Question

How many solutions does the equation A+B+C=5 have if A, B, and C are non negative integers?

Answer 1

`21`

Explanation:

If `A=0` there are `6` possible values for `B: [0,5]`
If `A=1` there are `5` possible values for `B: [0,4]`
If `A=2` there are `4` possible values for `B: [0,3]`
and so on.

Once choices have been made for `A` and `B` there is only ever `1` choice for `C`

Therefore the number of solutions is
`6+5+4+3+2+1 = 21`