If I evaluate degree of freedom and got some number $n$, then how can I know what are those $n$ independent coordinates? Using $3N-f=d$ we can evaluate the degree of freedom or independent coordinates of a system.
But how can we know which coordinates are actually independent? 
(Here $n$ = number of particles, $f$ = number of constraint equations and $d$ = degree of freedom or number of independent coordinates.)
If we take the case of double Atwood machine, we get 2 dof. So which two coordinates should be said to be independent? $x$ and $y$?
Update:
If I take the case of, "A particle falling under gravity", the dof will be 1. So there should be only one independent coordinate with which we can describe the situation. If I take the fall of the particle in $y$ direction, then that one independent coordinate will be $y$?
 A: If you have a system of $N$ particles with $f$ equations of constraints, then the effective degrees of freedom reduces from $3N$ to $3N-f$. This means you don't have to worry about all the $3N$ coordinates, but just focus on the $3N-f$ coordinates to study the dynamics of the system. This idea is based on a simple logic as I will try to explain it in terms of a circular motion.  
Suppose a free particle exhibits a circular motion in the $xy$-plane. The solution for the equation of motion will be in the form 
$$z(t)=0$$
$$x(t)^2+y(t)^2=r^2$$
where $x(t)$, $y(t)$ and $z(t)$ are the $x$, $y$ and $z$ coordinates at time $t$ and $r$ is the radius of the circular path and is assumed to be a constant. At first sight, a free particle has $3$ degrees of freedom. But the above equation of constraint($x(t)^2+y(t)^2=r^2$) and the restriction of the particle's motion confined to a plane ($z=0$) reduces the degree of freedom. Hence there is in effect $3-2=1$ degree of freedom of the system.  
This means you only need any of the coordinates- $x(t)$ or $y(t)$- to describe the mechanics of the system. The choice is yours. If you choose $x(t)$ as the independent coordinate, then since $r$ is a constant, once you fix a value of $x(t)$, the value of $y(t)$ get fixed automatically because of the constraint equation. In other words, $y(t)$ depends on $x(t)$ by the above equation.  
Now, you may ask that if you change the coordinate system from Cartesian to some other, say Spherical polar coordinates, then will the dof changes? No, it will not. The choice of a coordinate system will not affect the dynamics of the system. The above circular motion in plane polar coordinates can be written as:  
Put $x=rcos\theta$ and $y=rsin\theta$ in the previous equation and we get
$$\phi(t)=0$$
$$(rcos\theta)^2+(rsin\theta)^2=r^2$$  
Here, we have $r=\text{constant}$ and hence the only variable that changes with time is $\theta(t)$. So there is only one degree of freedom. You choose $\theta(t)$ as your independent coordinate. As you can see, the degree of freedom is still one. 
In the case of free fall of a particle, the solution is given by:  
$$y(t)=\frac{1}{2}gt^2$$  
where $y(t)$ is the position of the object in the $t^{th}$ second. Here, the degree of freedom is one. You only need $y$ to spot the particle at any time $t$.  

Update: Degree of freedom of Double Atwood Machine


In the case of a simple Atwood machine, there is only one degree of freedom. Now, you replace one of the masses by another Atwood machine to form a double Atwood machine (or sometimes called compound Atwood machine). Here the system has two degrees of freedom:
1. one is the freedom of mass $1$ (and the attached movable pulley) to move up and down about the fixed pulley, and
2.  one is the freedom of mass $2$ (and the attached mass $3$) to move up and down about the movable pulley  
How to do this in terms of constraints?  
To describe the configuration of the system, we need 3 coordinates each for masses $m_1$, $m_2$ and $m_3$ and another three for the movable pulley. i.e., a total of 12 coordinates. But thanks to the constraints present. There are $10$ constraints present here:   


*

*$8$ of which limit the motion of all the coordinates in a single direction ($x$ I'am taking here);   

*the remaining $2$ are given by
$(x_p+x_1)=l$ and $(x_2-x_p)+(x_3-x_p)=l'$  
where $x_p$ and $x_i$ are the vertical positions of the pulley and the masses $m_i$ respectively.   
Hence the effective degree of freedom of the system reduces to $2$.  
In a simple way, one degree of freedom is that of the mass $m_1$ and the other is that due to the mass $m_2$. As you can see from the figure, their motions are independent to each other. Hence $x$ and $x'$ are the independent coordinates here.
A: Technically, when you choose your $n$ generalized coordinates $q^1,\ldots,q^n,$ among the $3N$ position coordinates ${\bf r}_1,\ldots,{\bf r}_N,$ of $N$ point particles, with $n\leq 3N$, you should make sure that the $3N\times n$ rectangular matrix
$$ \frac{\partial {\bf r}_i}{\partial q^j}, \qquad i\in\{1,\ldots N\},\qquad  j\in \{1,\ldots n\}, $$
has maximal rank, i.e. has rank $n$.
