We use the fact that a number is divisible by 3 if and
only if the sum of its digits are divisible by 3.

Notice that 1+2+3+4+5 = 15, which is divisible by 3

The only other way we can have a sum of 5 digits 
divisible by 3 is to replace the 3 by the 0 making 
the sum 3 less:

1+2+0+4+5 = 12, which is divisible by 3

No other choice of 5 digits can have a sum divisible
by 3, because there is no other way to make the 
sum 12 or 15, and we certainly can't have a sum of
9 or 18. 

So the number of 5-digit numbers that can be formed 
from the digits {1,2,3,4,5} is 
choose the first digit 5 ways.
Choose the second digit 4 ways.
Choose the third digit 3 ways.
Choose the fourth digit 2 ways.
Choose the fifth digit 1 way.

5×4×3×2×1 = 5! = 120

And the number of 5-digit numbers that can be formed 
from the digits {1,2,0,4,5} is figured this way.

We choose the most restrictive digit first, which is the
first digit. It can be chosen 4 ways, so 

choose the first digit 4 ways. (It CANNOT be 0)
Choose the second digit 4 ways. (It CAN be 0)
Choose the third digit 3 ways.
Choose the fourth digit 2 ways.
Choose the fifth digit 1 way.

4×4×3×2×1 = 4×4! = 96

Answer: 120+96 = 216

Edwin