Revisiting combinatorics 1!

The number of values, where the first digit is odd (5 choices for each of the 6 digits, as there are 5 even and 5 odd digits, we can choose the position of the 2 even digits out of 5 (except the first digit)) :

${ 5 \choose 2} × 5^6 = 156250$

The number of values, where the first digit is even (4 choices for the first digit (as there are 4 even digits different from 0), 5 choices for the rest of the 5 digits each; the position of the other even digit can be chosen 5 ways):

$4 × 5 × 5^5 = 62500$

The total number of suitable values:

$156250+62500 = 218750$

Hence, our answer should be:

$\frac {218750}{10000} = 21.875 = \boxed { 22 \text { (0 d. p.) } }$