# 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?

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.

• Welcome to MathSE. This tutorial explains how to typeset mathematics on this site. – N. F. Taussig Feb 25 '19 at 1:10
• Hint: for the purpose of placing the blue balls only, red and green are equivalent. So when it comes to placing blue balls, you have 5 blue and 10 non-blue balls. Easier now? (Red and green are still distinct for the purposes of placing them) – smci Feb 25 '19 at 3:00

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.