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

Sum of the digits at tens place of the numbers formed by 3, 4, 5, 6 taken all at a time.

$$\begin{gathered} =(n-1)! (3+4+5+6) \\ =3! \left(18\right) \\ =6 \left(18\right) \\ =108 \end{gathered}$$