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?