The sum of the digits in the unit place of all numbers formed with the help of 3, 4, 5, 6 taken all at a time is

The correct option is C 108
Required sum $=3!(3+4+5+6)=6\times 18=108$.
[If we fix 3 of the unit place, other three digits can be arranged in $3!$ways similarly for 4, 5, 6.]