among the numbers-0,1,2,3,4,5,6,7,8,9 how many teams can be made so that it contains four numbers and the sum of them are divisible by 3.

Question

NOTE THAT,the sequence doesn't matter.such that (1,2,3,4) and (1,3,2,4) are same

Answer 1

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. 

With this surmise the answer is   210/3 = 70.