Determine the sum of all the numbers that can be formed with digits 1,2 and 3, each occurring only once in a number
possible numbers:
single digits:
1
2
3
double digits
12
13
23
21
31
32
triple digits:
123
132
213
231
312
321
Can you add them up ?
1470 is the answer.
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Here is the answer.
1470
To determine the sum of all the numbers that can be formed with the digits 1, 2, and 3, each occurring only once in a number, we need to consider all possible permutations of these digits.
The total number of permutations can be calculated using the formula for permutations of distinct objects, which is n!, where n is the number of objects. In this case, n = 3 (since there are three digits: 1, 2, and 3). So, the total number of permutations is 3! = 3 x 2 x 1 = 6.
To find all the different arrangements, let's list out the permutations:
1. 123
2. 132
3. 213
4. 231
5. 312
6. 321
Now, we can calculate the sum of these numbers:
1. 123 = 1 x 100 + 2 x 10 + 3 x 1 = 123
2. 132 = 1 x 100 + 3 x 10 + 2 x 1 = 132
3. 213 = 2 x 100 + 1 x 10 + 3 x 1 = 213
4. 231 = 2 x 100 + 3 x 10 + 1 x 1 = 231
5. 312 = 3 x 100 + 1 x 10 + 2 x 1 = 312
6. 321 = 3 x 100 + 2 x 10 + 1 x 1 = 321
Now, we add up all these numbers:
123 + 132 + 213 + 231 + 312 + 321 = 1332
Therefore, the sum of all the numbers that can be formed with the digits 1, 2, and 3, each occurring only once, is 1332.