This is my attempt so far:
Let $A_{1}$ be a set of 3-digit numbers that contain 8's, $A_{2}$ be a set of 3-digit numbers that contain 9's.
Then, by subtracting the number of 3-digit numbers that do not contain 8's from the total number of 3-digit numbers, I obtain $\left | A_{1} \right |=(9 \times 10 \times 10)-(8\times 9 \times 9)=252$.
Similarly, $\left | A_{2} \right |=(9 \times 10 \times 10)-(8\times 9 \times 9)=252$.
since $ A_{1} \cap A_{2}$ is the set of 3-digit numbers that contain both 8's and 9's, so $A_{1} \setminus (A_{1} \cap A_{2})$ is the set of 3-digit numbers such that '8' is the largest digit (which is the desired set for answering this question).
However, I am not sure how to find $ \left | A_{1} \cap A_{2} \right |$.
Can someone please help me on this? Or if you have other better alternatives in solving this question, please let me know.
This is a question under the chapter about the Inclusion-Exclusion Principle.
I suggest an alternative approach.
Let $A_1$ be the set of three-digit positive integers with largest digit $8$ that have an $8$ in the hundreds digit.
Let $A_2$ be the set of three-digit positive integers with largest digit $8$ that have an $8$ in the tens digit.
Let $A_3$ be the set of three-digit positive integers with largest digit $8$ that have an $8$ in the units digit.
Then, by the Inclusion-Exclusion Principle, we need to find $$|A_1 \cup A_2 \cup A_3| = |A_1| + |A_2| + |A_3| - |A_1 \cap A_2| - |A_1 \cap A_3| - |A_2 \cap A_3| + |A_1 \cap A_2 \cap A_3|$$
$|A_1|$
Since both the hundreds digit and the largest digit in the three-digit positive integer are equal to $8$, we have nine choices for the tens digit (since we cannot use $9$) and nine choices for the units digit (for the same reason). Hence, $|A_1| = 1 \cdot 9 \cdot 9 = 81$.
$|A_2|$
Since both the tens digit and the largest digit in the three-digit positive integer are equal to $8$, we have eight choices for the hundreds digit (as both $0$ and $9$ are excluded) and nine choices for the units digit (since $9$ is excluded). Hence, $|A_2| = 8 \cdot 1 \cdot 9 = 72$.
$|A_3|$
Since both the units digit and the largest digit in the three-digit positive integer are equal to $8$, we have eight choices for the hundreds digit (as both $0$ and $9$ are excluded) and nine choices for the tens digit (since $9$ is excluded). Hence, $|A_3| = 8 \cdot 9 \cdot 1 = 72$.
Can you take it from here?