Since each selection should have all the three colours, minimum 3 balls one

in each colour should be selected in $^3C_1\times ^4C_1\times ^2C_1$ ways.

Maximum 6 balls are to be selected.

$\therefore$ The required no. of ways =$^3C_1\times ^4C_1\times ^2C_1(^6C_0+^6C_1+^6C_2+^6C_3)$ ways

$=24(1+6+15+20)=42.4!$