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The sum of the digits in unit place of all the numbers formed with the help of 3, 4, 5 and 6 taken all at a time is
A. 432
B. 108
C. 36
D. 18

Answers (1)

The answer is the option (b)

If we fix 3 at the unit place, the remaining places can be filled in 3! Ways.

Thus ‘3’ appears in unit place in 3! Times.

 Similarly for each of the digits 4,5 and 6.  So, sum of digits in unit place = 3!(3+4+5+6) =18*6 =108

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