How many positive, three-digit integers contain at least one $3$ as a digit but do not contain a $5$ as a digit?
To count the number of such numbers, I need to consider how many ways I can choose the digits for each of the three places: hundreds, tens, and units. For the hundreds place, I can choose any digit from $1$ to $9$, except $5$, so that's $8$ possibilities. For the tens place, I can choose any digit from $0$ to $9$, except $5$, so that's $9$ possibilities. For the units place, I can choose any digit from $0$ to $9$, except $5$, so that's $9$ possibilities. However, this includes numbers that do not have any $3$'s, which I need to exclude. To count the numbers that do not have any $3$'s, I can repeat the same logic but also exclude $3$ as a possible digit. For the hundreds place, I can choose any digit from $1$ to $9$, except $3$ and $5$, so that's $7$ possibilities. For the tens place, I can choose any digit from $0$ to $9$, except $3$ and $5$, so that's $8$ possibilities. For the units place, I can choose any digit from $0$ to $9$, except $3$ and $5$, so that's $8$ possibilities. Multiplying these numbers, I get $7 \times 8 \times 8 = 448$ numbers that do not have any $3$'s. Therefore, to get the numbers that have at least one $3$, I need to subtract this from the total number of numbers that do not have any $5$'s. Multiplying the numbers for the total, I get $8 \times 9 \times 9 = 648$ numbers that do not have any $5$'s. Subtracting, I get $648 - 448 = 200$ numbers that have at least one $3$ but no $5$'s. 