The sum of the digits in unit place of all the numbers formed with the help of 3, 4, 5 and 6
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.

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