The solution could also be found using combinations with repetitions,

The solution is equal to a combination of putting 21 elements in 5 boxes(consider each variable a box).

Hence there are 21 elements and 4 borders x1 | x2 | x3 | x4 | x5 where the elements and borders could be combined in different ways.

as x3>=15, the solution could be simplified to 6 elements in 4 borders ( 5 boxes) ; for which the solution is C(10,6)

Now for the remaining conditions, subtract the case when x1>3 and when x2>3 , which is 2*C(6,2)

The final step is to subtract the case when x2=0 , which is tricky because when you take x2=0, there are now 6 elements to fill in 3 borders (4 boxes) and you also have to consider the case when x2=0 and x1>3, the value for which was already subtracted previously.

So in essence the final step would be C(9,6) - C(5,2) (All combinations when x2=1 - All combinations when x2=1 and x1>3)

The final solution is the C(10,6)-C(6,2)-C(6,2)-(C(9,6)-C(5,2)) = 106