From 8 red, 9 blue, and 10 green balls, how many ways are there to choose 2 red, 2 blue, and 2 green balls?

Correct answer: 45360

The governing concept is the product rule for independent combinations. The red selection can be made in C(8,2) = 8×7/2 = 28 ways. The blue selection can be made in C(9,2) = 9×8/2 = 36 ways, and the green selection can be made in C(10,2) = 10×9/2 = 45 ways. The choices from the three colours are independent, so each red pair can be combined with every blue pair and every green pair. Hence the total number of selections is 28 × 36 × 45 = 45,360. Therefore option C is correct. Permutations are inappropriate because the order in which balls are selected does not create a new selection. Selecting six balls without enforcing two of each colour would answer a different question.