a set includes 7 one digit numbers, 3 two digit numbers, and 4 three digit numbers. How many groups of 3 can you make so that each subset has a number of different digits.

84 groups of three elements so that each subset has a number of different digits.

Explanation:

This is a counting problem, which uses the Fundamental Counting Principle.

Assuming that the order does not matter, this is, that the groups {A, B, C}, {A, C, B}, {B, A, C}, {B, C, A}, {C, A, B}, and {C, B, A} are equivalent, you have that the number of groups of three different elements with one element (number) from each subset is equal to the product of the number of elements of each subset:

7 * 3 * 4 = 84 different groups.