OK
( use star and bar technique number of solution to x1 + x2 + x3 + .... xN = S = C( S + N - 1 , N - 1 ) where xi >= 0 )
First convert x3 >= 15 , into x3 >= 0
then x1 + x2 + x3 + x4 + x5 = 21 - 15 = 6
x1 + x2 + x3 + x4 + x5 = 6 where 0 <= x1 <= 3 , 1 <= x2 <= 3 , x3 >= 0 , x4 >= 0 , x5 >= 0
now x3 , x4 , x5 have no constraint on upper bound , we have to focus only on x1 and x2
x1 and x2 have 12 pairs : ( 0 , 1 ) ( 0 , 2 ) ( 0 , 3 ) ( 1 , 1 ) ( 1 , 2 ) ( 1 , 3 ) ( 2 , 1 ) ( 2 , 2 ) ( 2 , 3 ) ( 3 , 1 ) ( 3 ,2 ) ( 3 , 3 )
Min sum they can make is 1 and maximum is 6
Now for each sum find the number of ways they make sum S and multiply the number of ways x3,x4,x5 make sum (6 - S) where 1 <= S <= 6
numeber of ways for x1 and x2 for sum = 1 , 2 , 3 , 4 , 5 , 6 = ( 1 , 2 , 3 , 3 , 2 , 1 )
final result = 1 * ( x3 + x4 + x5 = 6 - 1 ) + 2 * ( x3 + x4 + x5 = 6 - 2 ) + 3 * ( x3 + x4 + x5 = 6 - 3 ) + 3 * ( x3 + x4 + x5 = 6 - 4 ) + 2 * ( x3 + x4 + x5 = 6 - 5 ) + 1 * ( x3 + x4 + x5 = 6 - 6 )
= 1 * C ( 5 + 3 - 1 , 3 - 1 ) + 2 * C ( 4 + 3 - 1 , 3 - 1 ) + 3 * C(3 + 3 - 1 , 3 - 1 ) + 3 * C ( 2 + 3 - 1 , 3 - 1 ) + 2 * C ( 1 + 3 - 1 , 3 - 1 ) + 1 * ( 0 + 3 - 1 , 3 - 1 )
= 1 * 21 + 2 * 15 + 3 * 10 + 3 * 6 + 2 * 3 + 1 * 1 = 21 + 30 + 30 + 18 + 6 + 1 = 106.