To find the sum of all the numbers that can be formed using the digits 2, 3, 4, 4, and 5, we will follow these steps:
### Step 1: Identify the digits and their frequency
The digits we have are: 2, 3, 4, 4, 5.
- The digit 4 is repeated twice, while the others occur once.
### Step 2: Calculate the total number of distinct permutations
The total number of distinct permutations of the digits can be calculated using the formula for permutations of multiset:
\[
\text{Total permutations} = \frac{n!}{p_1! \cdot p_2! \cdots p_k!}
\]
where \( n \) is the total number of items, and \( p_1, p_2, \ldots, p_k \) are the frequencies of the repeated items.
Here, \( n = 5 \) (total digits), and the frequency of 4 is 2. Thus, we have:
\[
\text{Total permutations} = \frac{5!}{2!} = \frac{120}{2} = 60
\]
### Step 3: Determine the contribution of each digit to each place value
Each digit will appear in each place (units, tens, hundreds, thousands, ten-thousands) an equal number of times. Since there are 60 total permutations and 5 places, each digit will appear in each place:
\[
\text{Occurrences of each digit in each place} = \frac{60}{5} = 12
\]
### Step 4: Calculate the contribution of each digit
The contribution of each digit to the total sum can be calculated by multiplying the digit by the number of times it appears in each place and the place value.
The place values for a 5-digit number are:
- Ten-thousands: \(10^4 = 10,000\)
- Thousands: \(10^3 = 1,000\)
- Hundreds: \(10^2 = 100\)
- Tens: \(10^1 = 10\)
- Units: \(10^0 = 1\)
The total contribution of each digit can be calculated as follows:
\[
\text{Total contribution} = \text{Occurrences} \times \text{Digit} \times \text{Sum of place values}
\]
Where the sum of the place values is:
\[
10,000 + 1,000 + 100 + 10 + 1 = 11,111
\]
### Step 5: Calculate the total contribution of all digits
Now, we calculate the total contribution for each digit:
- For digit 2: \(12 \times 2 \times 11,111 = 266,664\)
- For digit 3: \(12 \times 3 \times 11,111 = 399,996\)
- For digit 4: \(12 \times 4 \times 11,111 = 533,328\) (since 4 appears twice, we multiply this contribution by 2)
- For digit 5: \(12 \times 5 \times 11,111 = 666,660\)
Now, we sum these contributions:
\[
\text{Total sum} = 266,664 + 399,996 + 533,328 + 533,328 + 666,660
\]
Calculating this gives:
\[
\text{Total sum} = 2,399,976
\]
### Final Answer
Thus, the sum of all the numbers that can be formed by taking all digits 2, 3, 4, 4, 5 is **23,99,976**.
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