TITLE: In how many ways can we arrange $6$ red, $3$ blue and $4$ green balls in a row such that there is no adjacent green balls?
QUESTION [5 upvotes]: All ball of one color are identical.
My idea is to calculate first the total numbers to arrange the $13$ balls. It equals $\dfrac{13!}{6! 3! 4!}=60060.$ 
Then I want  remove the cases where all $4$ green ball are adjacent, then remove cases where $3$ green ball are adjacent and the cases  $2$ green ball are adjacent.
For the  first case (all $4$ green ball are adjacent) I have got $\dfrac{9!}{6! 3!}=84 $ cases,  but cant  calculate rest cases. Any   help please.

REPLY [7 votes]: There are $9$ non-green balls; line them all up in a row.  Now consider the $8$ spaces between the balls, and the $2$ spaces at the ends—$10$ in all.  We can put the $4$ green balls in any of those spaces (with no more than one in each space).  Thus the number of such arrangements is
$$
\binom{10}{4} = 210
$$
To this must be considered the multiple ways in which the $3$ blue balls can be placed amongst the $9$ non-green balls.  This is
$$
\binom{9}{3} = 84
$$
Thus the total number of possible arrangements is
$$
84 \cdot 210 = 17640
$$