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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
Using the digits 1,5,2,8 all possible four digit numbers are formed and the sum of all such numbers is between
Solution
Digits = 1,5,2,8
Number of arrangements = 4! = 24
Each digit appears 6 times in each place (unit, tens, hundreds, thousands).
Sum contributed by one digit = 6 × (1000 + 100 + 10 + 1) × digit
= 6 × 1111 × digit = 6666 × digit
Sum of digits = 1 + 5 + 2 + 8 = 16
Total sum = 6666 × 16 = 106656
This lies between 50000 and 100000.
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