Question: **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.

Answer: 
> 
> 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.
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.

[![enter image description here](https://i.stack.imgur.com/yQ4aA.png)](https://i.stack.imgur.com/yQ4aA.png)