Part of the solution is straightforward. The number of "teams" of four distinct digits is a combinatorial coefficient informally called " ten choose four" . This is equal to 10!/( 4! 6! ) = 210. However, not all of these 210 "teams" will sum to a number divisible by 3.
I do not have a rigorous approach to the next step which is to determine how many of the 210 actually sum to a number divisible by 3. However, a random set of four numbers has the property that the sum will be a multiple of three with probability 1/3. That is: Given a set of four numbers chosen at random , the sum will be a multiple of three 1/3 of the time.
For the problem at hand, this argument suggests that 1/3 of the 210 "teams" will sum to a multiple of three.