How many 4-digit numbers contain the digit 5 exactly once?

5 in the thousands place: other 3 digits from the 9 non-5 digits, $9^3=729$. 5 elsewhere (3 positions): thousands digit 1–9 except 5 (8 ways), other two from 9 each: $3\times 8\times 81=1944$. Total $729+1944=2673$.