.
The "ones" digit must be "1", or "3", or"5", or "7", or "9" -- in all, 5 options.
The "tens" digit then can be any of 8 remaining digits ("0" is excluded as the leading digit).
Therefore, 5*8 = 40 two-digit numbers are possible, satisfying given conditions.
Another solution is THIS :
In all, there are 90 two-digit numbers, from 10 to 99.
Of them, one half, i.e. 45 numbers, are even and the other half, i.e. 45 numbers, are odd.
Of 45 odd numbers, exclude these five numbers 11, 33, 55, 77 and 99, since they have two repeating digits.
The rest 40 = 45 - 5 two-digit numbers satisfy the problem's conditions.
