Key Concepts and Formulas

• Permutations with Repetitions: The number of permutations of $n$ objects where $n_1$ are of one kind, $n_2$ are of another kind, and so on, is given by $\frac{n!}{n_1!n_2!\dots}$.
• Sum of Digits at a Place Value: If $f_i$ is the frequency of digit $d_i$ at a specific place value, the sum of digits at that place value is $\sum (d_i \times f_i)$.
• Total Sum Calculation: The total sum of all numbers formed is the sum of (sum of digits at each place value $\times$ place value).

Step-by-Step Solution

Step 1: Identify Digits and Calculate Total Number of Distinct Numbers

We are given the digits 1, 2, 2, and 3. We want to find all distinct 4-digit numbers that can be formed using these digits. The total number of such numbers is given by the permutation formula with repetitions. Here, we have 4 digits in total, with the digit '2' repeating twice. $\text{Total Numbers} = \frac{4!}{2!} = \frac{4 \times 3 \times 2 \times 1}{2 \times 1} = 12$ There are 12 distinct 4-digit numbers.

Reasoning: This step calculates the total count of numbers we are considering. This is a necessary starting point to ensure the consistency of later frequency calculations.*

Step 2: Calculate Frequency of Each Digit at the Units Place

We need to find how many times each of the digits 1, 2, and 3 appears in the units place.

• Case A: '1' in the Units Place: If '1' is in the units place, the remaining digits are 2, 2, and 3. The number of ways to arrange these is: $\frac{3!}{2!} = \frac{3 \times 2 \times 1}{2 \times 1} = 3$ So, '1' appears 3 times in the units place.

• Case B: '2' in the Units Place: If '2' is in the units place, the remaining digits are 1, 2, and 3. The number of ways to arrange these is: $3! = 3 \times 2 \times 1 = 6$ So, '2' appears 6 times in the units place.

• Case C: '3' in the Units Place: If '3' is in the units place, the remaining digits are 1, 2, and 2. The number of ways to arrange these is: $\frac{3!}{2!} = \frac{3 \times 2 \times 1}{2 \times 1} = 3$ So, '3' appears 3 times in the units place.

  Verification: The sum of the frequencies ($3 + 6 + 3 = 12$) equals the total number of distinct numbers, confirming our calculations.*

Reasoning: This step determines how many times each unique digit contributes to the units place. We fix a digit in that position and then arrange the remaining digits. We must use the correct permutation formula (with or without repetitions) for the remaining digits.*

Step 3: Calculate the Sum of Digits at the Units Place

The sum of the digits in the units place is: $\text{Sum of Units Digits} = (1 \times 3) + (2 \times 6) + (3 \times 3) = 3 + 12 + 9 = 24$

Reasoning: This step translates the frequencies into a numerical sum for the units place. Each digit contributes its value multiplied by the number of times it appears in that position.*

Step 4: Calculate the Sum of Digits at all Place Values

Since any digit can occupy any position, the frequency of each digit will be the same for the tens, hundreds, and thousands places as it was for the units place. Therefore, the sum of the digits at each place is 24.

• Sum of digits at the tens place = 24
• Sum of digits at the hundreds place = 24
• Sum of digits at the thousands place = 24

Reasoning: We leverage the symmetry of permutations. The distribution of how many times each digit appears in any given position is identical. This avoids repeating the frequency calculation for each place value.*

Step 5: Calculate the Total Sum

The total sum of all the numbers is: $\text{Total Sum} = (24 \times 1) + (24 \times 10) + (24 \times 100) + (24 \times 1000)$ $\text{Total Sum} = 24 \times (1 + 10 + 100 + 1000) = 24 \times 1111 = 26664$

Reasoning: This final step aggregates all the contributions to arrive at the solution. The factorization simplifies the calculation.*

Common Mistakes & Tips

• Repeated digits: Remember to account for repeated digits when calculating permutations. Use the formula $\frac{n!}{n_1!n_2!\dots}$.
• Place value: Do not forget to multiply the sum of digits at each place value by the correct power of 10.

Summary

To find the sum of all distinct numbers formed by the digits 1, 2, 2, and 3, we calculated the total number of distinct numbers, the frequency of each digit at each place value, and then summed the contributions from each place value. The total sum is 26664.

Final Answer

The final answer is $\boxed{26664}$, which corresponds to option (A).