Question

Q 50. The sum of '5' digit numbers formed using all the digits 3, 4, 6, 7, 8 only once is
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Answer 1

The number of numbers of 5 digits, having 3 in unit's place = number of arrangement of the remaining 4 digits = ⁴P_4 =24

Similarly, for other digits also, each digit occurs 24 times in units place.

So the sum of digits in unit's place = 24×3+24×4+24×5+24×6+24×7=600

Similarly, the number of digits in each of the other places=600

Thus the required sum =

600×10000+600×1000+600×100+600×10+600×1= 6666600