Key Concepts and Formulas

• Permutations with Repetition: The number of distinct arrangements of $n$ objects, where $n_1$ are of one kind, $n_2$ are of another kind, and so on, is given by $\frac{n!}{n_1!n_2!...}$.
• Fundamental Principle of Counting: If there are $m$ ways to do one thing and $n$ ways to do another, then there are $m \times n$ ways to do both.
• Leading Zero Restriction: A three-digit number cannot have 0 as its first digit.

Step-by-Step Solution

Step 1: Define the problem and possible structures.

We need to find the number of three-digit numbers with exactly one digit repeated twice. This means the number will be of the form $AAB$, $ABA$, or $BAA$, where $A$ and $B$ are distinct digits.

Step 2: Case analysis based on the presence of '0'.

Since a three-digit number cannot start with '0', we'll divide the problem into two main cases based on whether the repeated digit is zero or not.

• Case 1: The repeated digit ($A$) is not zero.
• Case 2: The repeated digit ($A$) is zero.

Step 3: Analyze Case 1: The repeated digit ($A$) is not zero.

Since $A$ is not zero, $A$ can be any digit from 1 to 9. We need to further divide this case into two subcases based on whether the other digit ($B$) is zero or not.

• Subcase 1.1: The non-repeated digit ($B$) is not zero.
• Subcase 1.2: The non-repeated digit ($B$) is zero.

Step 4: Analyze Subcase 1.1: The repeated digit ($A$) is not zero, and the non-repeated digit ($B$) is not zero.

• Choosing the digits: $A$ can be chosen in 9 ways (1 to 9). Since $B$ must be different from $A$ and also non-zero, $B$ can be chosen in 8 ways. Therefore, there are $9 \times 8 = 72$ ways to choose the digits $A$ and $B$.
• Arranging the digits: For each choice of $A$ and $B$, we have digits $A, A, B$. The number of distinct arrangements is $\frac{3!}{2!} = 3$. Since neither $A$ nor $B$ is zero, all 3 arrangements are valid three-digit numbers.
• Total numbers in Subcase 1.1: $72 \times 3 = 216$.

Step 5: Analyze Subcase 1.2: The repeated digit ($A$) is not zero, and the non-repeated digit ($B$) is zero.

• Choosing the digits: $A$ can be chosen in 9 ways (1 to 9). $B$ is fixed as 0, so there is only 1 way to choose $B$. Therefore, there are $9 \times 1 = 9$ ways to choose the digits $A$ and $B$.
• Arranging the digits: For each choice of $A$ and $B$, we have digits $A, A, 0$. The number of distinct arrangements is $\frac{3!}{2!} = 3$. However, the arrangement $0AA$ is not a valid three-digit number. The valid arrangements are $AA0$ and $A0A$. Thus, there are only 2 valid arrangements.
• Total numbers in Subcase 1.2: $9 \times 2 = 18$.

Step 6: Analyze Case 2: The repeated digit ($A$) is zero.

Since $A$ is zero, the number will be of the form $00B$, $0B0$, or $B00$, where $B$ is a non-zero digit.

• Choosing the digits: $A$ is fixed as 0. $B$ can be chosen in 9 ways (1 to 9). Therefore, there are $1 \times 9 = 9$ ways to choose the digits $A$ and $B$.
• Arranging the digits: For each choice of $A$ and $B$, we have digits $0, 0, B$. The number of distinct arrangements is $\frac{3!}{2!} = 3$. However, $00B$ and $0B0$ are not valid three-digit numbers. Only $B00$ is a valid arrangement. So there is only 1 valid arrangement.
• Total numbers in Case 2: $9 \times 1 = 9$.

Step 7: Calculate the total number of three-digit numbers.

Total = (Subcase 1.1) + (Subcase 1.2) + (Case 2) = $216 + 18 + 9 = 243$.

Common Mistakes & Tips

• Forgetting the Leading Zero Restriction: Always remember that a three-digit number cannot start with zero. This often requires separate case analysis.
• Incorrectly Applying Permutations: Make sure to account for repeated digits when calculating permutations.
• Double-Counting: Ensure that each number is counted only once by using mutually exclusive cases.

Summary

By carefully considering the restriction that a three-digit number cannot start with zero and dividing the problem into cases based on the repeated digit and the other digit, we found that there are 243 three-digit numbers with one digit repeated exactly two times.

The final answer is $\boxed{243}$.