How many 3-digit numbers between 100 and 999 contain at least one digit equal to 7?

Count by cases so that each number is counted once according to the first place where 7 appears in the chapter's method: 7 in the units place gives $9\times 10\times 1=90$. 7 in the tens place but not units gives $9\times 1\times 9=81$. 7 in the hundreds place but neither tens nor units gives $1\times 9\times 9=81$. Total $=90+81+81=252$. Option A is for exactly one digit equal to 7, not at least one.