Um I know that there are $\large\frac{15!}{5!5!5!}$ combinations but I'm kinda stumped after that.
I tried doing the space thing and I got ${11 \choose 5}^2$ after my answer.
I don't really know what to do.
2026-03-30 12:24:14.1774873454
How many ways are there to arrange $5$ red, $5$ blue, and $5$ green balls in a row so that no two blue balls lie next to each other?
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Arrange the red and green balls first, which can be done in $\ {10\choose 5}\ $ ways. The blue balls can then only be placed one at either end of the row, or in a space between two of the red and green balls. There are thus exactly 11 places where they can be put, and this can be done in $\ {11\choose 5}\ $ ways. Therefore, there are $\ {10\choose 5}{11\choose 5}\ $ ways of arranging the balls so that no two blue ones lie next to each other.