Counting discrete math

Question

How many solutions are there to the equation a+b+c+d+e = 8, where all of {a,b,c,d,e} must be nonnegative integers?

Answer 1

4 +1+1+1+1= 8

3 +2 + 1+1+1 = 8

2+2+2 + 1 +1 = 8

3 basic solutions

but more when you assign values to a,b,c,d and e

a can equal 4 different values, same with b through e

only one solution if a=4, same with each of the other letters, for a total of 5 solutions using one four and four ones

if a=3 there are 4 solutions. multiply that times 5 = 20 solutions with a 3

if a=2 there are

if a= 1 there are

sum them up that's = 5+20+ + =