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.
Best Answer
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 = $\boxed {2\times7\binom94 = 1764}$