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

Question 1

(A) 432

(B) 108

(C) 36

(D) 18

Question 2

Answer is (B)

If the unit place is ‘3’ then remaining three places can be filled in 3! ways.

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

Similarly each digit appear in unit place 3! times.

So, sum of digits in unit place = 3!(3 + 4 + 5 + 6) = 18 x 6 = 108

kratos · Answer 1 · 2020-03-19T21:40:51+0000

Answer is (B)

If the unit place is ‘3’ then remaining three places can be filled in 3! ways.

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

Similarly each digit appear in unit place 3! times.

So, sum of digits in unit place = 3!(3 + 4 + 5 + 6) = 18 x 6 = 108

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