Arrangements of 10 red balls, 5 green balls and 5 blue balls where there can't be two adjacent balls of the same color In how many ways can 10 identical red balls, 5 identical green balls and 5 identical blue balls be arranged in a row, if there can't be two adjacent balls of the same color?
What I tried: The arrangement must be of the form:
red - another color - red - another color - ... etc.
or
another color - red - another color - red ... etc.
So the problem is equivalent to find the number of arrangements of 5 green balls and 5 blue balls and then multiply by 2.
the number of arrangements of 5 green balls and 5 blue balls is $\frac{10!}{5!5!}=252$.
So, there are $2 \times 252 = 504$ arrengements.
I'd like to know if this solution is correct. Thank you for your help.
 A: Consider the reds separated by $5$ blues and $4$ greens, with one green spare
$\color{red}{\Large\bullet}\color{blue}{\Large\bullet}\color{red}{\Large\bullet}\color{blue}{\Large\bullet}\color{red}{\Large\bullet}\color{blue}{\Large\bullet}\color{red}{\Large\bullet}\color{blue}{\Large\bullet}\color{red}{\Large\bullet}\color{blue}{\Large\bullet}\color{red}{\Large\bullet}\color{green}{\Large\bullet}\color{red}{\Large\bullet}\color{green}{\Large\bullet}\color{red}{\Large\bullet}\color{green}{\Large\bullet}\color{red}{\Large\bullet}\color{green}{\Large\bullet}\color{red}{\Large\bullet}$
The spare green has $7$ places to fit: $5$ right(say) of a blue, and $2$ ends, thus together with an interchange between blue/green,
Number of ways = $\boxed {2\times7\binom94 = 1764}$
A: There is another possibility you have not considered.  There could be both a green and a blue ball between two red balls.
Line up the ten red balls.  We must fill the nine spaces between successive red balls with at least one ball of another color.
If there is exactly one ball of another color between each red ball, then red balls must alternate with balls of another color.  Choosing whether such an arrangement begins or ends with a red ball determines which spaces are occupied by red balls.  In each case, choosing which five of the ten spaces reserved for balls of another color are occupied by the blue balls determines the arrangement.  Hence, as you found, there are
$$2\binom{10}{5}$$
such arrangements.
The other possibility is that one blue and one green ball are placed in the same space between successive red balls.  There are nine ways to choose this space and two ways to arrange the blue and green balls within this space.  That leaves eight spaces to fill with the remaining four blue and four green balls.  Choose which four of those eight spaces are filled with blue balls.  There are
$$\binom{9}{1}2!\binom{8}{4}$$
such arrangements.
That gives a total of
$$2\binom{10}{5} + \binom{9}{1}2!\binom{8}{4} = 504 + 1260 = 1764$$
admissible arrangements.
