Key Concepts and Formulas

• Permutations with Repetition: The number of permutations of $n$ objects, where $p_1$ are alike of one kind, $p_2$ are alike of another kind, and so on, is given by $\frac{n!}{p_1! p_2! ...}$.
• Place Value: In a number, each digit's value is determined by its position (units, tens, hundreds, thousands, etc.).
• Sum of Numbers Formed by Permutations: The sum of all numbers formed by permuting a set of digits can be calculated by considering the frequency of each digit at each place value.

Step-by-Step Solution

Step 1: Identify the Given Digits and Determine the Total Number of Distinct Arrangements

We are given the digits $1, 2, 2, 3$. The digit '2' is repeated twice. Our goal is to find the total number of distinct four-digit numbers that can be formed using these digits.

Using the formula for permutations with repetitions, where $n=4$ (total digits) and the digit '2' appears twice: $\text{Total numbers} = \frac{4!}{2!} = \frac{4 \times 3 \times 2 \times 1}{2 \times 1} = \frac{24}{2} = 12$ There are 12 unique four-digit numbers that can be formed.

Step 2: Calculate the Frequency of Each Distinct Digit at Each Place Value

We need to determine how many times each unique digit (1, 2, and 3) appears in the units, tens, hundreds, and thousands places. Due to symmetry, the frequency will be the same for each place value. We will calculate the frequency for the thousands place.

• Frequency of '1' in the Thousands Place: If '1' is fixed in the thousands place, the remaining digits are $2, 2, 3$. The number of arrangements of these remaining digits is: $\frac{3!}{2!} = \frac{3 \times 2 \times 1}{2 \times 1} = 3$ So, the digit '1' appears 3 times in the thousands place.

• Frequency of '2' in the Thousands Place: If '2' is fixed in the thousands place, the remaining digits are $1, 2, 3$. The number of arrangements of these remaining digits is: $3! = 3 \times 2 \times 1 = 6$ So, the digit '2' appears 6 times in the thousands place.

• Frequency of '3' in the Thousands Place: If '3' is fixed in the thousands place, the remaining digits are $1, 2, 2$. The number of arrangements of these remaining digits is: $\frac{3!}{2!} = \frac{3 \times 2 \times 1}{2 \times 1} = 3$ So, the digit '3' appears 3 times in the thousands place.

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

Step 3: Calculate the Sum of the Digits at Each Place Value

The sum of the digits in any specific place is: $\text{Sum} = (\text{Frequency of '1'} \times 1) + (\text{Frequency of '2'} \times 2) + (\text{Frequency of '3'} \times 3)$ $\text{Sum} = (3 \times 1) + (6 \times 2) + (3 \times 3) = 3 + 12 + 9 = 24$ This means the sum of the digits in the units, tens, hundreds, and thousands places is 24.

Step 4: Calculate the Total Sum of All Numbers

To find the total sum, we multiply the sum of the digits at each place value by its corresponding place value multiplier and add the results: $\text{Total Sum} = (24 \times 1) + (24 \times 10) + (24 \times 100) + (24 \times 1000)$ $\text{Total Sum} = 24 \times (1 + 10 + 100 + 1000) = 24 \times 1111 = 26664$

Common Mistakes & Tips

• Forgetting Repetitions: Always account for repeated digits when calculating the number of distinct arrangements and the frequency of each digit at each place value. Failing to do so will lead to an incorrect result.
• Incorrect Frequency Calculation: Ensure you are calculating the frequency of each unique digit correctly. Fix the digit in a particular place and permute the remaining digits, considering any repetitions among them.
• Arithmetic Errors: Double-check all calculations to avoid simple arithmetic mistakes that can significantly impact the final answer.

Summary

When calculating the sum of numbers formed by a set of digits with repetitions, it is crucial to account for the repetitions when determining the number of distinct arrangements and the frequency of each digit at each place value. By calculating the sum of the digits at each place and multiplying by the appropriate place value multipliers, we can determine the total sum of all possible numbers. The sum of all the four-digit numbers that can be formed using the digits 2, 1, 2, 3 is 26664.

The final answer is \boxed{26664}.