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Question
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 ______.
Options
432
108
36
18
Solution
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 108.
Explanation:
If we fix 3 at unit place, then the total possible numbers = 3!
If we fix 4, 5 and 6 at unit place, this is each case, total possible numbers are 3!
Required sum of unit digits of all such numbers is
= 3 × 3! + 4 × 3! + 5 × 3! + 6 × 3!
= (3 + 4 + 5 + 6) × 3!
= 18 × 3!
= 18 × 3 × 2 × 1
= 108
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