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