ red marbles, $4$ blue marbles, $2$ green marbles and
$2$ black marbles. How many different ways are there to arrange them?
Note: Two marbles of the same color are considered identical.
Answer:
Let $c$ be the count we seek.
\begin{align*}
c &= \dfrac{ 12! }{6!4!2!2!} \\
c &= \dfrac{ 12(11)(10)(9)(8)(7)}{ 4(3)(2)(2)(2)} \\
c &= \dfrac{ 11(10)(9)(8)(7)}{ 2(2)(2)} = 11(10)(9)(7) \\
c &= 6930
\end{align*}
Is my solution correct?

Your solution is correct.
We can choose six of the twelve positions for the red marbles in $\binom{12}{6}$ ways, four of the remaining positions for the blue marbles in $\binom{6}{4}$ ways, and two of the remaining four positions for the green marbles in $\binom{4}{2}$ ways.  Both of the remaining positions must be occupied by the two black marbles.  Hence, the number of distinguishable arrangements of $6$ red, $4$ blue, $2$ green, and $2$ black marbles is
$$\binom{12}{6}\binom{6}{4}\binom{4}{2} = \frac{12!}{6!6!} \cdot \frac{6!}{4!2!} \cdot \frac{4!}{2!2!} = \frac{12!}{6!4!2!2!} = 6930$$
as you found. The factors in the denominator represent the number of ways marbles of the same color could be permuted among themselves in a particular permutation without producing a permutation which is distinguishable from that permutation.
