Let's consider the information given:
⇒ we have a 3 digit number
⇒ it can only have one '1' within it
Let's consider the possibilities:
- first possibility: the digit '1' is in the hundreds place
⇒ 1 _ _
⇒ the tens digit has 9 choices of number 0 ⇔9 except 1
⇒ the ones digit has 9 choices of number 0⇔9 except 1
Total choices (in this case) = 9 * 9 = 81
- second possibility: the digit '1' is in the tens place
⇒ _ 1 _
⇒ the hundred digit has 8 choices '0 ⇔ 9' except 0 and 1
⇒ ones digit has 9 choices 0 ⇔ 9 except 1
Total choices (in this case) = 8 * 9 = 72
- third possibility: the digit '1' is in the ones place
⇒ _ _ 1
⇒ the hundreds digit has 8 choices 0 ⇔ 9 except 0 and 1
⇒ ones digit has 9 choices 0 ⇔ 9 except 1
Total choices (in this case) = 8 * 9 = 72
Overall Total choices = 81 + 72 + 72 = 81 + 144 = 225
Hope that helps!