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.

  • $\begingroup$ Welcome to MathSE. This tutorial explains how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Feb 25 '19 at 1:10
  • $\begingroup$ 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) $\endgroup$ – 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.

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  • $\begingroup$ Thanks what i did was 11 choose 5 squared not 10 choose 5 * 11 choose 5. Thank You! $\endgroup$ – Wesley Wang Feb 25 '19 at 0:13

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