I have the following question:
We have $n$ balls of color red, $n$ of color green and $n$ of color
blue (the balls with the same color are identical). We want to place
them in a sequence in such a way that no two blue balls are next to
each other. How many ways do we have to do it?
My approach is that once we pick the position for the red and green colors, then the position of the blue balls get uniquely determined. There are $\binom{3n}{n}$ ways to position red balls and $\binom{2n}{n}$ to position green balls. So, in total we have $\binom{3n}{n}\times \binom{2n}{n}$ ways to position red and green balls. From this, we have to somehow substract the cases where blue balls are next to each other. This is where I am stuck.
Best Answer
First arrange the red and green balls in $$ \frac{(2n)!}{n!n!} $$ ways to get a word like (for $n=3$) $$ RGRRGG $$ Next consider the $(2n+1)$ gaps determined by this word which are displayed by wedges in the above word. $$ \wedge R\wedge G\wedge R\wedge R\wedge G\wedge G \wedge $$ In order the fufill the requirement that blue balls not be together we choose $n$ of the gaps to be blue.
Hence there are
$$\frac{(2n)!}{n!n!}\times \binom{2n+1}{n}$$
ways to play the balls in a sequence without two blue balls next to one another.