The urn contains an unlimited number of red, blue, black and green balls. 3 balls are taken out of the urn.

The urn contains an unlimited number of red, blue, black and green balls. 3 balls are taken out of the urn. How many color options are there for these unordered triplets?

According to the condition of the problem, it is required to make a combination of k = 3 elements, which are selected with repetitions from objects of n = 4 types (red, blue, black and green balls).
This will be equal to the number of combinations with repetitions of four objects of 3, according to the formula for the number of combinations with repetitions, we have:
C` (4,3) = C (4 + 3 – 1, 3) = C (6,3) = 6! / (3! (6 – 3)) = 4 5 6 / (1 2 3) = 20;
Answer: There are 20 color options for unordered triplets.