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?
Best Answer
First, fill all places. $9\times 9\times 3=243$ possibilities. Now remove $0's$ which appear at the front, whose only possibilities are in the first two cases which are $9+9=18$. Hence $243-18=225$.