Edit 1: May be I should modify my question after getting the answers. I see why $(X_c, Y_c, Z_c, \theta, \phi)$ are legitimate Dof's of the dumb-bell, I never had any problem with that.
Please consider now any rigid body in general. It has $6$ Dof's, two examples of which are mentioned below:
$(X_c, Y_c, Z_c, \alpha, \beta, \gamma)$: $(X_c, Y_c, Z_c)$ are CoM co-ordinates, and $(\alpha, \beta, \gamma)$ are the Eulerian angles (or, the pitch, roll and the yaw) of the rigid body.
Take a line through any point $(x_1, y_1, z_1)$ inside the rigid body, fix a line through the point using two angles $\theta, \phi$ (or fix a unit vector along the line, using two components). The last $Dof$ is obtained by specifying an angle around this line as axis, $\theta'$ that a point not on this line makes with, say the horizontal.
My question is more about the constraint relations. There are $3N$ co-ordinates to begin with, and each constraint relation eliminates one redundant co-ordinate. If this is true, because a rigid body has $6$ Dof's, after considering all the constraint relations (minus all the redundant constraints), I will have only $6$ (out of the set of $3N$ co-ordinates to start with). So can I have the Dof's to be any $6$ co-ordinates out of the initial $3N$, for example $(x_1, y_2, z_3, x_4, x_5, z_6)$ - is that sufficient to locate the rigid body?
If not, the constraint relations only lowers the number of co-ordinates, doesn't eliminate (remove) though. The reduced set of co-ordinates doesn't have to be any of those before reduction, it could be any, just one less in number.