Numbers question - tricky one

IMO C.

Between 801 & 900, there are 100 digits of which 50 are even & 50 are odd. We need to subtract the numbers that have the units & tens digits same.
For the number 8 _ _ to be odd, the units digit should be odd (i.e. 1,3,5,7,9). The tens digit can be any number between 0-9, but should not be the same at the units & hundreds digit.

The odd units digit can be selected in 5 ways. The tens digit can then be selected in 9 ways. So, totally there are 5*9=45 odd numbers that don't have the same units & tens digit. Now, there are 881, 883, 885, 887 & 889 that were included in the count. So, totally there are 50-10 = 40 odd numbers between 800-900 that have all 3 digits different.

For the number 9_ _ to be odd, similar logic as above should be applied. There are 18 numbers that have either the hundreds-unit or hundred-tens or tens-units that have the same odd number. So, totally there are 50-18 = 32 odd numbers between 901-999 that have all 3 digits different.

Totally, there are 40+32 = 72 odd integers that have all 3 digits different.

It can be solved as an P&C:

First digit in two way
8 x x
9 x x

Last digit are: 1 3 5 7 9
for 8 x x , all 4 can be last digit
8 x 1/3/5/7/9: the x can be choosen in 8 ways
so a total of 5*8 = 40 ways

with 9: The last digit can be choosen from 1/3/5/7(9 is laready there)

9 x 1/3/5/7
so a total of 4*8 ways = 32 ways

total of 40+32 ways = 72