From 3 red, 4 yellow, and 5 blue balls, in how many ways can 1 red, 1 yellow, and 1 blue ball be selected?

Correct answer: 60

The governing idea is the multiplication principle applied to independent choices from separate colour groups. One red ball can be selected in C(3,1) = 3 ways, one yellow ball in C(4,1) = 4 ways, and one blue ball in C(5,1) = 5 ways. Every red choice can be paired with every yellow choice and every blue choice, so the total is 3 × 4 × 5 = 60. In combination notation this is C(3,1)C(4,1)C(5,1) = 60. Therefore option D is correct. The tempting expression C(12,3) is inappropriate because it permits arbitrary colour distributions, such as three red balls or two red and one yellow. The required one-of-each-colour condition must be enforced by separate choices and multiplication.