Sum Of Digits Permutations
Using all five digits 1, 2, 3, 4 and 5 exactly once to form five-digit numbers, the sum of all such distinct five-digit numbers that can be formed equals which value?
Select the correct option:
Solution
3999960
Summing all permutation-generated numbers uses the symmetry that each digit appears equally often in every place value, an elegant JEE Advanced counting argument. With 5 distinct digits, there are 5! = 120 numbers, and by symmetry each digit occupies each of the five positions in 120/5 = 24 of them. The sum of the digits is 1 + 2 + 3 + 4 + 5 = 15. In any fixed position, the total contribution from all numbers is 24 × 15 = 360 times the place value. Summing over place values 1, 10, 100, 1000, 10000 gives 360 × (11111) = 360 × 11111 = 3999960. Option 360000 stops at one place value. Option 1599984 miscounts the repetition factor. Option 3333300 uses an incorrect digit sum. Hence the total sum is 3999960. Plausibility check: the average five-digit number formed is 3999960/120 = 33333, which is exactly the repunit 11111 times the average digit 3, confirming the symmetric place-value reasoning. Counting geometric figures from points reduces to selecting the required number of vertices once degeneracies such as collinearity are accounted for. With no three points collinear every triple yields a genuine triangle, but when some points are collinear the count must subtract the degenerate selections, a refinement that frequently distinguishes routine from challenging versions of these problems.
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About This Question
- Subject
- mathematics
- Chapter
- permutations and combinations
- Topic
- sum of digits permutations
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
3999960
Summing all permutation-generated numbers uses the symmetry that each digit appears equally often in every place value, an elegant JEE Advanced counting argument. With 5 distinct digits, there are 5! = 120 numbers, and by symmetry each digit occupies each of the five positions in 120/5 = 24 of them. The sum of the digits is 1 + 2 + 3 + 4 + 5 = 15. In any fixed position, the total contribution from all numbers is 24 × 15 = 360 times the place value. Summing over place values 1, 10, 100, 1000, 10000 gives 360 × (11111) = 360 × 11111 = 3999960. Option 360000 stops at one place value. Option 1599984 miscounts the repetition factor. Option 3333300 uses an incorrect digit sum. Hence the total sum is 3999960. Plausibility check: the average five-digit number formed is 3999960/120 = 33333, which is exactly the repunit 11111 times the average digit 3, confirming the symmetric place-value reasoning. Counting geometric figures from points reduces to selecting the required number of vertices once degeneracies such as collinearity are accounted for. With no three points collinear every triple yields a genuine triangle, but when some points are collinear the count must subtract the degenerate selections, a refinement that frequently distinguishes routine from challenging versions of these problems.
This medium difficulty mathematics question is from the chapter permutations and combinations, covering the topic of sum of digits permutations. It appeared in the 2025 exam.
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