How many odd integers from 1000 through 9999 have distinct digits?

We know that
step 1
the number of elements in a list from n to m is (m-n)+1
therefore
from 1000 to 9999 there are
(9999-1000)+1 = 9000 elements
step 2
Since even and odd numbers alternate, half of our list is composed of odd numbers. Therefore, there are 4500 of them
step 3
To construct a 4 digit odd number, we have 5 choices for the last digit.
(1-3-5-7-9)
then
we have
8 choices for the first (not 0 or same as last digit),
8 choices for the second
7 choices for the third.
there are (8)(8)(7)(5) = 2240 

the answer is 2240 odd numbers with distinct digits in the list