Find the number of three-digit numbers in which exactly one digit $3$ is used? Find the number of three-digit numbers in which exactly one digit $3$ is used?
The number is of one of the forms $$\_\text{ }\_\text{ }3 \\ \_\text{ } 3 \text{ } \_ \\ 3\text{ }\_ \text{ }\_$$ There are $V_9^2-V_8^1=9\times8-8=64$ possibilities for each of the first $2$ forms, and $V_9^2$ for the third. This makes $2\times64+72=200$ possibilities in total. The given answer in my book is $225$. What am I missing?

We have $9$ digits other than $0$, but $0$ can't be the first digit. So for both the first and second, we have $8 \cdot 9=72$ possibilities each. For the last one we have $9 \cdot9=81$ possibilities.
Alternatively, there are $9 \times 10 \times 10 = 900$ three-digit numbers. $8 \times 9 \times 9 = 648$ of them have no $3$s as the first digit cannot be $0$ or $3$, and $1$ number has all three $3$s. For the numbers with two $3$s, they must be in the form _ 33, 3 _ 3, 33 _, which makes $8 + 9 + 9 = 26$ possibilities.
Thus there are $900 - 648 - 1 - 26 = 225$ three-digit numbers with exactly one $3$.
