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
(a) 432
(b) 108
(e) 36
(d) 18
Solution:
Out of 3, 4, 5 and 6
If the unit place is 3 (say), then remaining three place can be filled in 3! ways.
i.e 3 appears in unit place in 3! times.
Similarly each digit appear in unit place 3! times
So, sums of digits in unit place
= 3! (3 + 4 + 5 + 6)
= 3 × 2 (18)
= 108.