Answer to Question #336964 in Discrete Mathematics for Wende

Question

8) How many positive integers less than 1000 have at least one decimal digit equal to 9?

Answer 1

All positive integers less than $1000$ can be presented in the form: $a_1a_2a_3$, where digits $a_1,a_2,a_3$ take on values from to $9$. Integers less than $1000$, which have the last digit $9$, have the form: $a_1a_29$, where $a_1$ and $a_2$ take on values from to $9$. Using the multiplication principle of combinatorics, we receive that there are $100$ integers of this type. Integers that have the second digit $9$ have the form: $a_19a_3$, where $a_1$ takes on values from to $9$ and $a_3$ takes on values from to $8$. We omit the case $a_3=9$ because it is included in the integers of the previous type. We receive $90$ integers of this type using the multiplication principle of combinatorics. Integers that have the first digit $9$, have the form: $9a_2a_3$, where $a_2$ takes on values from to $8$ and $a_3$ takes on values from to $8$. We receive $81$ numbers of this type. Thus, we receive $100+90+81=271$ integers that have at least one digit $9$.

Answer: $271$ integers