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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?
- a)1233
- b)1322
- c)1323
- d)1332
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
Three-digit numbers are formed from the digits 1, 2, and 3 in such a w...
Sum = 12(10° + 101 + 102 )
= 111 × 12 = 1332
Most Upvoted Answer
Three-digit numbers are formed from the digits 1, 2, and 3 in such a w...
Solution:
To find the sum of all three-digit numbers formed from the digits 1, 2, and 3 without repetition, we need to consider all possible combinations of these digits.
Step 1: Determine the total number of possible combinations.
Since there are three digits available (1, 2, and 3), and each digit can only be used once, the total number of possible combinations is given by the permutation formula:
nPr = n! / (n - r)!
where n is the total number of items and r is the number of items chosen at a time.
In this case, n = 3 and r = 3, so the number of possible combinations is:
3P3 = 3! / (3 - 3)! = 3! / 0! = 3! = 3 x 2 x 1 = 6
Step 2: Find all the possible combinations.
The six possible combinations of the digits 1, 2, and 3 are:
123, 132, 213, 231, 312, 321
Step 3: Find the sum of the three-digit numbers.
To find the sum of these numbers, we need to add them together. The sum is:
123 + 132 + 213 + 231 + 312 + 321 = 1332
Therefore, the correct answer is option D) 1332.
To find the sum of all three-digit numbers formed from the digits 1, 2, and 3 without repetition, we need to consider all possible combinations of these digits.
Step 1: Determine the total number of possible combinations.
Since there are three digits available (1, 2, and 3), and each digit can only be used once, the total number of possible combinations is given by the permutation formula:
nPr = n! / (n - r)!
where n is the total number of items and r is the number of items chosen at a time.
In this case, n = 3 and r = 3, so the number of possible combinations is:
3P3 = 3! / (3 - 3)! = 3! / 0! = 3! = 3 x 2 x 1 = 6
Step 2: Find all the possible combinations.
The six possible combinations of the digits 1, 2, and 3 are:
123, 132, 213, 231, 312, 321
Step 3: Find the sum of the three-digit numbers.
To find the sum of these numbers, we need to add them together. The sum is:
123 + 132 + 213 + 231 + 312 + 321 = 1332
Therefore, the correct answer is option D) 1332.
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