ARML Practice Problem

One way of thinking about this problem is that there are 4 digits, each of which will receive some number of 1s, and there are only 7 1s to go around. In this view, we can use Stars and Bars to find the answer. First, we find the number of the ways the 7 1s can be distributed to the 4 digits. We have ${10 \choose 3}=120$ . However, this is including cases where the first digit is 0, like 0250. In these cases, the numbers are basically equivalent to a 3-digit number (or less). Therefore, we subtract ${9 \choose 2}=36$ from 120 which gives us $\boxed{84}$ .