Discussion

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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