How many 3 digit even numbers are there(No Repetition)? First find numbers ending with 0

So, 1's place-1 10's place-9 100's place-7 (2 digits are already consumed and 0 can't be used)

So 7*9*1.Im i doing the right thing?
First find the even numbers that are ending up in zero
so, for no. ending up with zero are
zero at one's place so 1 combination
now 9 at hundreds and 8 at tens
Now even numbers not ending with zero i.e. ending  in 2,4,6,8
so 4 at one's place only one is used so 8 at hundreds place since 0 can not be used and one number is 0 and now 8 are left for the tens place
so, total no. of digit sequences would be
=>9*8*1+8*8*4=72+256=328
When zero is used up there is no problem for using up tens or hundreds but when zero is not used up after putting one of 2,4,6,8 numbers at one's place we have to check if 0 is ending up in the most significant place or not. To avoid that we start with hundredth place. So we ensure this first.
The total 3 digit numbers are 999 including preceding zeros and there are 999/2 even numbers.
so total three digit even numbers are 499.