Three-digit numbers are formed from the digits 1, 2 and 3 in such a way that the digits are not repeated. What is the sum of such three-digit numbers?
1332
Calculating the Sum of Three-Digit Numbers Without Repetition
The question asks for the sum of all possible three-digit numbers that can be formed using the digits 1, 2, and 3, with the condition that the digits are not repeated in any number. Let's break down how to solve this problem.
Identifying the Three-Digit Numbers
We need to form three-digit numbers using the digits 1, 2, and 3 exactly once in each number. This is a permutation problem, as the order of the digits matters. The number of possible three-digit numbers is the number of permutations of 3 distinct items taken 3 at a time, which is \(P(3,3) = 3! = 3 \times 2 \times 1 = 6\).
The possible three-digit numbers are:
- 123
- 132
- 213
- 231
- 312
- 321
Calculating the Sum
We can find the sum by adding these six numbers directly:
\(123 + 132 + 213 + 231 + 312 + 321\)
Let's add them:
\(123 + 132 = 255\)
\(213 + 231 = 444\)
\(312 + 321 = 633\)
Total sum = \(255 + 444 + 633 = 699 + 633 = 1332\).
Alternative Method: Using Place Value
A more systematic way, especially useful with more digits, is to consider the contribution of each digit in each place value (hundreds, tens, units).
The digits available are 1, 2, and 3. The sum of these digits is \(1 + 2 + 3 = 6\).
Consider the hundreds place:
- How many numbers have 1 in the hundreds place? The remaining two digits (2 and 3) can be arranged in \(2! = 2\) ways in the tens and units places (123, 132).
- How many numbers have 2 in the hundreds place? Similarly, the remaining two digits (1 and 3) can be arranged in \(2! = 2\) ways (213, 231).
- How many numbers have 3 in the hundreds place? The remaining two digits (1 and 2) can be arranged in \(2! = 2\) ways (312, 321).
So, each digit (1, 2, and 3) appears in the hundreds place \( (3-1)! = 2 \) times.
The total value contributed by the hundreds place is the sum of the digits multiplied by the number of times each digit appears in the hundreds place, times 100:
\((1+2+3) \times 2 \times 100 = 6 \times 2 \times 100 = 12 \times 100 = 1200\)
Consider the tens place:
Similarly, each digit (1, 2, and 3) appears in the tens place \( (3-1)! = 2 \) times.
The total value contributed by the tens place is the sum of the digits multiplied by the number of times each digit appears in the tens place, times 10:
\((1+2+3) \times 2 \times 10 = 6 \times 2 \times 10 = 12 \times 10 = 120\)
Consider the units place:
Each digit (1, 2, and 3) appears in the units place \( (3-1)! = 2 \) times.
The total value contributed by the units place is the sum of the digits multiplied by the number of times each digit appears in the units place, times 1:
\((1+2+3) \times 2 \times 1 = 6 \times 2 \times 1 = 12 \times 1 = 12\)
The total sum of all the numbers is the sum of the contributions from each place value:
Total Sum = Hundreds Place Contribution + Tens Place Contribution + Units Place Contribution
Total Sum = \(1200 + 120 + 12 = 1332\)
This method gives the same result and is a good way to verify the calculation or handle problems with more digits.
| Place Value | Digits (1, 2, 3) | Sum of Digits | Number of times each digit appears in this place | Contribution to Total Sum |
|---|---|---|---|---|
| Hundreds (100) | 1, 2, 3 | \(1+2+3 = 6\) | \(2! = 2\) | \(6 \times 2 \times 100 = 1200\) |
| Tens (10) | 1, 2, 3 | \(1+2+3 = 6\) | \(2! = 2\) | \(6 \times 2 \times 10 = 120\) |
| Units (1) | 1, 2, 3 | \(1+2+3 = 6\) | \(2! = 2\) | \(6 \times 2 \times 1 = 12\) |
Total Sum = \(1200 + 120 + 12 = 1332\).
Conclusion
The sum of all three-digit numbers formed using the digits 1, 2, and 3 without repetition is 1332.
Revision Table: Sum of Three-Digit Numbers
| Concept | Description | Application in Problem |
|---|---|---|
| Permutation | Arrangement of objects where order matters. \(P(n, r) = n! / (n-r)!\) | Forming unique 3-digit numbers from 3 digits: \(3! = 6\) numbers. |
| Place Value | The value represented by a digit based on its position in a number (Units, Tens, Hundreds, etc.) | Used to calculate the contribution of digits based on their position in the formed numbers. |
| Sum of Digits | Adding the individual digits used. | Sum of digits 1, 2, 3 is \(1+2+3=6\). |
Additional Information: Sum of Permuted Numbers
For a set of \(n\) distinct non-zero digits, the sum of all \(n\)-digit numbers formed without repetition can be calculated using a general formula.
Let the digits be \(d_1, d_2, \dots, d_n\).
The number of \(n\)-digit numbers formed using these digits without repetition is \(n!\).
Each digit appears in each place value \((n-1)!\) times.
The sum of the digits is \(S = d_1 + d_2 + \dots + d_n\).
The sum of all such \(n\)-digit numbers is \(S \times (n-1)! \times (111\dots1)_{n \text{ times}}\).
In our case, digits are 1, 2, 3, so \(n=3\). Sum of digits \(S = 1+2+3=6\).
Number of permutations is \(3! = 6\).
Each digit appears in each place \((3-1)! = 2!\) = 2 times.
The "repetitive 1s" factor for a 3-digit number is 111.
Sum = \(6 \times 2 \times 111 = 12 \times 111 = 1332\).
This formula confirms the result obtained through the place value method.
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