Select one object from 7 objects for 4^th group in $7{\mathrm{C}}_1$ ways.

Remaining 6 objects to be divided into 3 groups containing 2 each in $\frac{6!}{2! 2! 2! 3!}$

Total no of ways 

$\begin{array}{rcl} & =& 7\times \frac{6!}{{(2!)}^{3}3!}\\ & =& 7\times \frac{720}{8\left(6\right)}=105\end{array}$