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Three variables x,y,z have a sum of 30. All three of them are non=negative integers. If any two variables don't have the same value and exactly one variable has a value of less than or equal to three, then find the number of possible solutions for the variables.
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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 

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