How many of the numbers 1, 2, 3, …, 1000 contain the digit 9 exactly once?

Pad every number below 1000 out to three digits — 007, 090, 481 — which changes nothing, because a leading zero is never a 9, and note that 1000 itself has no 9 at all. Now choose which of the three places holds the 9 (3 ways) and fill the other two places with any of the nine non-9 digits $(9\times 9=81$ ways), giving $3\times 81=243$. The trap 271 counts the numbers with AT LEAST one 9, which is $1000-9^3=1000-729$, and so also sweeps in 99, 199 and 999. Answering 300 uses $3\times 100$ and forgets to bar the digit 9 from the other two places.