Given ,
x + y + z = 30
where one of x,y,z will vary from 0 to 3..Hence the other two variables need to have values > 3 and these 3 values of x , y and z need to be distinct as well..
So to do this , we consider x to contain values from 0 to 3..
We fix x = 0 , then we have possible tuples for (y,z) as : (4,26) , (5,25) , .............(26 , 4) [ Except (15,15) as no two variables can have same value ]..
Hence possible tuple (y,z) for x = 0 = 22
Hence number of solutions with x = 0 : 22
Now similarly taking x = 1,2,3 , we get 22, 20 and 20 solutions respectively..
Hence fixing the values of x = 0 to 3 , we get number of solutions = 22 + 22 + 20 + 20
= 84
Similarly fixing the value of y from 0 to 3 and looking for (x,z) pair we get 84 solutions..So is the case with fixing the value of z from 0 to 3..
Hence total number of possible solutions = 84 + 84 + 84
= 252