What are the degrees of freedom of a dumbbell? 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.
 A: 
But can we really know the 6th coordinate?

You are right -- given knowledge of five coordinates, there is a discrete choice (corresponding to a reflection) for the remaining coordinate. However, by convention we typically (but not always) use the word "degree of freedom" to mean a continuously varying quantity. This expresses the idea that you aren't free to choose the 6-th coordinate arbitrarily as an initial condition; you have only finite number of choices for it. Once you have chosen your initial conditions however, the subsequent evolution from Newton's laws is completely fixed and there are only 5 functions of time you need to solve for.

If this is the effect of the constraint relation, then it doesn't really eliminate one redundant coordinate, but replaces the entire set, with one less coordinate.

I agree with your description that what is happening is "replacing the entire set, with one less coordinate." However, I also agree with the "eliminate one redundant coordinate" perspective. We started with $6$ coordinates and one constraint, and ended up with $5$ coordinates and no constraints; regardless of exactly how this was done, it seems fair to describe the net result as "eliminating" one of the 6 coordinates using the constraint.
A: "There is a new suggestion: $(X_c,Y_c,Z_c,θ,ϕ)$ where $(X_c,Y_c,Z_c)$ are the coordinates of the center-of-mass and θ and ϕ are the polar and the azimuthal angles. If this is the effect of the constraint relation, then it doesn't really eliminate one redundant coordinate, but replaces the entire set, with one less coordinate. Am I right?"
Not really. You can consider the set $(X_c,Y_c,Z_c,θ,ϕ,r)$, where $r$ is the distance between the two points. The constraint then simply becomes $r = \sqrt{6}$. This is a much cleaner approach because it addresses the ambiguity you posed in your first question.
If you want to use Cartesians as you suggested, then to address the ambiguity you need to apply a continuity or "no teleportation" condition. That is, if your choice is between $x_2 = 3, -1$, you pick the value which is closest to the value of $x_2$ at the preceding time step.
A: You are looking for 5 DOF. This can be established two ways

*

*6 DOF of a general rigid body, minus 1 rotational DOF due to symmetry

*4 DOF of an infinite line in space, plus 1 for locating the object along the line.

Your second approach is the correct one as three position coordinates and two angle coordinates are needed to fully describe the pose of the object.
Mathematically you need a little linear algebra and you can discover the position, velocity and acceleration of any point attached to the object.
Given the 5 DOF variables $(X_c,Y_c,Z_c,\varphi,\theta)$ and their derivatives here is what you do

*

*Position Kinematics
$$\begin{array}{l}
\text{Position vector for center of mass}\\
\boldsymbol{P}_{c}=\begin{pmatrix}X_{c}\\
Y_{c}\\
Z_{c}
\end{pmatrix}\\
\hline \text{Rotation matrix}\\
\mathbf{R}=\mathrm{RZ}(\varphi)\mathrm{RX}(\pi-\theta) \\ 
 =\begin{bmatrix}\cos\varphi & \sin\varphi\cos\theta & \sin\varphi\sin\theta\\
\sin\varphi & -\cos\varphi\cos\theta & -\cos\varphi\sin\theta\\
0 & \sin\theta & -\cos\theta
\end{bmatrix}\\
\hline \text{Position of point }\boldsymbol{p}=(x,y,z)\text{ local to body.}\\
\boldsymbol{P}=\boldsymbol{P}_{c}+\mathbf{R}\,\boldsymbol{p}
\end{array}$$


*Velocity Kinematics
$$\begin{array}{l}
\text{Velocity vector for center of mass}\\
\boldsymbol{V}_{c}=\begin{pmatrix}\dot{X}_{c}\\
\dot{Y}_{c}\\
\dot{Z}_{c}
\end{pmatrix}\\
\hline \text{Rotational velocity}\\
\boldsymbol{\Omega}=\boldsymbol{\hat{k}}\dot{\varphi}+{\rm RZ}(\varphi)\boldsymbol{\hat{i}}(-\dot{\theta})=\begin{pmatrix}-\dot{\theta}\cos\varphi\\
-\dot{\theta}\sin\varphi\\
\dot{\varphi}
\end{pmatrix}\\
\hline \text{Velocity of point }\boldsymbol{p}\\
\boldsymbol{V}=\boldsymbol{V}_{c}+\boldsymbol{\Omega}\times\left(\boldsymbol{P}-\boldsymbol{P}_{c}\right)
\end{array}$$

*

*Acceleation Kinematics
$$\begin{array}{l}
\text{Acceleration vector for center of mass}\\
\boldsymbol{A}_{c}=\begin{pmatrix}\ddot{X}_{c}\\
\ddot{Y}_{c}\\
\ddot{Z}_{c}
\end{pmatrix}\\
\hline \text{Rotational Acceleration}\\
\boldsymbol{\alpha}=\boldsymbol{\hat{k}}\ddot{\varphi}+{\rm RZ}(\varphi)\boldsymbol{\hat{i}}(-\ddot{\theta})+\boldsymbol{\Omega}\times\boldsymbol{\hat{k}}\dot{\varphi}\\
+\left(\boldsymbol{\Omega}+\boldsymbol{\hat{k}}\dot{\varphi}\right)\times{\rm RZ}(\varphi)\boldsymbol{\hat{i}}(-\dot{\theta})\\
\hline \text{Acceleration of point }\\
\boldsymbol{A}=\boldsymbol{A}_{c}+\boldsymbol{\alpha}\times\left(\boldsymbol{P}-\boldsymbol{P}_{c}\right)+\boldsymbol{\Omega}\times\left(\boldsymbol{V}-\boldsymbol{V}_{c}\right)
\end{array}$$
Nobody said 3D dynamics was easy. But if you are methodical with every step and pay attention to details you can figure it out.
A: For simplification I will take the 2D space .
The constraint equation is:
$$ \left( x_{{2}}-x_{{1}} \right) ^{2}+ \left( y_{{2}}-y_{{1}} \right) ^
{2}-{L}^{2}
=0\tag 1$$
thus you obtain three generalized coordinates
solving equation (1) for $x_2$
$$x_2=x_1\pm\sqrt{L^2-(y_2-y_1)^2}$$
you get two solution for $x_2$ but this doesn't affected the equations of motion . to solve the equations of motion the initial conditions must fulfilled the constraint equation. you get this two initial conditions configurations one with $+x_2~$ and one with $-x_2$ , but notion is wrong with this two solution.

