Consider five points in the plane, with no three of them collinear....

Case-I:
2 ways (corresponding to each color) As color of other sides got fixed Case-II:
Not possible $\therefore 0$ ways as color of other sides got fixed and there will be one triangle which will have all sides rud on block Case-III:
$\therefore \quad$ not possible $\therefore 0$ ways (same reason as above) Case-IV:
Only one way of colouring as color of other sides got fixed $\therefore \quad 2$ ways (corresponding to each color for shown figure) $\therefore \quad$ total number of ways $=2+{ }^5 C_1 \times 2=12$ ways ${ }^5 C_1$ ways of choosing 2 will active rides