What is the sum of all the 5-digit numbers that can be formed, if repetition of digits is not allowed?
Explanation:
If 2 is at the unit's place, the remaining digits 3, 4, 7 and 8 can be arranged in 4! ways. Hence, there are exactly 4! 5-digit numbers using the given digits.
Similarly, there are exactly 4! 5-digit numbers using the given digits. Similarly for other digits.
Hence the total sum of digits at unit's place for all numbers possible is 4! × (2 + 3 + 4 + 7 + 8) × 1
Similarly, the total sum of digits at ten's place for all numbers possible is 4! × (2 + 3 + 4 + 7 + 8) × 10
Similarly, the total sum at
hundred's place = 4! × (2 + 3 + 4 + 7 + 8) × 102
thousand's place = 4! × (2 + 3 + 4 + 7 + 8) × 103
ten thousand's place = 4! × (2 + 3 + 4 + 7 + 8) × 104
Hence, the required sum = 4! × (2 + 3 + 4 + 7 + 8) + 4! × (2 + 3 + 4 + 7 + 8) × 10 + 4! × (2 + 3 + 4 + 7 + 8) × 102 + 4! × (2 + 3 + 4 + 7 + 8) × 103 + 4! × (2 + 3 + 4 + 7 + 8) × 104
= 4! × (2 + 3 + 4 + 7 + 8) × (11111)
= 24 × 24 × 11111
= 63,99,936.
Hence, 6399936.
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