Partial Groupings

We can break this down into cases, looking separately at $d = 0,1,2,3,4,5.$

For $d = 0$ we then have $a + b + c = 20$ with $a,b,c$ non-negative integers, which is a stars and bars calculation with $\dbinom{22}{20} = 231$ solutions.

For $d = 1$ we then have $a + b + c = 16,$ which has $\dbinom{18}{16} = 153$ solutions. Continuing with this process, we have in total

$\dbinom{22}{20} + \dbinom{18}{16} + \dbinom{14}{12} + \dbinom{10}{8} + \dbinom{6}{4} + \dbinom{4}{0} =$

$231 + 153 + 91 + 45 + 15 + 1 = \boxed{536}$ solutions.

For d=0, a+b+c=20

therefore no. of ways in which the distribution can be done is (20+3-1)C(3-1) = 22(C)2 = 22*21/2 = 231

For d=1, a+b+c = 16

therefore no. of ways in which the distribution can be done is (16+3-1)C(3-1) = 18(C)2 = 18*17/2 = 153

For d=2, a+b+c = 12

therefore no. of ways in which the distribution can be done is (12+3-1)C(3-1) = 14(C)2 = 14*13/2 = 91

For d=3, a+b+c = 8

therefore no. of ways in which the distribution can be done is (8+3-1)C(3-1) = 10(C)2 = 10*9/2 = 45

For d=4, a+b+c = 4

therefore no. of ways in which the distribution can be done is (4+3-1)C(3-1) = 6(C)2 = 6*5/2 = 15

For d=5, a+b+c = 0

therefore no. of ways in which the distribution can be done is (0+3-1)C(3-1) = 2(C)2 = 1

Therefore, total no. of ways = 231+153+91+45+15+1 = 536.......this is the answer.