We are choosing 2 from 5, 2 from 3, and 2 from 2.

\(\displaystyle \frac{\binom{5}{2}\cdot\binom{3}{2}\cdot\binom{2}{2}}{\binom{10}{6}}\)

Or, we can do it this way:

\(\displaystyle \frac{6!}{2!2!2!}\underbrace{\left(\frac{5}{10}\right)\left(\frac{4}{9}\right)}_{\text{red}}\overbrace{\left(\frac{3}{8}\right)\left(\frac{2}{7}\right)}^{\text{blue}}\underbrace{\left(\frac{2}{6}\right)\left(\frac{1}{5}\right)}_{\text{green}}\)

We multiply by \(\displaystyle \frac{6!}{2!2!2!}=90\) because that is how many ways the choices can be arranged.

They may come out like RGRBGB or RRGGBB, and so on.