
Mike D. answered 04/17/21
Experienced high school discrete math teacher
Well as they are divisible by 10, the last digit must be 0.
To choose the first three digits you choose 3 digits sum is 10.
First cannot be zero.
So you can have
1 09 18 27 36 45 54 63 72 81 90
2 08 17 26 35 44 53 62 71 80
3 07 16 25 34 43 52 61 70
4 06 15 24 33 42 51 60
5 05 14 23 32 14 05
6 04 13 22 31 40
7 03 12 21 03
8 02 11 02
9 01 10
So 2+3+4+5+6+7+8+9+10 possible combinations with the last digit being zero