# How many integers between $1000$ and $9999$ inclusive have distinct digits and are odd numbers?

I know this question was already answered (How many integers between 1000 and 9999 inclusive consist of), and I understand the solution. However, I don't understand what's wrong with my approach. I did it from right to left. My reasoning:

Last digit has to be odd, so we have $$5$$ possibilities. In the tenths place, we can choose whatever we want except for the one already chosen, so we have $$9$$ possibilities. Similarly, in the hundredths place we have $$8$$ choices. In the thousands place, we can choose $$10 - 3 - 1$$ (not including $$0$$) $$= 6$$ possibilities, so in total we have $$5 * 9 * 8 * 6 = 2160$$. Why am I undercounting here?

Because if you've already chosen $$0$$ before (in the tens or hundreds place), then you're excluding $$0$$ twice in allowing only $$10-3-1=6$$ options for the thousands place.
To compensate this, you could count the number of cases where you have a $$0$$ in the tens or hundreds place. But as in the linked answer, it is easier to just start with the thousands place.